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Triangles

Mohit Sharma

14th January 2026

Mathematics test class 10th │ Maths test │ Coordinate Geometry test

2021-07-25T09:26:00.000+05:30

Mathematics test class 10th │ Maths test │ Coordinate Geometry test │ Class 10th Maths test

Class 10 Maths MCQs Chapter 7 Coordinate Geometry

1. The distance of the point P(2, 3) from the x-axis is
(a) 2
(b) 3
(c) 1
(d) 5

Answer/ Explanation

Answer: b
Explaination: Reason: The distance from x-axis is equal to its ordinate i.e., 3

2. The distance between the point P(1, 4) and Q(4, 0) is
(a) 4
(b) 5
(c) 6
(d) 3√3

Answer/ Explanation

Answer: b
Explaination: Reason: The required distance = \(\sqrt{(4-1)^{2}+(0-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5\)

3. The points (-5, 1), (1, p) and (4, -2) are collinear if
the value of p is
(a) 3
(b) 2
(c) 1
(d) -1

Answer/ Explanation

Answer: d
Explaination: Reason: The points are collinear if area of Δ = 0
= \(\frac{1}{2}\)[-5(p + 2) +l(-2 -1) + 4(1 – p)] – 0
⇒ -5 p -10-3 + 4-4p = 0
⇒ -9p = +9
∴ p = -1

4. The area of the triangle ABC with the vertices A(-5, 7), B(-4, -5) and C(4, 5) is
(a) 63
(b) 35
(c) 53
(d) 36

Answer/ Explanation

Answer: c
Explaination: Reason: Area of ΔABC = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[-5(-5 – 5) -4(5 – 7) + 4(7 – (-5))] = \(\frac{1}{2}\)[-5(-10) -4(-2) + 4(12)]
= \(\frac{1}{2}\)[50 + 8 + 48] = \(\frac{1}{2}\) × 106 = 53 sq. units

5. The distance of the point (α, β) from the origin is
(a) α + β
(b) α² + β²
(c) |α| + |β|
(d) \(\sqrt{\alpha^{2}+\beta^{2}}\)

Answer/ Explanation

Answer: d
Explaination: Reason: Distance of (α, β) from origin (0, 0) = \(\sqrt{(\alpha-0)^{2}+(\beta-0)^{2}}=\sqrt{\alpha^{2}+\beta^{2}}\)

6. The area of the triangle whose vertices are A(1, 2), B(-2, 3) and C(-3, -4) is
(a) 11
(b) 22
(c) 33
(d) 21

Answer/ Explanation

Answer: a
Explaination: Reason: Required area= \(\frac{1}{2}\)[1(3 + 4) -2(-4 – 2) -3(2 – 3)]
= \(\frac{1}{2}\)[7 + 12 + 3]
= \(\frac{1}{2}\) × 22 = 11

7. The line segment joining the points (3, -1) and (-6, 5) is trisected. The coordinates of point of trisection are
(a) (3, 3)
(b) (- 3, 3)
(c) (3, – 3)
(d) (-3,-3)

Answer/ Explanation

Answer: b
Explaination: Reason: Since the line segment AB is trisected

8. The line 3x + y – 9 = 0 divides the line joining the points (1, 3) and (2, 7) internally in the ratio
(a) 3 : 4
(b) 3 : 2
(c) 2 : 3
(d) 4 : 3

Answer/ Explanation

Answer: a
Explaination: Reason: Let the line 3x + y – 9 = 0 divide the line segment joining A(l, 3) ad B(2, 7) in the ratio K : 1 at point C.

9. The distance between A (a + b, a – b) and B(a – b, -a – b) is

Answer/ Explanation

Answer: c
Explaination:

10. If (a/3, 4) is the mid-point of the segment joining the points P(-6, 5) and R(-2, 3), then the value of ‘a’ is
(a) 12
(b) -6
(c) -12
(d) -4

Answer/ Explanation

Answer: c
Explaination:

11. If the distance between the points (x, -1) and (3, 2) is 5, then the value of x is
(a) -7 or -1
(b) -7 or 1
(c) 7 or 1
(d) 7 or -1

Answer/ Explanation

Answer: d
Explaination: Reason: We have \(\sqrt{(x-3)^{2}+(-1-2)^{2}}=5\)
⇒ (x – 3)² + 9 = 25
⇒ x² – 6x + 9 + 9 = 25
⇒ x² -6x – 7 = 0
⇒ (x – 7)(x + 1) = 0
⇒ x = 7 or x = -1

12. The points (1,1), (-2, 7) and (3, -3) are
(a) vertices of an equilateral triangle
(b) collinear
(c) vertices of an isosceles triangle
(d) none of these

Answer/ Explanation

Answer: b
Explaination: Reason: Let A(1, 1), B(-2, 7) and C(3, 3) are the given points, Then, we have

13. The coordinates of the centroid of a triangle whose vertices are (0, 6), (8,12) and (8, 0) is
(a) (4, 6)
(b) (16, 6)
(c) (8, 6)
(d) (16/3, 6)

Answer/ Explanation

Answer: d
Explaination: Reason: The co-ordinates of the centroid of the triangle is

14. Two vertices of a triangle are (3, – 5) and (- 7,4). If its centroid is (2, -1), then the third vertex is
(a) (10, 2)
(b) (-10,2)
(c) (10,-2)
(d) (-10,-2)

Answer/ Explanation

Answer: c
Explaination: Reason: Let the coordinates of the third vertex be (x, y)

15. The area of the triangle formed by the points A(-1.5, 3), B(6, -2) and C(-3, 4) is
(a) 0
(b) 1
(c) 2
(d) 3/2

Answer/ Explanation

Answer: a
Explaination: Reason: Area of ΔABC = \(\frac{1}{2}\) [-1.5(-2 – 4) + 6(4 – 3) + (-3) (3 + 2)] = \(\frac{1}{2}\) [9 + 6 – 15] = 0. It is a straight line.

16. If the points P(1, 2), B(0, 0) and C(a, b) are collinear, then
(a) 2a = b
(b) a = -b
(c) a = 2b
(d) a = b

Answer/ Explanation

Answer: a
Explaination: Reason: Area of ΔPBC = 0
⇒ \(\frac{1}{2}\)[1(0 – b) + 0(6 – 1) + a(2 – 0)] = 0
⇒ \(\frac{1}{2}\)[-6 + 2a] = 0
⇒ -b + 2a = 0
∴ 2a = b

Mathematics test class 10th │ Maths test │Coordinate Geometry Test

2021-07-25T09:02:00.001+05:30

Mathematics test class 10th │ Maths test │Coordinate Geometry Test

Mathematics test class 10th │ Maths test │Coordinate Geometry Test │ Class 10th Maths test

Mathematics test class 10th │ Maths test │ Free maths test │ Class 10th Maths test

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Mathematics Test

Hello everyone. This is Mohit Sharma. I want to inform you that this is a practice test paper of Mathematics. First read all the guidelines then after that you can attempt this short quiz. Here a link is given to you at last. Click on the link and attempt this short quiz.

सभी को नमस्कार। मेरा नाम मोहित शर्मा है । मैं आपको सूचित करना चाहता हूं कि यह Maths का एक अभ्यास परीक्षण पेपर है। सभी दिशानिर्देशों को पढ़ें इसके बाद आप इस संक्षिप्त प्रश्नोत्तरी का प्रयास कर सकते हैं। यहां पर आपको आखिरी में एक लिंक दिया गया है। लिंक पर क्लिक करें और इस संक्षिप्त प्रश्नोत्तरी का प्रयास करें।

ALL THE BEST. ✌

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This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Mathematics
Topic: Coordinate Geometry

Link: https://forms.gle/EvEssTLS7PLazJTf8

After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Mathematics test class 10th │ Maths test │ Free maths test │ Class 10th Maths test

2021-07-13T06:27:00.000+05:30

Mathematics test class 10th │ Maths test │ Free maths test │ Class 10th Maths test

Mathematics test class 10th │ Maths test │ Free maths test │ Class 10th Maths test

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Mathematics TEST

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Mathematics TEST

Mathematics Test

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Guidelines:

This is a quiz test. You have to attempt all questions.
Total Questions: 5
Max. marks: 10 (Each question carry 2 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

Link: https://forms.gle/p8JvuLbTwcB8FRj68

After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

2021-07-13T06:09:00.000+05:30

Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

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Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

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Computer TEST

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Computer Test

Hello everyone. This is Mohit Sharma. I want to inform you that this is a practice test paper of Computer. First read all the guidelines then after that you can attempt this short quiz. Here a link is given to you at last. Click on the link and attempt this short quiz.

सभी को नमस्कार। मेरा नाम मोहित शर्मा है । मैं आपको सूचित करना चाहता हूं कि यह COMPUTER का एक अभ्यास परीक्षण पेपर है। सभी दिशानिर्देशों को पढ़ें इसके बाद आप इस संक्षिप्त प्रश्नोत्तरी का प्रयास कर सकते हैं। यहां पर आपको आखिरी में एक लिंक दिया गया है। लिंक पर क्लिक करें और इस संक्षिप्त प्रश्नोत्तरी का प्रयास करें।

ALL THE BEST. ✌

Guidelines:

This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

Link: https://forms.gle/Q7YTSHd1X5CFxzHd9

After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

2021-07-13T05:50:00.002+05:30

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

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Computer TEST

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Computer TEST

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Computer Test

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Guidelines:

This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

Link: https://forms.gle/6hHpq8U3frSBWpri8

After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

2021-07-07T05:55:00.003+05:30

Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

Computer test class 7th │ Computer test │ Free computer test │ class 6th Computer test

Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

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Computer TEST

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Computer TEST

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Computer Test

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Guidelines:

This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

Link: https://forms.gle/yEcXk3w4vMbetx976

After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

2021-07-07T05:40:00.000+05:30

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

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Computer TEST

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Computer TEST

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Computer Test

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Guidelines:

This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

Link: https://forms.gle/yvVVMrNvKajvYLoZ9

After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Mathematics test class 10th │ Maths test │ Free maths test │ Class 10th Maths test

2021-07-02T11:29:00.004+05:30

Mathematics test class 10th │ Maths test │ Free maths test │ Class 10th Maths test

Mathematics test class 10th │ Maths test │ Free maths test │ Class 10th Maths test

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Mathematics Test

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This is a quiz test. You have to attempt all questions.
Total Questions: 5
Max. marks: 10 (Each question carry 2 mark)
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Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

2021-07-02T11:07:00.003+05:30

Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

Computer test class 7th │ Computer test │ Free computer test │ class 6th Computer test

Computer test class 7th │ Computer test │ Free computer test │ Class 7th Computer test

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Computer TEST

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Computer TEST

Computer Test

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Guidelines:

This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

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After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

2021-07-02T10:27:00.001+05:30

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

Computer test class 6th │ Computer test │ Free computer test │ class 6th Computer test

To give test, Click on:

Computer TEST

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Computer TEST

Computer Test

ALL THE BEST. ✌

Guidelines:

This is a quiz test. You have to attempt all questions.
Total Questions: 10
Max. marks: 10 (Each question carry 1 mark)
Subject: Computer
School: J.P. PUBLIC ACADEMY, KASGANJ

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After attempting short quiz you will get your SCORE on screen with correct answers if you have done any question incorrect.

Trigonometry | Trigonometry hand trick | Trigonometry Class 10 | Trigonometry tricks | त्रिकोणमिति

2021-06-18T14:52:00.000+05:30

Trigonometry | Trigonometry hand trick | Trigonometry Class 10 | Trigonometry tricks | त्रिकोणमिति

Quadratic Formula | द्विघात समीकरण | Quadratic Equation Tricks | How To Solve Quadratic Equations

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Trigonometry | Trigonometry hand trick | Trigonometry Class 10 | Trigonometry tricks | त्रिकोणमिति

Trigonometry | Trigonometry hand trick | Trigonometry Class 10 | Trigonometry tricks | त्रिकोणमिति About this video: Solve Trigonometry questions in 3 seconds. Trigonometry shortcut for JEE/NDA/NA/AIRFORCE/RAILWAYS/ BANKING/SSC-CGL/MBA/GRE/SAT/CETs./Competitive exams. Hey guys This is Mohit Sharma. Welcome to our youtube channel Style stardum MS. SUBSCRIBE( it's free): Style stardum MS- https://www.youtube.com/channel/UCMzOdlcCIVacd9Bw7CM5Pnw?view_as=subscriber maths study- https://www.youtube.com/channel/UCQNspKbWcNnibwaIH1vY_yA?view_as=subscriber ___________________________________________________________________________________ Our apps: maths study: https://play.google.com/store/apps/details?id=com.app.mathematicsnew maths study 2: https://play.google.com/store/apps/details?id=in.co.mathsstudy.ms __________________________________________________________________________________ Our Social links- ●Youtube- https://www.youtube.com/channel/UCMzOdlcCIVacd9Bw7CM5Pnw?view_as=subscriber ●Facebook- https://www.facebook.com/mohitsharma.ksj ●Twitter- https://www.twitter.com/MohitKsj ●Instagram- https://www.instagram.com/mohitsharma_163/ ●Website- https://www.mathsstudy.co.in ● Other youtube official channel of maths- https://www.youtube.com/channel/UCQNspKbWcNnibwaIH1vY_yA ●Pinterest- https://pin.it/66WthEK ●Quora - Come and Join.... https://www.quora.com/q/zjrbytriwtrjpsjt?invite_code=u75iHFSa8qMSb74a5w5s Email- mohitsharma.ksj@gmail.com ___________________________________________________________________________________ @Style stardum MS @Dear Sir @SSC MAKER @Unacademy UPSC Hindi @wifistudy @Vedantu Class 9 & 10 @BYJU'S @Magnet Brains Trigonometry,Trigonometry hand trick ,Trigonometry Class 10/11/12,Trigonometry tricks in hindi,Style stardum MS,trigonometry class 10th in hindi,class 10 chapter 8,trigonometry,trigonometry formulas,trigonometry trick,trigonometry class 10,trigonometry class 11,trigonometry class 12,trigonometry table,trigonometry ratios,trigonometry tricks,trigonometry basics,math shortcut and tricks,ssc-cgl math tricks,jee tricks maths, #TRIGONMETRYTRICKS #TRIGONOMETRYTRICKS #math #mathematics #maths #NEHAMATHTRICKS ___________________________________________________________________________________ trigonometry tricks,trigonometry shortcuts,math shortcut and tricks,trigonometry cbse isc class xii 12th,trigonometry cbse isc class xi 11th,mathematically inclined,neha agarwal maths,jee tricks math,nda/na math tricks,eamcet/gujcet/mhcet/kcet/comedk math tricks,sat / gre/mba tricks,banking/railways/airforce math tricks,ssc-cgl math tricks,trigonometry,trigonometry basics,trigonometry tricks,basic trigonometry,pythagorean theorem,trigonometry trick,trigonometry table,sin cos tan - basic trigonometry in hindi,trigonometry ssc cgl,trigonometry concept,trigonometry introduction,math,maths,mathematics,trigonometry class 10 chapter 8,trigonometry tricks,class 10 maths,maths class 10,trigonometry table,right triangle trigonometry, trigonometry class 11,trigo class 11,class 11 maths chapter 3,trigonometric functions class 11,radian to degree,trigonometry class 11 important questions,trigonometry shortcut tricks,trigonometry tricks class 11,trigonometry basics,trigonometric functions,trigonometry,trigonometry formulas,trigonometry trick,trigonometry table,trigonometry ratios,trigonometry functions,chapter 3 trigo class 11,chapter 3 trigonometry class 11,trigonometry ssc cgl, _______________________________________________________________________________________ Trigonometry Class 10 Maths | Trigonometry Formulas/Table Tricks | Trigonometry Questions Tricks Solve Trigonometry in 10 Seconds | IIT JEE Short Cuts & Tricks | JEE Main 2021 | Score Quick Marks Math Shortcuts for JEE Mains | Advanced | BITSAT | VITEEE | WBJEE. Learn the super awesome Tricks that will blow your mind. jee mains maths | jee main mock test | jee mains 2021 | jee mains mathematics | jee mock test | jee 2021 | jee maths | jee main mock test 2021| iit jee 2021 | iitmaths | jee mains mock test free online | jee exam 2021 | iit jee maths | jee main online mock test | jee mains exam | iit jee maths lectures | jee maths lectures CBSE Class 11 Math Trigonometric Functions | Class 11 NCERT chapter 1 | Domain Range | Sin, Cos, Tan, Cosec, Sec, Cot | Formula Short cut| Memory tip | Trigonometry Short Tricks |Trigonometry Identities | Trigonometry Equation | General and Principal Solution | Important Practice Questions | Angle Radian Degree | Crash Course Class 11 Trigonometric Functions| Sets Complete Chapter in 1 video | Formulas Trick | Trigonometry Class 11 | Trigonometry ssc cgl _______________________________________________________________________________________

Quadratic Formula | द्विघात समीकरण | Quadratic Equation Tricks | How To Solve Quadratic Equations

2021-06-10T11:59:00.006+05:30

Quadratic Formula | द्विघात समीकरण | Quadratic Equation Tricks | How To Solve Quadratic Equations

_________________________________________________

Youtube Video: Quadratic Formula

An example of a Quadratic Equation:

The function makes nice curves like this one:

Name

The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x²).

It is also called an "Equation of Degree 2" (because of the "2" on the x)

Youtube Video: Quadratic Formula

Standard Form

The Standard Form of a Quadratic Equation looks like this:

a, b and c are known values. a can't be 0.

"x" is the variable or unknown (we don't know it yet).

Here are some examples:

2x² + 5x + 3 = 0

In this one a=2, b=5 and c=3

x² − 3x = 0

This one is a little more tricky:

Where is a? Well a=1, as we don't usually write "1x²"
b = −3
And where is c? Well c=0, so is not shown.

5x − 3 = 0

Oops! This one is not a quadratic equation: it is missing x²
(in other words a=0, which means it can't be quadratic)

Youtube Video: Quadratic Formula

Hidden Quadratic Equations!

As we saw before, the Standard Form of a Quadratic Equation is

ax² + bx + c = 0

But sometimes a quadratic equation doesn't look like that!

For example:

In disguise		In Standard Form	a, b and c
x² = 3x − 1	Move all terms to left hand side	x² − 3x + 1 = 0	a=1, b=−3, c=1
2(w² − 2w) = 5	Expand (undo the brackets), and move 5 to left	2w² − 4w − 5 = 0	a=2, b=−4, c=−5
z(z−1) = 3	Expand, and move 3 to left	z² − z − 3 = 0	a=1, b=−1, c=−3

About the Quadratic Formula

Plus/Minus

First of all what is that plus/minus thing that looks like ± ?

The ± means there are TWO answers:

x = −b + √(b²− 4ac)2a

x = −b − √(b²− 4ac)2a

Here is an example with two answers:

But it does not always work out like that!

Imagine if the curve "just touches" the x-axis.
Or imagine the curve is so high it doesn't even cross the x-axis!

This is where the "Discriminant" helps us ...

Youtube Video: Quadratic Formula

Discriminant

Do you see b² − 4ac in the formula above? It is called the Discriminant, because it can "discriminate" between the possible types of answer:

when b² − 4ac is positive, we get two Real solutions
when it is zero we get just ONE real solution (both answers are the same)
when it is negative we get a pair of Complex solutions

Complex solutions? Let's talk about them after we see how to use the formula.

Youtube Video: Quadratic Formula

Using the Quadratic Formula

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x² + 6x + 1 = 0

Coefficients are:a = 5, b = 6, c = 1

Quadratic Formula:x = −b ± √(b²− 4ac)2a

Put in a, b and c:x = −6 ± √(6²− 4×5×1)2×5

Solve:x = −6 ± √(36− 20)10

x = −6 ± √(16)10

x = −6 ± 410

x = −0.2 or −1

Answer: x = −0.2 or x = −1

And we see them on this graph.

Check -0.2:		5×(−0.2)² + 6×(−0.2) + 1 = 5×(0.04) + 6×(−0.2) + 1 = 0.2 − 1.2 + 1 = 0
Check -1:		5×(−1)² + 6×(−1) + 1 = 5×(1) + 6×(−1) + 1 = 5 − 6 + 1 = 0

Example: Solve x² − 4x + 6.25 = 0

Coefficients are:a=1, b=−4, c=6.25

Note that the Discriminant is negative:b² − 4ac = (−4)² − 4×1×6.25
= −9

Use the Quadratic Formula:x = −(−4) ± √(−9)2

√(−9) = 3i
(where i is the imaginary number √−1)

So:x = 4 ± 3i2

Answer: x = 2 ± 1.5i

The graph does not cross the x-axis. That is why we ended up with complex numbers.

BUT an upside-down mirror image of our equation does cross the x-axis at 2 ± 1.5 (note: missing the i).

Just an interesting fact for you!

____________________________________________________________

Youtube Video: Quadratic Formula

A quadratic equation is a second-order polynomial equation in a single variable

(1)

with . Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions. These solutions may be both real, or both complex.

Among his many other talents, Major General Stanley in Gilbert and Sullivan's operetta the Pirates of Penzance impresses the pirates with his knowledge of quadratic equations in "The Major General's Song" as follows: "I am the very model of a modern Major-General, I've information vegetable, animal, and mineral, I know the kings of England, and I quote the fights historical, From Marathon to Waterloo, in order categorical; I'm very well acquainted too with matters mathematical, I understand equations, both the simple and quadratical, About binomial theorem I'm teeming with a lot o' news-- With many cheerful facts about the square of the hypotenuse."

The roots can be found by completing the square,

(2)

src="https://mathworld.wolfram.com/images/equations/QuadraticEquation/NumberedEquation3.gif" style="border: 0px;" width="217" />

(3)

(4)

Solving for then gives

(5)

This equation is known as the quadratic formula.

Here are the steps required to solve a quadratic using the quadratic formula:

Youtube Video: Quadratic Formula

Adding Polynomials | Polynomials Class 10/9 | Class 10 Maths Chapter 2 | Basics of Polynomial

2021-06-04T03:18:00.006+05:30

Adding Polynomials | Polynomials Class 10/9 | Class 10 Maths Chapter 2 | Basics of Polynomial

Polynomial addition and subtraction

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Adding Polynomials | Polynomials Class 10/9 | Class 10 Maths Chapter 2 | Basics of Polynomial

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Algebraic Expression│Class 7 RS Aggarwal │Introduction of chapter 6

2021-06-04T02:41:00.005+05:30

Algebraic Expression│Class 7 RS Aggarwal │Introduction of chapter 6

Algebraic Expression│Class 7 RS Aggarwal │Introduction of chapter 6│Class 7 chapter 6 in hindi

Youtube Video:- Algebraic Expressions

What is an Algebraic Expression?

An algebraic expression in mathematics is an expression which is made up of variables and constants, along with algebraic operations (addition, subtraction, etc.). Expressions are made up of terms. They are also termed algebraic equations.

Examples

3x + 4y – 7, 4x – 10, etc.

These expressions are represented with the help of unknown variables, constants and coefficients. The combination of these three (as terms) is said to be an expression. It is to be noted that, unlike the algebraic equation, an algebraic expression has no sides or equal to sign. Some of its examples include

3x + 2y – 5
x – 20
2x2 − 3xy + 5

Variables, Coefficient & Constant

In Algebra we work with Variable, Symbols or Letters whose value is unknown to us.

In the above expression (i.e. 5x – 3),

x is a variable, whose value is unknown to us which can take any value.
5 is known as the coefficient of x, as it is a constant value used with the variable term and is well defined.
3 is the constant value term which has a definite value.

The whole expression is known to be the Binomial term, as it has two unlikely terms.

Types of Algebraic expression

There are 3 main types of algebraic expressions which include:

Monomial Expression
Binomial Expression
Polynomial Expression

Monomial Expression

An algebraic expression which is having only one term is known as a monomial.

Examples of monomial expression include 3x4, 3xy, 3x, 8y, etc.

Binomial Expression

A binomial expression is an algebraic expression which is having two terms, which are unlike.

Examples of binomial include 5xy + 8, xyz + x3, etc.

Polynomial Expression

In general, an expression with more than one terms with non-negative integral exponents of a variable is known as a polynomial.

Examples of polynomial expression include ax + by + ca, x3 + 2x + 3, etc.

Other Types of Expression:

Apart from monomial, binomial and polynomial types of expressions, an algebraic expression can also be classified into two additional types which are:

Numeric Expression
Variable Expression

Numeric Expression

A numeric expression consists of numbers and operations, but never include any variable. Some of the examples of numeric expressions are 10 + 5, 15 ÷ 2, etc.

Variable Expression

A variable expression is an expression which contains variables along with numbers and operation to define an expression. A few examples of a variable expression include 4x + y, 5ab + 33, etc.

Algebraic expression for Class 7

In Class 7, students will come across the terms of algebraic equations such as:

Coefficient of a term
Variables
Constant
Factors of a term
Terms of equations
Like and Unlike terms

Example of using these terms are given below.

If 2x2+3xy+4x+7 is an algebraic expression.

Then, 2x2, 3xy, 4x and 7 are the terms

Coefficient of term x2 = 2

Constant term = 7

Example of like and unlike terms:

Like terms: 2x and 3x

Unlike terms: 2x and 3y

Factors of a term:

If 3xy is a term, then its factors are 3, x and y.

Monomial, Binomial & Trinomial

Also, in grade 7 we will learn about types of expressions, such as monomial, binomial and trinomial. Let us see examples of each.

Monomial: 2x

Binomial: 2x+3y

Trinomial: 2x+3y+9

Addition and Subtraction of Algebraic Expressions

We can add and subtract like terms easily.

Example: Add 3x + 5y – 6z and x – 4y + 2z.

By adding both the expressions we get;

(3x + 5y – 6z) + (x – 4y + 2z)

Separating the like terms and adding them together:

(3x + x) + (5y – 4y) + (-6z + 2z)

4x + y – 4z

Formulas

The general algebraic formulas we use to solve the expressions or equations are:

(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
a2 – b2 = (a – b)(a + b)
(a + b)3 = a3 + b3 + 3ab(a + b)
(a – b)3 = a3 – b3 – 3ab(a – b)
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)

Solved Problem

Example: Simplify the given expressions by combining the like terms and write the type of Algebraic expression.

(i) 3xy3 + 9x2 y3 + 5y3x

(ii) 7ab2 c2 + 2a3 b2 − 3abc – 5ab2 c2 – 2b2 a3 + 2ab

(iii) 50x3 – 20x + 8x + 21x3 – 3x + 15x – 41x3

Solution:

Creating a table to find the solution:

S.no	Term	Simplification	Type of Expression
1	3xy3 + 9x2 y3 + 5y3x	8xy3 + 9x2y3	Binomial
2	7ab2 c2 + 2a3 b2 − 3abc – 5ab2 c2 – 2b2 a3 + 2ab	2ab2 c2 − 3abc + 2ab	Trinomial
3	50x3 – 20x + 8x + 21x3 – 3x + 15x – 41x3	30x³	Monomial

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Scientific notation │Scientific Notation in hindi │What is Scientific Notation?

2021-06-03T22:32:00.009+05:30

Scientific notation │Scientific Notation in hindi │What is Scientific Notation?

Scientific notation is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10. For example, 650,000,000 can be written in scientific notation as 6.5 ✕ 10^8.

Youtube Video:- Scientific Notation

Scientific Notation (also called Standard Form in Britain) is a special way of writing numbers:

Like this:

Or this:

It makes it easy to use big and small values.

OK, How Does it Work?

Example: 700

Why is 700 written as 7 × 10² in Scientific Notation ?

700 = 7 × 100

and 100 = 10² (see powers of 10)

so 700 = 7 × 10²

Both 700 and 7 × 10² have the same value, just shown in different ways.

Example: 4,900,000,000

1,000,000,000 = 10⁹ ,
so 4,900,000,000 = 4.9 × 10⁹ in Scientific Notation

The number is written in two parts:

Just the digits, with the decimal point placed after the first digit, followed by
× 10 to a power that puts the decimal point where it should be
(i.e. it shows how many places to move the decimal point).

In this example, 5326.6 is written as 5.3266 × 10³,
because 5326.6 = 5.3266 × 1000 = 5.3266 × 10³

Other Ways of Writing It

3.1 × 10^8

We can use the ^ symbol (above the 6 on a keyboard), as it is easy to type.

Example: 3 × 10^4 is the same as 3 × 10⁴

3 × 10^4 = 3 × 10 × 10 × 10 × 10 = 30,000

Calculators often use "E" or "e" like this:

Example: 6E+5 is the same as 6 × 10⁵

6E+5 = 6 × 10 × 10 × 10 × 10 × 10 = 600,000

Example: 3.12E4 is the same as 3.12 × 10⁴

3.12E4 = 3.12 × 10 × 10 × 10 × 10 = 31,200

How to Do it

To figure out the power of 10, think "how many places do I move the decimal point?"

	When the number is 10 or greater, the decimal point has to move to the left, and the power of 10 is positive.

	When the number is smaller than 1, the decimal point has to move to the right, so the power of 10 is negative.

Example: 0.0055 is written 5.5 × 10^-3

Because 0.0055 = 5.5 × 0.001 = 5.5 × 10^-3

Example: 3.2 is written 3.2 × 10⁰

We didn't have to move the decimal point at all, so the power is 10⁰

But it is now in Scientific Notation

Check!

After putting the number in Scientific Notation, just check that:

The "digits" part is between 1 and 10 (it can be 1, but never 10)
The "power" part shows exactly how many places to move the decimal point

Why Use It?

Because it makes it easier when dealing with very big or very small numbers, which are common in Scientific and Engineering work.

Example: it is easier to write (and read) 1.3 × 10^-9 than 0.0000000013

It can also make calculations easier, as in this example:

Example: a tiny space inside a computer chip has been measured to be 0.00000256m wide, 0.00000014m long and 0.000275m high.

What is its volume?

Let's first convert the three lengths into scientific notation:

width: 0.000 002 56m = 2.56×10^-6
length: 0.000 000 14m = 1.4×10^-7
height: 0.000 275m = 2.75×10^-4

Then multiply the digits together (ignoring the ×10s):

2.56 × 1.4 × 2.75 = 9.856

Last, multiply the ×10s:

10^-6 × 10^-7 × 10^-4 = 10^-17 (easier than it looks, just add −6, −4 and −7 together)

The result is 9.856×10^-17 m³

It is used a lot in Science:

Example: Suns, Moons and Planets

The Sun has a Mass of 1.988 × 10³⁰ kg.

Easier than writing 1,988,000,000,000,000,000,000,000,000,000 kg
(and that number gives a false sense of many digits of accuracy.

It can also save space! Here is what happens when you double on each square of a chess board:

Values are rounded off, so 53,6870,912 is shown as just 5×10⁸

That last value, shown as 9×10¹⁸ is actually 9,223,372,036,854,775,808

Engineering Notation

Engineering Notation is like Scientific Notation, except that we only use powers of ten that are multiples of 3 (such as 10³, 10^-3, 10¹² etc).

Examples:

2,700 is written 2.7 × 10³
27,000 is written 27 × 10³
270,000 is written 270 × 10³
2,700,000 is written 2.7 × 10⁶

Example: 0.00012 is written 120 × 10^-6

Notice that the "digits" part can now be between 1 and 1,000 (it can be 1, but never 1,000).

The advantage is that we can replace the ×10s with Metric Numbers. So we can use standard words (such as thousand or million), prefixes (such as kilo, mega) or the symbol (k, M, etc)

Example: 19,300 meters is written 19.3 × 10³ m, or 19.3 km

Example: 0.00012 seconds is written 120 × 10^-6 s, or 120 microseconds

Class 10 Maths NCERT Ch 3 Linear Equations in two variables | CBSE Class 10 Maths

2021-05-28T02:18:00.003+05:30

Class 10 Maths NCERT Ch 3 Linear Equations in two variables | CBSE Class 10 Maths

Access Answers of Maths NCERT Class 10 Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.4

1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

(iv) x/2+ 2y/3 = -1 and x-y/3 = 3

Solutions:

(i) x + y = 5 and 2x – 3y = 4

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get,

5y = 6

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

By the method of substitution.

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

When the values are substituted in equation (v), we get:

x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

By the method of elimination.

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:

7x = 14

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2 and y = 1

By the method of Substitution

From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10

7y = 7

y = 1

When y = 1 is substituted in equation (v) we get,

A = 1 + 1 = 2

Therefore, A = 2 and B = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,

9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii) we get,

13y = -5

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,

3x +25/13 −4=0

3x = 27/13

x =9/13

∴x = 9/13 and y = -5/13

By the method of Substitution:

From the equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0

13y = -5

y = -5/13

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3

x = 9/13

∴x = 9/13, y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

By the method of Elimination.

3x + 4y = -6 …………………………. (i)

x-y/3 = 3

3x – y = 9 ……………………………. (ii)

When the equation (ii) is subtracted from equation (i) we get,

-5y = -15

y = 3 ………………………………….(iii)

When the equation (iii) is substituted in (i) we get,

3x – 12 = -6

3x = 6

x = 2

Hence, x = 2 , y = -3

By the method of Substitution:

From the equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6

5y = -15

y = -3

When y = -3 is substituted in equation (v) we get,

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?

Solution:

Let the fraction be a/b

According to the given information,

(a+1)/(b-1) = 1

=> a – b = -2 ………………………………..(i)

a/(b+1) = 1/2

=> 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii) we get,

a = 3 …………………………………………………..(iii)

When a = 3 is substituted in equation (i) we get,

3 – b = -2

-b = -5

b = 5

Hence, the fraction is 3/5.

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution:

Let us assume, present age of Nuri is x

And present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

Now,

x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get,

y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,

x – 3.20 = -10

x – 60 = -10

x = 50

Therefore,

Age of Nuri is 50 years

Age of Sonu is 20 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number (n) = 10B + A

N after reversing order of the digits = 10A + B

According to the given information, A + B = 9…………………….(i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 ………………………………………………………….. (ii)

Adding the equations (i) and (ii) we get,

9B = 9

B = 1……………………………………………………………………….(3)

Substituting this value of B, in the equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

(iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.

Solution:

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

According to the given information,

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting the equation (iii) from the equation (ii) we get,

50B = 750

B = 15

Substituting in the equation (i) we get,

A = 10

Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs.15

And the Charge per day is Rs.3

Class 10 Maths NCERT Ch 3 Linear Equations in two variables | CBSE Class 10 Maths

2021-05-22T14:14:00.002+05:30

Class 10 Maths NCERT Ch 3 Linear Equations in two variables | CBSE Class 10 Maths

#_____________________________________________________

Class 10 Maths NCERT Ch 3 Linear Equations in two variables | CBSE Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables will help the students in understanding how the problems under this concept are solved. Maths is one subject that requires a lot of practice. The students appearing for the 10th grade board examinations can turn to the NCERT Solutions Class 10 for reference. These solutions of the Chapter Pair of Linear Equations in Two Variables lay out step-wise answers to all the Maths problems in the NCERT textbook. An equation that can be of the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y. The chapter also lets the students understand the fact that a solution of such an equation is a pair of values, one for x and the other for y, which makes the two sides of the equation equal. Students also learn that every solution of the equation is a point on the line representing it. Students get to learn the problem-solving method of the questions present in Chapter 3 Pair of Linear Equations in Two Variables, revolving around the concepts mentioned above and more, by practising the NCERT Solutions provided underneath. These Solutions of NCERT are extremely beneficial from the examination perspective.

Youtube video: https://youtu.be/JiUOPIdaTrQ

Youtube video: Class 10 Maths NCERT Ch 3 Linear Equations in two variables | CBSE Class 10 Maths

Access Answers of Maths NCERT class 10 Chapter 3 – Pair of Linear Equations in Two Variables

Exercise 3.1 Page: 44

1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Solutions:Let the present age of Aftab be ‘x’.

And, the present age of his daughter be ‘y’.

Now, we can write, seven years ago,

Age of Aftab = x-7

Age of his daughter = y-7

According to the question,

x−7 = 7(y−7)

⇒x−7 = 7y−49

⇒x−7y = −42 ………………………(i)

Also, three years from now or after three years,

Age of Aftab will become = x+3.

Age of his daughter will become = y+3

According to the situation given,

x+3 = 3(y+3)

⇒x+3 = 3y+9

⇒x−3y = 6 …………..…………………(ii)

Subtracting equation (i) from equation (ii) we have

(x−3y)−(x−7y) = 6−(−42)

⇒−3y+7y = 6+42

⇒4y = 48

⇒y = 12

The algebraic equation is represented by

x−7y = −42

x−3y = 6

For, x−7y = −42 or x = −42+7y

The solution table is

For, x−3y = 6 or x = 6+3y

The solution table is

The graphical representation is:

2. The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later, she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.

Solutions: Let us assume that the cost of a bat be ‘Rs x’

And,the cost of a ball be ‘Rs y’

According to the question, the algebraic representation is

3x+6y = 3900

And x+3y = 1300

For, 3x+6y = 3900

Or x = (3900-6y)/3

The solution table is

For, x+3y = 1300

Or x = 1300-3y

The solution table is

The graphical representation is as follows.

3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.

Solutions:Let the cost of 1 kg of apples be ‘Rs. x’

And, cost of 1 kg of grapes be ‘Rs. y’

According to the question, the algebraic representation is

2x+y = 160

And 4x+2y = 300

For, 2x+y = 160 or y = 160−2x, the solution table is;

For 4x+2y = 300 or y = (300-4x)/2, the solution table is;

The graphical representation is as follows;

Exercise 3.2 Page: 49

1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

Solution:

(i)Let there are x number of girls and y number of boys. As per the given question, the algebraic expression can be represented as follows.

x +y = 10

x– y = 4

Now, for x+y = 10 or x = 10−y, the solutions are;

For x – y = 4 or x = 4 + y, the solutions are;

The graphical representation is as follows;

From the graph, it can be seen that the given lines cross each other at point (7, 3). Therefore, there are 7 girls and 3 boys in the class.

(ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y.

According to the question, the algebraic expression cab be represented as;

5x + 7y = 50

7x + 5y = 46

For, 5x + 7y = 50 or x = (50-7y)/5, the solutions are;

For 7x + 5y = 46 or x = (46-5y)/7, the solutions are;

Hence, the graphical representation is as follows;

From the graph, it is can be seen that the given lines cross each other at point (3, 5).

So, the cost of a pencil is 3/- and cost of a pen is 5/-.

2. On comparing the ratios a1/a2 , b1/b2 , c1/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Solutions:

(i) Given expressions;

5x−4y+8 = 0

7x+6y−9 = 0

Comparing these equations with a1x+b1y+c1 = 0

And a2x+b2y+c2 = 0

We get,

a1 = 5, b1 = -4, c1 = 8

a2 = 7, b2 = 6, c2 = -9

(a1/a2) = 5/7

(b1/b2) = -4/6 = -2/3

(c1/c2) = 8/-9

Since, (a1/a2) ≠ (b1/b2)

So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(ii) Given expressions;

9x + 3y + 12 = 0

18x + 6y + 24 = 0

Comparing these equations with a1x+b1y+c1 = 0

And a2x+b2y+c2 = 0

We get,

a1 = 9, b1 = 3, c1 = 12

a2 = 18, b2 = 6, c2 = 24

(a1/a2) = 9/18 = 1/2

(b1/b2) = 3/6 = 1/2

(c1/c2) = 12/24 = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2)

So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) Given Expressions;

6x – 3y + 10 = 0

2x – y + 9 = 0

Comparing these equations with a1x+b1y+c1 = 0

And a2x+b2y+c2 = 0

We get,

a1 = 6, b1 = -3, c1 = 10

a2 = 2, b2 = -1, c2 = 9

(a1/a2) = 6/2 = 3/1

(b1/b2) = -3/-1 = 3/1

(c1/c2) = 10/9

Since (a1/a2) = (b1/b2) ≠ (c1/c2)

So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

3. On comparing the ratio, (a1/a2) , (b1/b2) , (c1/c2) find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

(iii)(3/2)x+(5/3)y = 7; 9x – 10y = 14

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

(v)(4/3)x+2y = 8 ; 2x + 3y = 12

Solutions:

(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0

and 2x – 3y = 7 or 2x – 3y -7 = 0

Comparing these equations with a1x+b1y+c1 = 0

And a2x+b2y+c2 = 0

We get,

a1 = 3, b1 = 2, c1 = -5

a2 = 2, b2 = -3, c2 = -7

(a1/a2) = 3/2

(b1/b2) = 2/-3

(c1/c2) = -5/-7 = 5/7

Since, (a1/a2) ≠ (b1/b2)

So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9

Therefore,

a1 = 2, b1 = -3, c1 = -8

a2 = 4, b2 = -6, c2 = -9

(a1/a2) = 2/4 = 1/2

(b1/b2) = -3/-6 = 1/2

(c1/c2) = -8/-9 = 8/9

Since , (a1/a2) = (b1/b2) ≠ (c1/c2)

So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

(iii)Given (3/2)x + (5/3)y = 7 and 9x – 10y = 14

Therefore,

a1 = 3/2, b1 = 5/3, c1 = -7

a2 = 9, b2 = -10, c2 = -14

(a1/a2) = 3/(2×9) = 1/6

(b1/b2) = 5/(3× -10)= -1/6

(c1/c2) = -7/-14 = 1/2

Since, (a1/a2) ≠ (b1/b2)

So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent.

(iv) Given, 5x – 3y = 11 and – 10x + 6y = –22

Therefore,

a1 = 5, b1 = -3, c1 = -11

a2 = -10, b2 = 6, c2 = 22

(a1/a2) = 5/(-10) = -5/10 = -1/2

(b1/b2) = -3/6 = -1/2

(c1/c2) = -11/22 = -1/2

Since (a1/a2) = (b1/b2) = (c1/c2)

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

(v)Given, (4/3)x +2y = 8 and 2x + 3y = 12

a1 = 4/3 , b1= 2 , c1 = -8

a2 = 2, b2 = 3 , c2 = -12

(a1/a2) = 4/(3×2)= 4/6 = 2/3

(b1/b2) = 2/3

(c1/c2) = -8/-12 = 2/3

Since (a1/a2) = (b1/b2) = (c1/c2)

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i) x + y = 5, 2x + 2y = 10

(ii) x – y = 8, 3x – 3y = 16

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solutions:

(i)Given, x + y = 5 and 2x + 2y = 10

(a1/a2) = 1/2

(b1/b2) = 1/2

(c1/c2) = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2)

∴The equations are coincident and they have infinite number of possible solutions.

So, the equations are consistent.

For, x + y = 5 or x = 5 – y

For 2x + 2y = 10 or x = (10-2y)/2

So, the equations are represented in graphs as follows:

From the figure, we can see, that the lines are overlapping each other.

Therefore, the equations have infinite possible solutions.

(ii) Given, x – y = 8 and 3x – 3y = 16

(a1/a2) = 1/3

(b1/b2) = -1/-3 = 1/3

(c1/c2) = 8/16 = 1/2

Since, (a1/a2) = (b1/b2) ≠ (c1/c2)

The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.

(iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0

(a1/a2) = 2/4 = ½

(b1/b2) = 1/-2

(c1/c2) = -6/-4 = 3/2

Since, (a1/a2) ≠ (b1/b2)

The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent.

Now, for 2x + y – 6 = 0 or y = 6 – 2x

And for 4x – 2y – 4 = 0 or y = (4x-4)/2

So, the equations are represented in graphs as follows:

From the graph, it can be seen that these lines are intersecting each other at only one point,(2,2).

(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0

(a1/a2) = 2/4 = ½

(b1/b2) = -2/-4 = 1/2

(c1/c2) = 2/5

Since, a1/a2 = b1/b2 ≠ c1/c2

Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solutions:Let us consider.

The width of the garden is x and length is y.

Now, according to the question, we can express the given condition as;

y – x = 4

and

y + x = 36

Now, taking y – x = 4 or y = x + 4

For y + x = 36, y = 36 – x

The graphical representation of both the equation is as follows;

From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) Parallel lines

(iii) Coincident lines

Solutions:

(i) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;

(a1/a2) ≠ (b1/b2)

Thus, another equation could be 2x – 7y + 9 = 0, such that;

(a1/a2) = 2/2 = 1 and (b1/b2) = 3/-7

Clearly, you can see another equation satisfies the condition.

(ii) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;

(a1/a2) = (b1/b2) ≠ (c1/c2)

Thus, another equation could be 6x + 9y + 9 = 0, such that;

(a1/a2) = 2/6 = 1/3

(b1/b2) = 3/9= 1/3

(c1/c2) = -8/9

Clearly, you can see another equation satisfies the condition.

(iii) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;

(a1/a2) = (b1/b2) = (c1/c2)

Thus, another equation could be 4x + 6y – 16 = 0, such that;

(a1/a2) = 2/4 = 1/2 ,(b1/b2) = 3/6 = 1/2, (c1/c2) = -8/-16 = 1/2

Clearly, you can see another equation satisfies the condition.

7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution: Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.

For, x – y + 1 = 0 or x = 1+y

For, 3x + 2y – 12 = 0 or x = (12-2y)/3

Hence, the graphical representation of these equations is as follows;

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

Exercise 3.3 Page: 53

1. Solve the following pair of linear equations by the substitution method

(i) x + y = 14

x – y = 4

(ii) s – t = 3

(s/3) + (t/2) = 6

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) √2 x+√3 y = 0

√3 x-√8 y = 0

(vi) (3x/2) – (5y/3) = -2

(x/3) + (y/2) = (13/6)

Solutions:

(i) Given,

x + y = 14 and x – y = 4 are the two equations.

From 1st equation, we get,

x = 14 – y

Now, substitute the value of x in second equation to get,

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Or y = 5

By the value of y, we can now find the exact value of x;

∵ x = 14 – y

∴ x = 14 – 5

Or x = 9

Hence, x = 9 and y = 5.

(ii) Given,

s – t = 3 and (s/3) + (t/2) = 6 are the two equations.

From 1st equation, we get,

s = 3 + t ________________(1)

Now, substitute the value of s in second equation to get,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

⇒5t = 30

⇒t = 6

Now, substitute the value of t in equation (1)

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

(iii) Given,

3x – y = 3 and 9x – 3y = 9 are the two equations.

From 1st equation, we get,

x = (3+y)/3

Now, substitute the value of x in the given second equation to get,

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

⇒ 9 = 9

Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.

(iv) Given,

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations.

From 1st equation, we get,

x = (1.3- 0.3y)/0.2 _________________(1)

Now, substitute the value of x in the given second equation to get,

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

⇒ y = 3

Now, substitute the value of y in equation (1), we get,

x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

(v) Given,

√2 x + √3 y = 0 and √3 x – √8 y = 0

are the two equations.

From 1st equation, we get,

x = – (√3/√2)y __________________(1)

Putting the value of x in the given second equation to get,

√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0

⇒ y = 0

Now, substitute the value of y in equation (1), we get,

x = 0

Therefore, x = 0 and y = 0.

(vi) Given,

(3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations.

From 1st equation, we get,

(3/2)x = -2 + (5y/3)

⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1)

Putting the value of x in the given second equation to get,

((-12+10y)/9)/3 + y/2 = 13/6

⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6

Now, substitute the value of y in equation (1), we get,

(3x/2) – 5(3)/3 = -2

⇒ (3x/2) – 5 = -2

⇒ x = 2

Therefore, x = 2 and y = 3.

2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11…………………………..(I)

2x – 4y = -24………………………… (II)

From equation (II), we get

x = (11-3y)/2 ………………….(III)

Substituting the value of x in equation (II), we get

2(11-3y)/2 – 4y = 24

11 – 3y – 4y = -24

-7y = -35

y = 5……………………………………..(IV)

Putting the value of y in equation (III), we get

x = (11-3×5)/2 = -4/2 = -2

Hence, x = -2, y = 5

Also,

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Therefore the value of m is -1.

3. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Solution:

Let the two numbers be x and y respectively, such that y > x.

According to the question,

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of (1) into (2), we get

3x – x = 26

x = 13 ……………. (3)

Substituting (3) in (1), we get y = 39

Hence, the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:

Let the larger angle by xo and smaller angle be yo.

We know that the sum of two supplementary pair of angles is always 180o.

According to the question,

x + y = 180o……………. (1)

x – y = 18o ……………..(2)

From (1), we get x = 180o – y …………. (3)

Substituting (3) in (2), we get

180o – y – y =18o

162o = 2y

y = 81o ………….. (4)

Using the value of y in (3), we get

x = 180o – 81o

= 99o

Hence, the angles are 99o and 81o.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.

Solution:

Let the cost a bat be x and cost of a ball be y.

According to the question,

7x + 6y = 3800 ………………. (I)

3x + 5y = 1750 ………………. (II)

From (I), we get

y = (3800-7x)/6………………..(III)

Substituting (III) in (II). we get,

3x+5(3800-7x)/6 =1750

⇒3x+ 9500/3 – 35x/6 = 1750

⇒3x- 35x/6 = 1750 – 9500/3

⇒(18x-35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/3

⇒-17x = -8500

x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get

y = (3800-7 ×500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Solution:

Let the fixed charge be Rs x and per km charge be Rs y.

According to the question,

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1), we get x = 105 – 10y ………………. (3)

Substituting the value of x in (2), we get

105 – 10y + 15y = 155

5y = 50

y = 10 …………….. (4)

Putting the value of y in (3), we get

x = 105 – 10 × 10 = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution:

Let the fraction be x/y.

According to the question,

(x+2) /(y+2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

(x+3) /(y+3) = 5/6

6x + 18 = 5y +15

6x – 5y = -3 ………………. (2)

From (1), we get x = (-4+9y)/11 …………….. (3)

Substituting the value of x in (2), we get

6(-4+9y)/11 -5y = -3

-24 + 54y – 55y = -33

-y = -9

y = 9 ………………… (4)

Substituting the value of y in (3), we get

x = (-4+9×9 )/11 = 7

Hence the fraction is 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solutions:

Let the age of Jacob and his son be x and y respectively.

According to the question,

(x + 5) = 3(y + 5)

x – 3y = 10 …………………………………….. (1)

(x – 5) = 7(y – 5)

x – 7y = -30 ………………………………………. (2)

From (1), we get x = 3y + 10 ……………………. (3)

Substituting the value of x in (2), we get

3y + 10 – 7y = -30

-4y = -40

y = 10 ………………… (4)

Substituting the value of y in (3), we get

x = 3 x 10 + 10 = 40

Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.

Exercise 3.4 Page: 56

1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

(iv) x/2+ 2y/3 = -1 and x-y/3 = 3

Solutions:

(i) x + y = 5 and 2x – 3y = 4

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get,

5y = 6

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

By the method of substitution.

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

When the values are substituted in equation (v), we get:

x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

By the method of elimination.

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:

7x = 14

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2 and y = 1

By the method of Substitution

From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10

7y = 7

y = 1

When y = 1 is substituted in equation (v) we get,

A = 1 + 1 = 2

Therefore, A = 2 and B = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,

9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii) we get,

13y = -5

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,

3x +25/13 −4=0

3x = 27/13

x =9/13

∴x = 9/13 and y = -5/13

By the method of Substitution:

From the equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0

13y = -5

y = -5/13

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3

x = 9/13

∴x = 9/13, y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

By the method of Elimination.

3x + 4y = -6 …………………………. (i)

x-y/3 = 3

3x – y = 9 ……………………………. (ii)

When the equation (ii) is subtracted from equation (i) we get,

-5y = -15

y = 3 ………………………………….(iii)

When the equation (iii) is substituted in (i) we get,

3x – 12 = -6

3x = 6

x = 2

Hence, x = 2 , y = -3

By the method of Substitution:

From the equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6

5y = -15

y = -3

When y = -3 is substituted in equation (v) we get,

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?

Solution:

Let the fraction be a/b

According to the given information,

(a+1)/(b-1) = 1

=> a – b = -2 ………………………………..(i)

a/(b+1) = 1/2

=> 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii) we get,

a = 3 …………………………………………………..(iii)

When a = 3 is substituted in equation (i) we get,

3 – b = -2

-b = -5

b = 5

Hence, the fraction is 3/5.

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution:

Let us assume, present age of Nuri is x

And present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

Now,

x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get,

y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,

x – 3.20 = -10

x – 60 = -10

x = 50

Therefore,

Age of Nuri is 50 years

Age of Sonu is 20 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number (n) = 10B + A

N after reversing order of the digits = 10A + B

According to the given information, A + B = 9…………………….(i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 ………………………………………………………….. (ii)

Adding the equations (i) and (ii) we get,

9B = 9

B = 1……………………………………………………………………….(3)

Substituting this value of B, in the equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

(iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.

Solution:

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

According to the given information,

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting the equation (iii) from the equation (ii) we get,

50B = 750

B = 15

Substituting in the equation (i) we get,

A = 10

Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.

Solution:

Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs.15

And the Charge per day is Rs.3

Exercise 3.5 Page: 62

1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8

(iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

Solutions:

(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0

a1/a2=1/3 , b1/b2= -3/-9 =1/3, c1/c2=-3/-2 = 3/2

(a1/a2) = (b1/b2) ≠ (c1/c2)

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

(ii) Given, 2x + y = 5 and 3x +2y = 8

a1/a2 = 2/3 , b1/b2 = 1/2 , c1/c2 = -5/-8

(a1/a2) ≠ (b1/b2)

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

x/(b1c2-c1b2) = y/(c1a2 – c2a=) = 1/(a1b2-a2b1)

x/(-8-(-10)) = y/(15+16) = 1/(4-3)

x/2 = y/1 = 1

∴ x = 2 and y =1

(iii) Given, 3x – 5y = 20 and 6x – 10y = 40

(a1/a2) = 3/6 = 1/2

(b1/b2) = -5/-10 = 1/2

(c1/c2) = 20/40 = 1/2

a1/a2 = b1/b2 = c1/c2

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(a1/a2) = 1/3

(b1/b2) = -3/-3 = 1

(c1/c2) = -7/-15

a1/a2 ≠ b1/b2

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)

x/24 = y/ -6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

∴ x = 4 and y = 1.

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k – 1) x + (k – 1) y = 2k + 1

Solution:

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a1/a2 = 2/(a-b) , b1/b2 = 3/(a+b) , c1/c2 = -7/-(3a + b -2)

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4 ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a -5 x 1= 0

a = 5

Thus at a = 5 and b = 1 the given equations will have infinite solutions.

(ii) 3x + y -1 = 0

(2k -1)x + (k-1)y – 2k -1 = 0

a1/a2 = 3/(2k -1) , b1/b2 = 1/(k-1), c1/c2 = -1/(-2k -1) = 1/( 2k +1)

For no solutions

a1/a2 = b1/b2 ≠ c1/c2

3/(2k-1) = 1/(k -1) ≠ 1/(2k +1)

3/(2k –1) = 1/(k -1)

3k -3 = 2k -1

k =2

Therefore, for k = 2 the given pair of linear equations will have no solution.

3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get

x = (4 – 2y )/ 3 ……………………. (3)

Using this value in equation 1, we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

∴ x = -2 and y =5.

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solutions:

(i) Let x be the fixed charge and y be the charge of food per day.

According to the question,

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from (ii) we get

6y = 180

y = Rs.30

Using this value in equation (ii) we get

x = 1180 -26 x 30

x= Rs.400.

Therefore, fixed charges is Rs.400 and charge per day is Rs.30.

(ii) Let the fraction be x/y.

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4 => 4x –y =8 ………………..(2)

Subtracting equation (1) from (2) , we get

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

(4×5)– y = 8

y= 12

Therefore, the fraction is 5/12.

(iii) Let the number of right answers is x and number of wrong answers be y

According to the given question;

3x−y=40……..(1)

4x−2y=50

⇒2x−y=25…….(2)

Subtracting equation (2) from equation (1), we get;

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25

Or y = 5

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, total number of questions = 20

(iv) Let x km/h be the speed of car from point A and y km/h be the speed of car from point B.

If the car travels in the same direction,

5x – 5y = 100

x – y = 20 …………………………………(i)

If the car travels in the opposite direction,

x + y = 100………………………………(ii)

Solving equation (i) and (ii), we get

x = 60 km/h………………………………………(iii)

Using this in equation (i), we get,

60 – y = 20

y = 40 km/h

Therefore, the speed of car from point A = 60 km/h

Speed of car from point B = 40 km/h.

(v) Let,

The length of rectangle = x unit

And breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0……………………………(1)

(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

Therefore, x = 17 and y = 9.

Hence, the length of rectangle = 17 units

And breadth of the rectangle = 9 units

Exercise 3.6 Page: 67

1. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Solution:

Let us assume 1/x = m and 1/y = n , then the equation will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0…………………….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0……………………….(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2

4/√x + 9/√y = -1

Solution:

Substituting 1/√x = m and 1/√y = n in the given equations, we get

2m + 3n = 2 ………………………..(i)

4m – 9n = -1 ………………………(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 ………………….…..(iii)

Adding equation (ii) and (iii), we get

10m = 5

m = 1/2…………………………….…(iv)

Now by putting the value of ‘m’ in equation (i), we get

2×1/2 + 3n = 2

3n = 1

n = 1/3

m =1/√x

½ = 1/√x

x = 4

n = 1/√y

1/3 = 1/√y

y = 9

Hence, x = 4 and y = 9

(iii) 4/x + 3y = 14

3/x -4y = 23

Solution:

Putting in the given equation we get,

So, 4m + 3y = 14 => 4m + 3y – 14 = 0 ……………..…..(1)

3m – 4y = 23 => 3m – 4y – 23 = 0 ……………………….(2)

By cross-multiplication, we get,

m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

-m/125 = y/50 = -1/ 25

-m/125 = -1/25 and y/50 = -1/25

m = 5 and b = -2

m = 1/x = 5

So , x = 1/5

y = -2

(iv) 5/(x-1) + 1/(y-2) = 2

6/(x-1) – 3/(y-2) = 1

Solution:

Substituting 1/(x-1) = m and 1/(y-2) = n in the given equations, we get,

5m + n = 2 …………………………(i)

6m – 3n = 1 ……………………….(ii)

Multiplying equation (i) by 3, we get

15m + 3n = 6 …………………….(iii)

Adding (ii) and (iii) we get

21m = 7

m = 1/3

Putting this value in equation (i), we get

5×1/3 + n = 2

n = 2- 5/3 = 1/3

m = 1/ (x-1)

⇒ 1/3 = 1/(x-1)

⇒ x = 4

n = 1/(y-2)

⇒ 1/3 = 1/(y-2)

⇒ y = 5

Hence, x = 4 and y = 5

(v) (7x-2y)/ xy = 5

(8x + 7y)/xy = 15

Solution:

(7x-2y)/ xy = 5

7/y – 2/x = 5…………………………..(i)

(8x + 7y)/xy = 15

8/y + 7/x = 15…………………………(ii)

Substituting 1/x =m in the given equation we get,

– 2m + 7n = 5 => -2 + 7n – 5 = 0 ……..(iii)

7m + 8n = 15 => 7m + 8n – 15 = 0 ……(iv)

By cross-multiplication method, we get,

m/(-105-(-40)) = n/(-35-30) = 1/(-16-49)

m/(-65) = n/(-65) = 1/(-65)

m/-65 = 1/-65

m = 1

n/(-65) = 1/(-65)

n = 1

m = 1 and n = 1

m = 1/x = 1 n = 1/x = 1

Therefore, x = 1 and y = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

Solution:

6x + 3y = 6xy

6/y + 3/x = 6

Let 1/x = m and 1/y = n

=> 6n +3m = 6

=>3m + 6n-6 = 0…………………….(i)

2x + 4y = 5xy

=> 2/y + 4/x = 5

=> 2n +4m = 5

=> 4m+2n-5 = 0……………………..(ii)

3m + 6n – 6 = 0

4m + 2n – 5 = 0

By cross-multiplication method, we get

m/(-30 –(-12)) = n/(-24-(-15)) = 1/(6-24)

m/-18 = n/-9 = 1/-18

m/-18 = 1/-18

m = 1

n/-9 = 1/-18

n = 1/2

m = 1 and n = 1/2

m = 1/x = 1 and n = 1/y = 1/2

x = 1 and y = 2

Hence, x = 1 and y = 2

(vii) 10/(x+y) + 2/(x-y) = 4

15/(x+y) – 5/(x-y) = -2

Solution:

Substituting 1/x+y = m and 1/x-y = n in the given equations, we get,

10m + 2n = 4 => 10m + 2n – 4 = 0 ………………..…..(i)

15m – 5n = -2 => 15m – 5n + 2 = 0 ……………………..(ii)

Using cross-multiplication method, we get,

m/(4-20) = n/(-60-(20)) = 1/(-50 -30)

m/-16 = n/-80 = 1/-80

m/-16 = 1/-80 and n/-80 = 1/-80

m = 1/5 and n = 1

m = 1/(x+y) = 1/5

x+y = 5 …………………………………………(iii)

n = 1/(x-y) = 1

x-y = 1……………………………………………(iv)

Adding equation (iii) and (iv), we get

2x = 6 => x = 3 …….(v)

Putting the value of x = 3 in equation (3), we get

y = 2

Hence, x = 3 and y = 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4

1/2(3x+y) – 1/2(3x-y) = -1/8

Solution:

Substituting 1/(3x+y) = m and 1/(3x-y) = n in the given equations, we get,

m + n = 3/4 …………………………….…… (1)

m/2 – n/2 = -1/8

m – n = -1/4 …………………………..…(2)

Adding (1) and (2), we get

2m = 3/4 – 1/4

2m = 1/2

Putting in (2), we get

1/4 – n = -1/4

n = 1/4 + 1/4 = 1/2

m = 1/(3x+y) = 1/4

3x + y = 4 …………………………………(3)

n = 1/( 3x-y) = 1/2

3x – y = 2 ………………………………(4)

Adding equations (3) and (4), we get

6x = 6

x = 1 ……………………………….(5)

Putting in (3), we get

3(1) + y = 4

y = 1

Hence, x = 1 and y = 1

2. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solutions:

(i) Let us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

Adding both the eq.1 and 2, we get,

2x = 12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

Speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii) Let us consider,

Number of days taken by women to finish the work = x

Number of days taken by men to finish the work = y

Work done by women in one day = 1/x

Work done by women in one day = 1/y

As per the question given,

4(2/x + 5/y) = 1

(2/x + 5/y) = 1/4

And, 3(3/x + 6/y) = 1

(3/x + 6/y) = 1/3

Now, put 1/x=m and 1/y=n, we get,

2m + 5n = 1/4 => 8m + 20n = 1…………………(1)

3m + 6n =1/3 => 9m + 18n = 1………………….(2)

Now, by cross multiplication method, we get here,

m/(20-18) = n/(9-8) = 1/ (180-144)

m/2 = n/1 = 1/36

m/2 = 1/36

m = 1/18

m = 1/x = 1/18

or x = 18

n = 1/y = 1/36

y = 36

Therefore,

Number of days taken by women to finish the work = 18

Number of days taken by men to finish the work = 36.

(iii) Let us consider,

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

60/x + 240/y = 4 …………………(1)

100/x + 200/y = 25/6 …………….(2)

Put 1/x=m and 1/y=n, in the above two equations;

60m + 240n = 4……………………..(3)

100m + 200n = 25/6

600m + 1200n = 25 ………………….(4)

Multiply eq.3 by 10, to get,

600m + 2400n = 40 ……………………(5)

Now, subtract eq.4 from 5, to get,

1200n = 15

n = 15/1200 = 1/80

Substitute the value of n in eq. 3, to get,

60m + 3 = 4

m = 1/60

m = 1/x = 1/60

x = 60

And y = 1/n

y = 80

Therefore,

Speed of the train = 60 km/h

Speed of the bus = 80 km/h

Exercise 3.7 Page: 68

1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution:

The age difference between Ani and Biju is 3 yrs.

Either Biju is 3 years older than that of Ani or Ani is 3 years older than Biju. From both the cases we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.

Let the ages of Ani and Biju be A and B respectively.

Therefore, the age of Dharam = 2 x A = 2A yrs.

And the age of Biju sister Ann B/2 yrs

By using the information that is given,

Case (i)

When Ani is older than that of Biju by 3 yrs then A – B = 3 – – – – – – – – (1)

2A−B/2 = 30

4A – B = 60 – – – – – – – – – – – (2)

By subtracting the equations (1) and (2) we get,

3A = 60 – 3 = 57

A = 57/3 = 19

Therefore, the age of Ani = 19 yrs

And the age of Biju is 19 – 3 = 16 yrs.

Case (ii)

When Biju is older than Ani,

B – A = 3 – – – – – – – – – (1)

2A − B/2 = 30

4A – B = 60 – – – – – – – – – (2)

Adding the equation (1) and (2) we get,

3A = 63

A = 21

Therefore, the age of Ani is 21 yrs

And the age of Biju is 21 + 3 = 24 yrs.

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

Solution:

Let Sangam have Rs A with him and Reuben have Rs B with him.

Using the information that is given we get,

A + 100 = 2(B – 100) ⇒ A + 100 = 2B – 200

Or A – 2B = -300 – – – – – – – (1)

And

6(A – 10) = ( B + 10 )

Or 6A – 60 = B + 10

Or 6A – B = 70 – – – – – – (2)

When equation (2) is multiplied by 2 we get,

12A – 2B = 140 – – – – – – – (3)

When equation (1) is subtracted from equation (3) we get,

11A = 140 + 300

11A = 440

⇒ A = 440/11 = 40

Using A =40 in equation (1) we get,

40 – 2B = -300

40 + 300 = 2B

2B = 340

B = 170

Therefore, Sangam had Rs 40 and Reuben had Rs 170 with them.

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution:

Let the speed of the train be A km/hr and the time taken by the train to travel a distance be N hours and the distance to travel be X hours.

Speed of the train = Distance travelled by train / Time taken to travel that distance

A = N (distance)/ X (time)

Or, N = AX – – – – – – – – – – – (1)

Using the information that is given, we get:

(A+10) = X/(N-2)

(A + 10) (N – 2) = X

AN + 10N – 2A – 20 = X

By using the equation (1) we get,

– 2A + 10N = 20 – – – – – – – – – – (2)

(A-10) = X/(N+3)

(A – 10) (N + 3) = X

AN – 10N + 3A – 30 = X

By using the equation (1) we get,

3A – 10N = 30 – – – – – – – – – (3)

Adding equation (2) and equation (3) we get,

A = 50

Using the equation (2) we get,

(-2) x (50) + 10N = 20

-100 +10N = 20

=> 10N = 120

N = 12hours

From the equation (1) we get,

Distance travelled by the train, X = AN

= 50 x 12

= 600 km

Hence, the distance covered by the train is 600km.

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let the number of rows be A and the number of students in a row be B.

Total number of students = Number of rows x Number of students in a row

=AB

Using the information, that is given,

First Condition:

Total number of students = (A – 1) ( B + 3)

Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3

Or 3A – B – 3 = 0

Or 3A – Y = 3 – – – – – – – – – – – – – (1)

Second condition:

Total Number of students = (A + 2 ) ( B – 3 )

Or AB = AB + 2B – 3A – 6

Or 3A – 2B = -6 – – – – – – – – – (2)

When equation (2) is subtracted from (1)

(3A – B) – (3A – 2B) = 3 – (-6)

-B + 2B = 3 + 6B = 9

By using the equation (1) we get,

3A – 9 =3

3A = 9+3 = 12

A = 4

Number of rows, A = 4

Number of students in a row, B = 9

Number of total students in a class = AB = 4 x 9 = 36

5. In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠A + ∠ B). Find the three angles.

Solution:

Given,

∠C = 3 ∠B = 2(∠B + ∠A)

∠B = 2 ∠A+2 ∠B

∠B=2 ∠A

∠A – ∠B= 0- – – – – – – – – – – – (i)

We know, the sum of all the interior angles of a triangle is 180O.

Thus, ∠ A +∠B+ ∠C = 180O

∠A + ∠B +3 ∠B = 180O

∠A + 4 ∠B = 180O– – – – – – – – – – – – – – -(ii)

Multiplying 4 to equation (i) , we get

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii) we get

9 ∠A = 180O

∠A = 20O

Using this in equation (ii), we get

20O+ 4∠B = 180O

∠B = 40O

3∠B =∠C

∠C = 3 x 40 = 120O

Therefore, ∠A = 20O

∠B=40O

∠C = 120O

6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Solutions:

Given,

5x – y = 5

=> y = 5x – 5

Its solution table will be.

Also given,3x – y = 3

y = 3x – 3

The graphical representation of these lines will be as follows:

From the above graph we can see that the triangle formed is ∆ABC by the lines and the y axis. Also the coordinates of the vertices are A(1,0) , C(0,-5) and B(0,-3).

7. Solve the following pair of linear equations:

(i) px + qy = p – q

qx – py = p + q

(ii) ax + by = c

bx + ay = 1 + c

(iii) x/a – y/b = 0

ax + by = a2 + b2

(iv) (a – b)x + (a + b) y = a2 – 2ab – b2

(a + b)(x + y) = a2 + b2

(v) 152x – 378y = – 74

–378x + 152y = – 604

Solutions:

(i) px + qy = p – q……………(i)

qx – py = p + q……………….(ii)

Multiplying p to equation (1) and q to equation (2), we get

p2x + pqy = p2 − pq ………… (iii)

q2x − pqy = pq + q2 ………… (iv)

Adding equation (iii) and equation (iv),we get

p2x + q2 x = p2 + q2

(p2 + q2 ) x = p2 + q2

x = (p2 + q2)/ p2 + q2 = 1

From equation (i), we get

p(1) + qy = p – q

qy = p-q-p

qy = -q

y = -1

(ii) ax + by= c…………………(i)

bx + ay = 1+ c………… ..(ii)

Multiplying a to equation (i) and b to equation (ii), we obtain

a2x + aby = ac ………………… (iii)

b2x + aby = b + bc…………… (iv)

Subtracting equation (iv) from equation (iii),

(a2 – b2) x = ac − bc– b

x = (ac − bc– b)/ (a2 – b2)

x = c(a-b) –b / (a2+b2)

From equation (i), we obtain

ax +by = c

a{c(a−b)−b)/ (a2 – b2)} +by=c

ac(a−b)−ab/ (a2 – b2)+by=c

by=c–ac(a−b)−ab/(a2 – b2)

by=abc – b2 c+ab/a2-b2

y = c(a-b)+a/a2-b2

(iii) x/a – y/b = 0

ax + by = a2 + b2

x/a – y/b = 0

=> bx − ay = 0 ……. (i)

ax + by = a2 + b2 …….. (ii)

Multiplying a and b to equation (i) and (ii) respectively, we get

b2x − aby = 0 …………… (iii)

a2x + aby = a 3 + ab3 …… (iv)

Adding equations (iii) and (iv), we get

b2x + a2x = a 3 + ab2

x (b2 + a2) = a (a2 + b2) x = a

Using equation (i), we get

b(a) − ay = 0

ab − ay = 0

ay = ab,

y = b

(iv) (a – b)x + (a + b) y = a2 – 2ab – b2

(a + b)(x + y) = a2 + b2

(a + b) y + (a – b) x = a2− 2ab − b2 …………… (i)

(x + y)(a + b) = a 2 + b2

(a + b) y + (a + b) x = a2 + b2 ………………… (ii)

Subtracting equation (ii) from equation (i), we get

(a − b) x − (a + b) x = (a 2 − 2ab − b 2) − (a2 + b2)

x(a − b − a − b) = − 2ab − 2b2

− 2bx = − 2b (b + a)

x = b + a

Substituting this value in equation (i), we get

(a + b)(a − b) +y (a + b) = a2− 2ab – b2

a2 − b2 + y(a + b) = a2− 2ab – b2

(a + b) y = − 2ab

y = -2ab/(a+b)

(v) 152x − 378y = − 74

76x − 189y = − 37

x =(189y-137)/76……………..…(i)

− 378x + 152y = − 604

− 189x + 76y = − 302 ………….. (ii)

Using the value of x in equation (ii), we get

−189(189y−37/76)+76y=−302

− (189)2y + 189 × 37 + (76)2 y = − 302 × 76

189 × 37 + 302 × 76 = (189)2 y − (76)2y

6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

Using equation (i), we get

x = (189-37)/76

x = 152/76 = 2

8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Solution:

It is known that the sum of the opposite angles of a cyclic quadrilateral is 180o

Thus, we have

∠C +∠A = 180

4y + 20− 4x = 180

− 4x + 4y = 160

x − y = − 40 ……………(1)

And, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 ………..(2)

Multiplying 3 to equation (1), we get

3x − 3y = − 120 ………(3)

Adding equation (2) to equation (3), we get

− 7x + 3x = 180 – 120

− 4x = 60

x = −15

Substituting this value in equation (i), we get

x − y = − 40

-y−15 = − 40

y = 40-15

= 25

∠A = 4y + 20 = 20+4(25) = 120°

∠B = 3y − 5 = − 5+3(25) = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D = 5-7x

∠D= 5− 7(−15) = 110°

Hence, all the angles are measured.

NCERT Solutions for Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables

Chapter 3- Pair of Linear Equations in Two Variables holds a weightage of 11 marks in the examinations. This chapter gives an account of the various topics related to the linear equations in two variables. The topics discussed in the chapter are mentioned below:

3.1 Introduction
In earlier classes, you have studied Linear Equations in Two Variables. You have also studied that a Linear Equation in Two Variables has infinitely many solutions. In this chapter, the knowledge of Linear Equations in Two Variables shall be recalled and extended to that of Pair of Linear Equations In Two Variables.
3.2 Pair of Linear Equations in Two Variables
An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not zeros, is called a linear equation in two variables x and y. The solution of such a problem is a pair of values, one for x and the other for y, which makes the two sides of the equation equal. The topic also discusses the geometrical representation of Pair of Linear Equations in Two Variables along with suitable examples.
3.3 Graphical Method of Solution of a Pair of Linear Equations
In the previous section, you have seen how we can graphically represent a pair of linear equations as two lines. You have also seen that the lines may intersect, or may be parallel, or may coincide. In this section, you will know how to solve it in each case from the geometrical point of view.
3.4 Algebraic Methods of Solving a Pair of Linear Equations
In the previous section, we discussed how to solve a pair of linear equations graphically. In some of the cases, the graphical method is not convenient. In this topic, we shall discuss various algebraic methods such as the Substitution Method, Elimination Method, and Cross – Multiplication Method. Each subtopic is explained elaborately with suitable examples, for better understanding.
3.5 Equations Reducible to a Pair of Linear Equation in Two Variables
In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. The process is explained through some examples related to the subtopic.
3.6 Summary
The Summary section consists of the overall important points that need to be memorized while solving the exercise questions of the chapter Pair of Linear Equations in Two Variables. The points mentioned in this section will help you to revise all the concepts mentioned in the chapter.
Two linear equations in the same two variables are called pair of linear equations in two variables. The pair of linear equations in two variables can be represented graphically and algebraically. The graph can be represented by two lines:

If the lines intersect at a point, the pair of equations is said to be consistent.
If the lines coincide, the pair of equations is dependent.
If the lines are parallel, the pair of equations is inconsistent.

Algebraically the following methods can be used to solve the pair of linear equations in two variables:

Substitution method
Elimination method
Cross-multiplication method

Key Features of NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables

NCERT Solutions are created by subject experts.
The answers are provided after a lot of brainstorming and are accurate.
These contain questions related to all the important topics.
NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables includes the solutions to the exercises given in the textbook as well.

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 3

What are the important topics from an exam perspective present in NCERT Solutions for Class 10 Maths Chapter 3?

The important topics present in NCERT Solutions for Class 10 Maths Chapter 3 are substitution method, elimination method and cross-multiplication method of pair of linear equations in two variables. By solving problems based on these concepts students can score well in Class 10 board exams.

Is it necessary to learn all three methods to solve pairs of linear equations in two variables of NCERT Solutions for Class 10 Maths Chapter 3?

Yes, it is compulsory to learn all three methods to solve a pair of linear equations in two variables of NCERT Solutions for Class 10 Maths Chapter 3. These topics will continue in higher studies as well as it may come in their finals of Class 10. These concepts are explained by BYJU’S flawlessly. Hence, the main aim of these solutions formulated by BYJU’S experts is to provide knowledge on the fundamental aspects of Maths, which in turn, helps students to understand every concept clearly.

How to start reading NCERT Solutions for Class 10 Maths Chapter 3?

Very first learn and understand the definition of linear equations by the help of BYJU’S website. Then go through the solutions provided by BYJU’S experts at their site. By solving these exercises one can be thorough in all the concepts present in NCERT Solutions for Class 10 Maths Chapter 3.

NCERT academic calendar English for primary | Alternative Academic Calendar

2021-05-16T09:54:00.004+05:30

NCERT academic calendar English for primary | Alternative Academic Calendar

Alternative Academic Calendar

In this period of Covid-19, which is declared as a global pandemic, our teachers, parents, and students have to remain at homes to prevent its spread in the community. In this situation, it is our responsibility to provide them with multiple alternative ways of learning at home through interesting activities. It is necessary because in the present environment of stress we have to not only keep our children busy but also to maintain continuity of their learning in their new classes. In this context, NCERT has developed an Alternative Academic Calendar for all the stages of school education.

Polynomial Division | Division of Polynomial | Division Algorithm for Polynomials in hindi

2021-05-04T21:56:00.002+05:30

Polynomial Division | Division of Polynomial | Division Algorithm for Polynomials in hindi

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Polynomial Division | Division of Polynomial | Division Algorithm for Polynomials in hindi

Polynomial Division | Division of Polynomial | Division Algorithm for Polynomials in hindi Class: 10th Subject: Maths Chapter: Polynomials Topic Name: Division Algorithm for Polynomials Points covered in this video:- From Euclid’s division lemma, we know that Dividend = (Divisor x Quotient) + Remainder Applying the same to polynomials, If f(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that f(x) = q(x) x g(x) + r(x), Degree of r(x) less than Degree of g(x) If r(x) = 0, then polynomial g(x) is a factor of polynomial f(x). Class 10 Maths Chapter 2 Polynomials | Division Algorithm for Polynomials class 10 maths chapter 2 polynomials,division algorithm for polynomials, maths polynomials,polynomials class 10,polynomial division in hindi,division algorithm for polynomials class 10 maths,division algorithm for polynomials class 9,class 10 maths chapter 2,polynomials in maths,polynomials in 9 maths,polynomials,polynomials in class 10,polynomials division algorithm class 10 maths,polynomials division algorithm class 10,division algorithm class 10,Style stardum MS, algebric expressions class 8,division of polynomials,division of polynomials class 8,division of algebraic expressions,algebra division,algebra divide,division of algebraic expressions class 8,algebraic expressions class 7,how to divide polynomials,division of polynomial by another polynomial,बहुपद का भाग,long division of polynomials,introduction algebraic expressions,polynomial long division, division of polynomials class 8 Algebraic expressions class 8 division of algebraic expressions class 8 algebra divide class 8 division of polynomials how to divide algebraic expressions algebra division Polynomial division algebra ka bhag division of polynomial division method how to divide polynomials algebraic expressions algebraic expressions class 7 algebra class 8 polynomial divide polynomial division class 9 polynomial class 9 #polynomial #maths #division #mathematics #Class10

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Algebraic Identities ll algebraic Expressions ll CBSE ICSE maths class 8 class 9 class 10 in hindi

2021-05-03T09:41:00.000+05:30

Algebraic Identities ll algebraic Expressions ll CBSE ICSE maths class 8 class 9 class 10 in hindi

Algebraic Identities ll algebraic Expressions ll CBSE ICSE maths class 8 class 9 class 10 in hindiा

Algebraic Identities ll algebraic Expressions ll CBSE ICSE maths class 8 class 9 class 10 in hindi

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The algebraic equations which are valid for all values of variables in them are called algebraic identities. They are also used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials. You have already learned about a few of them in the junior grades. In this article, we will recall them and introduce you to some more standard algebraic identities, along with examples.

Standard Algebraic Identities List

All the standard Algebraic Identities are derived from the Binomial Theorem, which is given as:

(a+b)n=nC0.an.b0+nC1.an−1.b1+……..+nCn−1.a1.bn−1+nCn.a0.bn

Some Standard Algebraic Identities list are given below:

Identity I: (a + b)2 = a2 + 2ab + b2

Identity II: (a – b)2 = a2 – 2ab + b2

Identity III: a2 – b2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab

Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)

Identity VIII: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Example 1: Find the product of (x + 1)(x + 1) using standard algebraic identities.

Solution: (x + 1)(x + 1) can be written as (x + 1)2. Thus, it is of the form Identity I where a = x and b = 1. So we have,

(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x2 + 2x + 1

Example 2: Factorise (x4 – 1) using standard algebraic identities.

Solution: (x4 – 1) is of the form Identity III where a = x2 and b = 1. So we have,

(x4 – 1) = ((x2)2– 12) = (x2 + 1)(x2 – 1)

The factor (x2 – 1) can be further factorised using the same Identity III where a = x and b = 1. So,

(x4 – 1) = (x2 + 1)((x)2 –(1)2) = (x2 + 1)(x + 1)(x – 1)

Eample 3: Factorise 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities.

Solution: 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx is of the form Identity V. So we have,

16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx = (4x)2 + (-2y)2 + (-3z)2 + 2(4x)(-2y) + 2(-2y)(-3z) + 2(-3z)(4x)= (4x – 2y – 3z)2 = (4x – 2y – 3z)(4x – 2y – 3z)

Example 4: Expand (3x – 4y)3 using standard algebraic identities.

Solution: (3x– 4y)3 is of the form Identity VII where a = 3x and b = 4y. So we have,

(3x – 4y)3 = (3x)3 – (4y)3– 3(3x)(4y)(3x – 4y) = 27x3 – 64y3 – 108x2y + 144xy2

Example 5: Factorize (x3 + 8y3 + 27z3 – 18xyz) using standard algebraic identities.

Solution: (x3 + 8y3 + 27z3 – 18xyz)is of the form Identity VIII where a = x, b = 2y and c = 3z. So we have,

(x3 + 8y3 + 27z3 – 18xyz) = (x)3 + (2y)3 + (3z)3 – 3(x)(2y)(3z)= (x + 2y + 3z)(x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)

Frequently Asked Questions on Algebraic Identities

What are the three algebraic identities in Maths?

The three algebraic identities in Maths are:

Identity 1: (a+b)^2 = a^2 + b^2 + 2ab

Identity 2: (a-b)^2 = a^2 + b^2 – 2ab

Identity 3: a^2 – b^2 = (a+b) (a-b)

What is the difference between an algebraic expression and identities?

An algebraic expression is an expression which consists of variables and constants. In expressions, a variable can take any value. Thus, the expression value can change if the variable values are changed. But algebraic identity is equality which is true for all the values of the variables.

How to verify algebraic identity?

The algebraic identities are verified using the substitution method. In this method, substitute the values for the variables and perform the arithmetic operation. Another method to verify the algebraic identity is the activity method. In this method, you would need a prerequisite knowledge of Geometry and some materials are needed to prove the identity.

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NCERT Class 9 Maths Chapter 8 – Quadrilaterals

2021-01-25T01:09:00.004+05:30

NCERT Class 9 Maths Chapter 8 – Quadrilaterals

Exercise 8.2 Page: 150

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that:
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution:

(i) In ΔDAC,

R is the mid point of DC and S is the mid point of DA.

Thus by mid point theorem, SR || AC and SR = ½ AC

(ii) In ΔBAC,

P is the mid point of AB and Q is the mid point of BC.

Thus by mid point theorem, PQ || AC and PQ = ½ AC

also, SR = ½ AC

, PQ = SR

(iii) SR || AC ———————- from question (i)

and, PQ || AC ———————- from question (ii)

⇒ SR || PQ – from (i) and (ii)

also, PQ = SR

, PQRS is a parallelogram.

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution:

Given in the question,

ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

To Prove,

PQRS is a rectangle.

Construction,

Join AC and BD.

Proof:

In ΔDRS and ΔBPQ,

DS = BQ (Halves of the opposite sides of the rhombus)

∠SDR = ∠QBP (Opposite angles of the rhombus)

DR = BP (Halves of the opposite sides of the rhombus)

, ΔDRS ≅ ΔBPQ [SAS congruency]

RS = PQ [CPCT]———————- (i)

In ΔQCR and ΔSAP,

RC = PA (Halves of the opposite sides of the rhombus)

∠RCQ = ∠PAS (Opposite angles of the rhombus)

CQ = AS (Halves of the opposite sides of the rhombus)

, ΔQCR ≅ ΔSAP [SAS congruency]

RQ = SP [CPCT]———————- (ii)

Now,

In ΔCDB,

R and Q are the mid points of CD and BC respectively.

⇒ QR || BD

also,

P and S are the mid points of AD and AB respectively.

⇒ PS || BD

⇒ QR || PS

, PQRS is a parallelogram.

also, ∠PQR = 90°

Now,

In PQRS,

RS = PQ and RQ = SP from (i) and (ii)

∠Q = 90°

, PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:

Given in the question,

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

Construction,

Join AC and BD.

To Prove,

PQRS is a rhombus.

Proof:

In ΔABC

P and Q are the mid-points of AB and BC respectively

, PQ || AC and PQ = ½ AC (Midpoint theorem) — (i)

In ΔADC,

SR || AC and SR = ½ AC (Midpoint theorem) — (ii)

So, PQ || SR and PQ = SR

As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.

, PS || QR and PS = QR (Opposite sides of parallelogram) — (iii)

Now,

In ΔBCD,

Q and R are mid points of side BC and CD respectively.

, QR || BD and QR = ½ BD (Midpoint theorem) — (iv)

AC = BD (Diagonals of a rectangle are equal) — (v)

From equations (i), (ii), (iii), (iv) and (v),

PQ = QR = SR = PS

So, PQRS is a rhombus.

Hence Proved

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC.

Solution:

Given that,

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.

To prove,

F is the mid-point of BC.

Proof,

BD intersected EF at G.

In ΔBAD,

E is the mid point of AD and also EG || AB.

Thus, G is the mid point of BD (Converse of mid point theorem)

Now,

In ΔBDC,

G is the mid point of BD and also GF || AB || DC.

Thus, F is the mid point of BC (Converse of mid point theorem)

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.

Solution:

Given that,

ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.

To show,

AF and EC trisect the diagonal BD.

Proof,

ABCD is a parallelogram

, AB || CD

also, AE || FC

Now,

AB = CD (Opposite sides of parallelogram ABCD)

⇒½ AB = ½ CD

⇒ AE = FC (E and F are midpoints of side AB and CD)

AECF is a parallelogram (AE and CF are parallel and equal to each other)

AF || EC (Opposite sides of a parallelogram)

Now,

In ΔDQC,

F is mid point of side DC and FP || CQ (as AF || EC).

P is the mid-point of DQ (Converse of mid-point theorem)

⇒ DP = PQ — (i)

Similarly,

In ΔAPB,

E is midpoint of side AB and EQ || AP (as AF || EC).

Q is the mid-point of PB (Converse of mid-point theorem)

⇒ PQ = QB — (ii)

From equations (i) and (i),

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.

Hence Proved.

6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution:

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Now,

In ΔACD,

R and S are the mid points of CD and DA respectively.

, SR || AC.

Similarly we can show that,

PQ || AC,

PS || BD and

QR || BD

, PQRS is parallelogram.

PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB

Solution:

(i) In ΔACB,

M is the midpoint of AB and MD || BC

, D is the midpoint of AC (Converse of mid point theorem)

(ii) ∠ACB = ∠ADM (Corresponding angles)

also, ∠ACB = 90°

, ∠ADM = 90° and MD ⊥ AC

(iii) In ΔAMD and ΔCMD,

AD = CD (D is the midpoint of side AC)

∠ADM = ∠CDM (Each 90°)

DM = DM (common)

, ΔAMD ≅ ΔCMD [SAS congruency]

AM = CM [CPCT]

also, AM = ½ AB (M is midpoint of AB)

Hence, CM = MA = ½ AB

NCERT Class 9 Maths Chapter 8 – Quadrilaterals

2021-01-25T01:04:00.002+05:30

NCERT Class 9 Maths Chapter 8 – Quadrilaterals

https://www.mathsstudy.co.in/2021/01/ncertclass9.html

NCERT Class 9 Maths Chapter 8 – Quadrilaterals

Exercise 8.1 Page: 146

1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution:

Let the common ratio between the angles be = x.

We know that the sum of the interior angles of the quadrilateral = 360°

Now,

3x+5x+9x+13x = 360°

⇒ 30x = 360°

⇒ x = 12°

, Angles of the quadrilateral are:

3x = 3×12° = 36°

5x = 5×12° = 60°

9x = 9×12° = 108°

13x = 13×12° = 156°

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:

Given that,

AC = BD

To show that, ABCD is a rectangle if the diagonals of a parallelogram are equal

To show ABCD is a rectangle we have to prove that one of its interior angles is right angled.

Proof,

In ΔABC and ΔBAD,

BC = BA (Common)

AC = AD (Opposite sides of a parallelogram are equal)

AC = BD (Given)

Therefore, ΔABC ≅ ΔBAD [SSS congruency]

∠A = ∠B [Corresponding parts of Congruent Triangles]

also,

∠A+∠B = 180° (Sum of the angles on the same side of the transversal)

⇒ 2∠A = 180°

⇒ ∠A = 90° = ∠B

, ABCD is a rectangle.

Hence Proved.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

Given that,

OA = OC

OB = OD

and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°

To show that,

if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

i.e., we have to prove that ABCD is parallelogram and AB = BC = CD = AD

Proof,

In ΔAOB and ΔCOB,

OA = OC (Given)

∠AOB = ∠COB (Opposite sides of a parallelogram are equal)

OB = OB (Common)

Therefore, ΔAOB ≅ ΔCOB [SAS congruency]

Thus, AB = BC [CPCT]

Similarly we can prove,

BC = CD

CD = AD

AD = AB

, AB = BC = CD = AD

Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.

, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

Hence Proved.

4. Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:

Let ABCD be a square and its diagonals AC and BD intersect each other at O.

To show that,

AC = BD

AO = OC

and ∠AOB = 90°

Proof,

In ΔABC and ΔBAD,

AB = BA (Common)

∠ABC = ∠BAD = 90°

BC = AD (Given)

ΔABC ≅ ΔBAD [SAS congruency]

Thus,

AC = BD [CPCT]

diagonals are equal.

Now,

In ΔAOB and ΔCOD,

∠BAO = ∠DCO (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite)

AB = CD (Given)

, ΔAOB ≅ ΔCOD [AAS congruency]

Thus,

AO = CO [CPCT].

, Diagonal bisect each other.

Now,

In ΔAOB and ΔCOB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

, ΔAOB ≅ ΔCOB [SSS congruency]

also, ∠AOB = ∠COB

∠AOB+∠COB = 180° (Linear pair)

Thus, ∠AOB = ∠COB = 90°

, Diagonals bisect each other at right angles

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:

Given that,

Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O.

To prove that,

The Quadrilateral ABCD is a square.

Proof,

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite)

OB = OD (Diagonals bisect each other)

, ΔAOB ≅ ΔCOD [SAS congruency]

Thus,

AB = CD [CPCT] — (i)

also,

∠OAB = ∠OCD (Alternate interior angles)

⇒ AB || CD

Now,

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD (Vertically opposite)

OD = OD (Common)

, ΔAOD ≅ ΔCOD [SAS congruency]

Thus,

AD = CD [CPCT] — (ii)

also,

AD = BC and AD = CD

⇒ AD = BC = CD = AB — (ii)

also, ∠ADC = ∠BCD [CPCT]

and ∠ADC+∠BCD = 180° (co-interior angles)

⇒ 2∠ADC = 180°

⇒∠ADC = 90° — (iii)

One of the interior angles is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

Hence Proved.

6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Solution:

(i) In ΔADC and ΔCBA,

AD = CB (Opposite sides of a parallelogram)

DC = BA (Opposite sides of a parallelogram)

AC = CA (Common Side)

, ΔADC ≅ ΔCBA [SSS congruency]

Thus,

∠ACD = ∠CAB by CPCT

and ∠CAB = ∠CAD (Given)

⇒ ∠ACD = ∠BCA

Thus,

AC bisects ∠C also.

(ii) ∠ACD = ∠CAD (Proved above)

⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)

Also, AB = BC = CD = DA (Opposite sides of a parallelogram)

Thus,

ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Solution:

Given that,

ABCD is a rhombus.

AC and BD are its diagonals.

Proof,

AD = CD (Sides of a rhombus)

∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)

also, AB || CD

⇒∠DAC = ∠BCA (Alternate interior angles)

⇒∠DCA = ∠BCA

, AC bisects ∠C.

Similarly,

We can prove that diagonal AC bisects ∠A.

Following the same method,

We can prove that the diagonal BD bisects ∠B and ∠D.

8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:

(i) ABCD is a square

(ii) Diagonal BD bisects ∠B as well as ∠D.

Solution:

(i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)

⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)

also, CD = AB (Opposite sides of a rectangle)

,AB = BC = CD = AD

Thus, ABCD is a square.

(ii) In ΔBCD,

BC = CD

⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)

also, ∠CDB = ∠ABD (Alternate interior angles)

⇒ ∠CBD = ∠ABD

Thus, BD bisects ∠B

Now,

∠CBD = ∠ADB

⇒ ∠CDB = ∠ADB

Thus, BD bisects ∠D

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Solution:

(i) In ΔAPD and ΔCQB,

DP = BQ (Given)

∠ADP = ∠CBQ (Alternate interior angles)

AD = BC (Opposite sides of a parallelogram)

Thus, ΔAPD ≅ ΔCQB [SAS congruency]

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,

BQ = DP (Given)

∠ABQ = ∠CDP (Alternate interior angles)

AB = CD (Opposite sides of a parallelogram)

Thus, ΔAQB ≅ ΔCPD [SAS congruency]

(iv) As ΔAQB ≅ ΔCPD

AQ = CP [CPCT]

(v) From the questions (ii) and (iv), it is clear that APCQ has equal opposite sides and also has equal and opposite angles. , APCQ is a parallelogram.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

Solution:

(i) In ΔAPB and ΔCQD,

∠ABP = ∠CDQ (Alternate interior angles)

∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars)

AB = CD (ABCD is a parallelogram)

, ΔAPB ≅ ΔCQD [AAS congruency]

(ii) As ΔAPB ≅ ΔCQD.

, AP = CQ [CPCT]

11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).

Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ΔABC ≅ ΔDEF.

Solution:

(i) AB = DE and AB || DE (Given)

Two opposite sides of a quadrilateral are equal and parallel to each other.

Thus, quadrilateral ABED is a parallelogram

(ii) Again BC = EF and BC || EF.

Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.

⇒ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)

, AD = CF.

Also, AD || BE and BE || CF (Opposite sides of a parallelogram are parallel)

, AD || CF

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.

(v) Since ACFD is a parallelogram

AC || DF and AC = DF

(vi) In ΔABC and ΔDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

, ΔABC ≅ ΔDEF [SSS congruency]

12. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ΔABC ≅ ΔBAD

(iv) diagonal AC = diagonal BD

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Solution:

To Construct: Draw a line through C parallel to DA intersecting AB produced at E.

(i) CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

, BC = CE

⇒∠CBE = ∠CEB

also,

∠A+∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)

∠B +∠CBE = 180° ( As Linear pair)

⇒∠A = ∠B

(ii) ∠A+∠D = ∠B+∠C = 180° (Angles on the same side of transversal)

⇒∠A+∠D = ∠A+∠C (∠A = ∠B)

⇒∠D = ∠C

(iii) In ΔABC and ΔBAD,

AB = AB (Common)

∠DBA = ∠CBA

AD = BC (Given)

, ΔABC ≅ ΔBAD [SAS congruency]

(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

____________________________________________________________________________________

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