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Pythagorean Theorem and Right Triangles

2011-05-24T00:44:00.000-07:00

This post is a discussion of the calculation of the sides of right triangles using Pythagorean theorem. It consists of illustrated problems involving right triangles, and those right triangles formed by the following:

the diagonal and the two consecutive sides of a square
the diagonal and the two consecutive sides of a rectangle
the altitude and the two sides of an equilateral triangle
the altitude and the two legs of an isosceles triangle

The Pythagorean theorem states that for any right triangle the square of the length of its hypotenuse is equal to the sum of the squares of the lengths of its two legs.

Mathematically, the theorem is expressed as follows

c² = a² + b²

where a and b are the lengths of the legs and c is the length of the hypotenuse of the right triangle.

If any two sides are known, then the third side can be calculated

a² = c² - b²

b² = c² - a²

In addition, the following properties also hold true for all right triangles:

the hypotenuse is always the longest side and is always opposite the right angle
the greater of the two acute angles is always opposite the longer leg

These properties will be of great help when checking your computations.

For a quick check of your calculations, refer to the Pythagorean triples given below. They are sets of three numbers that satisfy the Pythagorean theorem c² = a² + b². The table is arranged in ascending order by the hypotenuse.

a	b	c
3	4	5
6	8	10
5	12	13
9	12	15
8	15	17
12	16	20
7	24	25
15	20	25
24	7	25
10	24	26
20	21	29
18	24	30
16	30	34
21	28	35
12	35	37
15	36	39
24	32	40
9	40	41
27	36	45
14	48	50
30	40	50
24	45	51
20	48	52
28	45	53
33	44	55
40	42	58
36	48	60
60	11	61
56	33	65
60	25	65
60	32	68
56	42	70
55	48	73
60	45	75

When solving right triangles, there are two special right triangles that you might want to remember for their properties. Knowledge of their properties can simplify much of the calculations.

These are the:

45°-45°-90° right triangle
30°-60°-90° right triangle

An example of the 45°-45°-90° right triangle is given below.

Given isosceles right triangle ABC with two known and equal legs, find the length of its hypotenuse.

Applying the Pythagorean theorem, the hypotenuse can be calculated thus

c² = a² + b²

c² = 20² + 20²

c² = 400 + 400

√c² = √800

√c² = √400 x √2

c = 20√2

Take note that the length of the hypotenuse is equal to the length of a leg multiplied by √2.

The following problem deals with right triangles formed from a square.

Since any two consecutive sides of a square are equal, an isosceles 45°-45°-90° right triangle is also formed by the diagonal and two consecutive sides of a square as can be seen in the next diagram.

Given square ABCD, right triangle BCD is formed by BC and the two sides BD and CD. In this illustrated problem, the length of the diagonal d is given while the length of the sides is unknown.

To get the length of the sides of square ABCD, the Pythagorean theorem will be applied.

With side BC as the hypotenuse and the sides BD and CD as the legs, we have

BD² + CD² = BC²

Since BD = CD = s and BC = 10√2, the expression is reduced to an equation with one unknown.

s² + s² = (10√2)²

2s² = 100 x 2

s² = 200/2

s² = 100

√s² = √100

s = 10

Observe that the length of the hypotenuse is equal to the length of a leg times √2.

This gives us the general rule:

For any 45°-45°-90° right triangle, the length of its hypotenuse is always equal to the product of its leg and √2.

Another special right triangle that deserves attention is the 30°-60°-90° right triangle.

This kind of triangle is usually encountered in such a problem involving an equilateral triangle with unknown altitude (see the illustration below).

Here, triangle ABC is an equilateral triangle whose altitude h, drawn from vertex C to the side AB, bisects vertex angle C and side AB thus forming two 30°-60°-90° right triangles.

In triangle ACD above, AC is the hypotenuse, CD and AD are the legs. Side CD, which is also the altitude of triangle ACD, is unknown.

To find the altitude, we use the Pythagorean theorem

c² = a² + b²

AC² = CD² + AD²

Since CD bisects AB, AB/2 = AD = BD = 12.

24² = h² + 12²

h² = 24² - 12²

h² = 576 - 144

√h² = √432

√h² = √144 x √3

h = 12√3

The length of the altitude is one-half the length of the hypotenuse times √3.

Let's take a look at another example of a 30°-60°-90° right triangle.

Given triangle ABC with c=28 and b=14, what is the altitude of the triangle?

c² = a² + b²

28² = a² + 14²

a² = 28² - 14²

a² = 784 - 196

√a² = √588

√a² = √196 x √3

a = 14√3

These examples confirm a general rule about a 30°-60°-90° right triangle: its altitude is the product of one half the length of its hypotenuse and √3.

For any other right triangles, the unknown side is calculated using the Pythagorean theorem as shown in the next three illustrated problems.

Given isosceles triangle ABC, find its altitude.

Its altitude CD bisects the base AB and forms two right triangles with its two congruent sides.

(For a discussion of the properties of an isosceles triangle and its altitude see Parts Of A Triangle And Their Graphs.)

Take note that the right triangles formed from this triangle are different from 30°-60°-90° right triangles formed from an equilateral triangle.

Remember, an equilateral triangle is always an isosceles triangle but an isosceles triangle is not always an equilateral triangle.

Applying the Pythagorean theorem in triangle ACD,

c² = a² + b²

AC² = CD² + AD²

15² = h² + 12²

h² = 15² - 12²

h² = 225 - 144

√h² = √81

h = 9

In rectangle ABCD shown below, the length of its diagonal BC is unknown.

Since diagonal BC forms a right triangle with sides BD and CD, we use the Pythagorean theorem to get the length of the diagonal.

c² = a² + b²

BC² = BD² + CD²

BC² = 10² + 24²

BC² = 100 + 576

√BC² = √676

BC = 26

In this last problem, given rhombus ABCD with the lengths of its diagonals AD=12 and BC=16, find the length of its sides.

We know that the diagonals of a rhombus are perpendicular bisector of each other and that the two consecutive sides of a rhombus are equal.

Given these properties, we can take any of the four right triangles formed by the diagonals in order to obtain the length of one side of rhombus ABCD.

With CD as the hypotenuse of one of the right triangles, we find its length by applying the Pythagorean theorem

c² = a² + b²

CD² = 8² + 6²

CD² = 64 + 36

√CD² = √100

CD = 10

For a discussion of the properties of squares, rectangles rhombi and their diagonals, see the following topics:

The Exterior Angle Bisectors and the Excenters of a Triangle

2011-04-19T13:20:00.000-07:00

The exterior angle bisectors of a triangle intersect at three points called excenters.

These points of intersection are the centers of the circles tangent to the sides and the side extensions of the triangle.

This post is the last part of a series of posts about the points of concurrency of a triangle's parts.

The Problem

Given an isosceles triangle ABC with vertices at A(-2, 2), B(2, 0) and C(0, -2), find the points of intersection of its three exterior angle bisectors. Find the equations of the circles tangent to the sides of the triangle.

Discussion

In the given illustration above, line segments DE, EF and DF are the bisectors of the exterior angles of isosceles triangle ABC.

Their points of intersection (point D, point E and point F) are called excenters because it is the centers of the circles tangent to the sides and side extensions of triangle ABC.

Points M to U are the points of tangency.

Solution

Centers of the Circles

circle D (1756/1511, 7800/1511)
circle E (18132000/7594697, -18129800/7594697)
circle F (- 36000/6973, - 8108/6973)

Radii of the Circles

circle D : radius DM = radius DN = radius DO = √ 18
circle E : radius EP = radius EQ = radius ER = √ 38/10
circle F : radius FS = radius FT = radius FU = √ 18

Points of Tangency

point M (- 92/35, 114/35)
point N (- 5554/7555, 10332/7555)
point O (6289/1511, 3267/1511)
point P (123976994/37973485, - 24015012/37973485)
point Q (690527/690427, - 690327/690427)
point R (24012812/37973485, -123972594/37973485)
point S (-15081/6973, - 29027/6973)
point T (- 47676/34865, 25622/34865)
point U (- 113838/34865, 91784/34865)

Distance Between Two Points

side_AB = side_AC = √ 20
side_BC = √ 8
line segment_AM = distance_AN = √ 2
distance_AT = distance_AU = √ 2
distance_BN = distance_BO = √ 9
distance_BP = distance_BQ = √ 2
distance_CQ = distance_CR = √ 2
distance_CS = distance_CT = √ 9

Equations of the Line Segments

line AB: x + 2y - 2 = 0
line BC: x - y - 2 = 0
line AC: 2x + y + 2 = 0
line DF: x - y + 4 = 0
line DE: 1300x + 211y - 2600 = 0
line EF: 973x + 6000y + 12000 = 0

Equation of the Circles

Circle D: ( x - 1756/1511 )² + ( y - 7800/1511 )² = 18
Circle E: ( x - 18132000/7594697 )² + ( y + 18129800/7594697 )² = 38/10
Circle F: ( x + 36000/6973 )² + ( y + 8108/6973 )² = 18

Slopes of the Line Segments

m_AB = - 1/2
m_BC = 1
m_AC = - 2
m_DE = - 1300/211
m_DF = 1
m_EF = - 973/6000

Interior Angles of the Triangle

angle NAT = 36.87°
angle NBQ = 71.56°
angle QCT = 71.56°

Exterior Angles of the Triangle

angle MAN = angle TAU = 143.13°
angle NBO = angle PBQ = 108.44°
angle QCR = angle SCT = 108.44°
½ angle MAN = ½ angle TAU = angle DAM = angle DAN = angle FAT = angle FAU
½ angle NBO = ½ angle PBQ = angle DBN = angle DBO = angle EBP = angle EBQ
½ angle QCR = ½ angle SCT = angle ECQ = angle ECR = angle FCS = angle FCT

The Perpendicular Bisectors and the Circumcenter of a Triangle

2011-03-24T07:23:00.000-07:00

The perpendicular bisectors of the sides of a triangle are concurrent at a point called circumcenter.

This point of concurrency is the center of the circle circumscribing the triangle and is equidistant from the vertices of the triangle.

This post is the second to the last part of a series of posts about the points of concurrency of a triangle's parts.

Here's a list of my previous posts about concurrent parts of a triangle:

orthocenter: the point of concurrency of a triangle's altitudes (see Altitudes and Orthocenter of a Triangle)
centroid: the point of concurrency of a triangle's medians (see Medians And Centroid Of A Triangle)
incenter: the point of concurrency of a triangle's angle bisectors (see The Angle Bisectors and the Incenter of a Triangle)

The Problem

Given the vertices A(-5, 2), B(2, 3) and C(-1, -6) of triangle ABC, find the point of concurrency of the perpendicular bisectors of its sides. Find the equation of the circle circumscribed about the triangle.

Discussion

Referring to the given illustration above, line segments OP, OQ and OR are the perpendicular bisectors of the sides of scalene triangle ABC.

Their point of concurrency, point O, is called circumcenter because it is the center of the circle circumscribing the triangle.

Line segments AO, BO and CO are the radii of circle O.

The problem illustrates the fact that any triangle can always be circumscribed by a circle.

In other words, a set of three noncollinear points which are on the same circle (concyclic) can always be found.

Solution

Coordinates of Circumcenter: (-1, -1)

Midpoints of the Sides

side AB: P(- 3/2, 5/2)
side BC: Q(1/2, - 3/2)
side AC: R(-3, -2)

Distance Between Two Points

distance_AB = √ 50
distance_BC = √ 90
distance_AC = √ 80
distance_AO = distance_BO = distance_CO = √ 25
distance_AP = distance_BP = √ 25/2
distance_BQ = distance_CQ = √ 45/2
distance_AR = distance_CR = √ 20

Equations of the Line Segments

side AB: x - 7y + 19 = 0; x-intercept = -19; y-intercept = 19/7
side BC: 3x - y - 3 = 0; x-intercept = 1; y-intercept = -3
side AC: 2x + y + 8 = 0; x-intercept = - 4; y-intercept = - 8
perpendicular bisector OP: 7x + y + 8 = 0; x-intercept = - 8/7; y-intercept = - 8
perpendicular bisector OQ: x + 3y + 4 = 0; x-intercept = - 4; y-intercept = - 4/3
perpendicular bisector OR: x - 2y - 1 = 0; x-intercept = 1; y-intercept = - 1/2
radius AO: 3x + 4y + 7 = 0; x-intercept = - 7/3; y-intercept = - 7/4
radius BO: 4x - 3y + 1 = 0; x-intercept = - 1/4; y-intercept = 1/3
radius CO: x = - 1; x-intercept = - 1

Equation of the Circle:

( x + 1 )² + ( y + 1 )² = (√ 25)²

x² + y² + 2x + 2y - 23 = 0

Slopes of the Line Segments

m_AB = 1/7
m_BC = 3
m_AC = - 2
m_AO = - 3/4
m_BO = 4/3
m_OP = - 7
m_OQ = - 1/3
m_OR = 1/2

Interior Angles of the Triangle

angle A = 71.6°
angle B = 63.4°
angle C = 45°

The Angle Bisectors and the Incenter of a Triangle

2011-03-09T11:55:00.000-08:00

The problem presented here serves as an illustration of the properties of the following:

angle bisectors of a triangle
lines tangent to a circle
a circle inscribed in a triangle

The Problem

Given the vertices A(0, 2), B(3, 5) and C(4, -2) of right triangle ABC, show analytically that the bisectors of this triangle's angles are concurrent at a point. Show also that this point of concurrency is the center of the inscribed circle.

Discussion

In the illustration above, line segments AO, BO and CO are the angle bisectors of right triangle ABC.

Their point of concurrency, point O, is called incenter because it is the center of the inscribed circle, circle O.

Circle O is tangent to sides AB, BC and AC at points D, E and F, respectively.

Since line segments DO, EO and FO are the radii of circle O and are perpendicular to the three sides of the triangle, point O is equidistant from the three sides of triangle ABC.

As for the tangent lines, there is a theorem that states: from a point outside the circle, lines tangent to the circle are congruent.

Hence,

line segment AD = line segment AF
line segment BD = line segment BE
line segment CE = line segment CF

Solution

Coordinates of Incenter: (2, 2)

Distance Between Two Points

distance_AB = √ 18
distance_BC = √ 50
distance_AC = √ 32
distance_DO = distance_EO = distance_FO = √ 2
distance_AD = distance_AF = √ 2
distance_BD = distance_BE = √ 8
distance_CE = distance_CF = √ 18

Equations of the Line Segments

side AB: x - y + 2 = 0; x-intercept = - 2; y-intercept = 2
side BC: 7x + y - 26 = 0; x-intercept = 26/7; y-intercept = 26
side AC: x + y - 2 = 0; x-intercept = 2; y-intercept = 2
angle bisector AO: y - 2 = 0; x-intercept = 2
angle bisector BO: 3x - y - 4 = 0; x-intercept = 4/3; y-intercept = - 4
angle bisector CO: 2x + y - 6 = 0; x-intercept = 3; y-intercept = 6
radius DO: x + y - 4 = 0; x-intercept = 4; y-intercept = 4
radius EO: x - 7y + 12 = 0; x-intercept = - 12; y-intercept = 12/7
radius FO: x - y = 0; x-intercept = 0; y-intercept = 0

Equation of the Circle:

( x - 2 )² + ( y - 2 )² = (√ 2)²

x² + y² - 4x - 4y + 6 = 0

Slopes of the Line Segments

m_AB = 1
m_BC = - 7
m_AC = - 1
m_AO = 0
m_BO = 3
m_CO = - 2
m_DO = - 1
m_EO = 1/7
m_FO = 1

Interior Angles of the Triangle

angle A = 90.0°
angle B = 53.13°
angle C = 36.87°
angle BAO = angle CAO = 45.0°
angle ABO = angle CBO = 26.57°
angle BCO = angle ACO = 18.43°

Altitudes and Orthocenter of a Triangle

2011-02-24T08:36:00.000-08:00

The altitude of a triangle is the line that passes through a vertex and is perpendicular to the side (or to the extension of that side) opposite to that vertex.

The problem presented here is about the point of concurrency of the altitudes of a triangle. This point is referred to as the orthocenter of the triangle.

For a right triangle, this point of concurrency of the altitudes is at one of its vertices. This is so because two of its altitudes coincide with its legs. See illustration below.

For an obtuse triangle, this point of concurrency of the altitudes is outside of the triangle. Two of its altitudes are drawn to the respective extensions of its opposite sides. See illustration below.

The Problem

Given triangle ABC with its vertices at A(-5, 0), B(3, 4) and C(0, -5), find the point of concurrency of its altitudes.

Discussion

The altitudes of a triangle are concurrent at a point.This point of concurrency is called the orthocenter of a triangle.

In the illustrated triangle above, triangle ABC is a scalene triangle. Line segments AT, BU and CS are its altitudes. Point O is its orthocenter.

Solution

Point of Concurrency: O (-2, -1)

Distance Between Two Points

distance_AB = √ 80
distance_BC = √ 90
distance_AC = √ 50
distance_AT = √ 40
distance_BU = √ 72
distance_CS = √ 45

Equations of the Line Segments

side AB: x - 2y + 5 = 0; x-intercept = - 5; y-intercept = 5/2
side BC: 3x - y - 5 = 0; x-intercept = 5/3; y-intercept = - 5
side AC: x + y + 5 = 0; x-intercept = - 5; y-intercept = - 5
altitude AT: x + 3y + 5 = 0; x-intercept = - 5; y-intercept = - 5/3
altitude BU: x - y + 1 = 0; x-intercept = -1; y-intercept = 1
altitude CS: 2x + y + 5 = 0; x-intercept = - 5/2; y-intercept = - 5

Slopes of the Line Segments

m_AB = 1/2
m_BC = 3
m_AC = - 1
m_AT = - 1/3
m_BU = 1
m_CS = - 2

Medians And Centroid Of A Triangle

2011-02-17T12:23:00.000-08:00

This is a continuation of my previous post. The problems presented here are about the medians of a triangle and their point of concurrency.

The following theorems are applied in the given problems below:

In a right triangle,the length of the median to the hypotenuse is one half the length of the hypotenuse (see Problem 1).
In an isosceles triangle, the median to the base is perpendicular to the base itself (see Problem 2).
The medians of a triangle have a point of concurrency (see Problem 2).

Problem 1

Given a right triangle ABC with vertices at A(-2, 5), B(2, -5) and C(-5, 2) and its median that intersects the hypotenuse, find the length of hypotenuse AB if median CY = √29.

Discussion

Point Y is the midpoint of hypotenuse AB.

Line segment CY is a median, a line drawn from a vertex to the midpoint of the opposite side.

Solution

Distance Between Two Points

distance_AC = √ 18
distance_AB = √ 116
distance_BC = √ 98
distance_CY = distance_AY = distance_BY = √ 29

Equations of the Line Segments

line segment AB: 5x + 2y = 0; x-intercept = 0; y-intercept = 0
line segment BC: x + y + 3 = 0; x-intercept = -3; y-intercept = -3
line segment AC: x - y + 7 = 0; x-intercept = -7; y-intercept = 7
line segment CY: 2x + 5y = 0; x-intercept = 0; y-intercept = 0

Slopes of the Line Segments

m_AB = - 5/2
m_BC = - 1
m_AC = 1
m_CY = - 2/5

Problem 2

Given isosceles triangle ABC with its medians AE, BF and CD, find their point of concurrency. The vertices of the triangle are A(-6, 5), B(4, 1) and C(-2, -5).

Discussion

In the above figure, the medians of isosceles triangle ABC intersect at point X which is called centroid and is the center of gravity of the triangle.

This point is the trisection point of each median.

Taking median AE, line segment AX = 2/3 AE = 2 EX and line segment EX = 1/3 AE. This property holds true for the other two medians as well.

Solution

Point of Concurrency of the Medians: X (- 4/3, 1/3)

Distance Between Two Points

distance_AB = distance_AC = √ 116
distance_BC = √ 72
distance_AE = √ 98
distance_BF = distance_CD = √ 65
distance_AX = 2/3 distance = √ 392/9
distance_EX = 1/3 distance_AE = √ 98/9
distance_BX = 2/3 distance_BF = √ 260/9
distance_FX = 1/3 distance_BF = √ 65/9
distance_CX = 2/3 distance_CD = √ 260/9
distance_DX = 1/3 distance_CD = √ 65/9

Midpoints of the Sides

midpoint of side AB: D(-1, 3)
midpoint of side BC: E(1, -2)
midpoint of side AC: F(-4, 0)

Equations of the Line Segments

line segment AB: 2x + 5y - 13 = 0; x-intercept = 13/2; y-intercept = 13/5
line segment BC: x - y - 3 = 0; x-intercept = 3; y-intercept = - 3
line segment AC: 5x + 2y + 20 = 0; x-intercept = - 4; y-intercept = - 10
line segment AE: x + y + 1 = 0; x-intercept = - 1; y-intercept = - 1
line segment BF: x - 8y + 4 = 0; x-intercept = - 4; y-intercept = 1/2
line segment CD: 8x - y + 11 = 0; x-intercept = - 11/8; y-intercept = 11

Slopes of the Line Segments

m_AB = - 2/5
m_BC = 1
m_AC = - 5/2
m_AE = -1

Parts Of A Triangle

2011-02-11T08:30:00.000-08:00

Aside from its vertices, sides (legs and base), interior and exterior angles, a triangle has also the following parts:

median
altitude
perpendicular bisector of its sides
angle bisector

These parts are discussed in the following problem.

The Problem

Given isosceles triangle ABC with vertices at A(-3, 4), B(5, 0) and C(1, -4), show analytically that the perpendicular bisector of its base passes through its vertex angle.

Discussion

The illustration above shows isosceles triangle ABC with point X as the midpoint of side BC which is the base.

Given the definitions of the following parts:

The median is the line formed by a triangle's vertex and the midpoint of the side opposite to that vertex.
The altitude is the perpendicular line drawn from a triangle's vertex to the side opposite to that vertex (or to the extension of the side opposite to that vertex).
The perpendicular bisector of a triangle's side divides that side into two equal line segments.
The angle bisector is a line that divides an angle into two equal angles.

line segment AX is all of the above.

The solution to the above problem serves as a coordinate proof of the theorem: the perpendicular bisector of the base of an isosceles triangle divides the vertex angle into two equal angles.

Solution

In order to show that the perpendicular bisector of the base passes through the vertex angle, it must be shown that line segment AX:

is perpendicular to the base AB;
divides base AB into two equal line segments;
divides angle A into two equal angles.

Distance Between Two Points

distance_AB = distance_AC = √ 80
distance_BC = √ 32
distance_BX = distance_CX = √ 8

Equations of the Line Segments

line segment AB: x + 2y - 5 = 0; x-intercept = 5; y-intercept = 5/2
line segment BC: x - y - 5 = 0; x-intercept = 5; y-intercept = - 5
line segment AC: 2x + y + 2 = 0; x-intercept = - 1; y-intercept = - 2
line segment AX: x + y - 1 = 0; x-intercept = 1; y-intercept = 1

Slopes of the Line Segments

m_AB = - 1/2
m_BC = 1
m_AC = - 2
m_AX = - 1

Interior Angles of the Triangle

vertex angle A = 36.87°
base angle B = base angle C = 71.56°
angle CAX = angle BAX = 18.43°

Medians And Altitudes of Isosceles Triangles

Given the same isosceles triangle ABC:

point D is the midpoint of leg AB;
point E is the midpoint of leg AC;
point F is the intersection of leg AB and altitude CF;
point G is the intersection of leg AC and altitude BG.

Line segments CD and BE are medians to the legs of triangle ABC.

The graph above shows that for any isosceles triangle, its medians to the legs are congruent.

Distance Between Two Points

distance_AD = distance_BE = distance_CD = distance_CE = √ 20
distance_BE = distance_CD = √ 36
distance_BG = distance_CF = √ 720/25

Equations of the Medians And the Altitudes

median CD: x = 1
median BE: y = 0
altitude CF: 2x - y - 6 = 0
altitude BG: x - 2y - 5 = 0

Coordinates of:

point F: (17/5, 4/5)
point G: (1/5, - 12/5)

This last graph illustrates that for any isosceles triangle, its altitudes to the legs are congruent.

Plotting The Points of a Linear Equation

2011-01-29T12:02:00.000-08:00

For any given linear equation, constructing its graph requires plotting at least two points of the given line.

When choosing points for construction, the following suggestions may be considered:

the points must be wide apart to include as many points as desired;
it's desirable to plot points whose x-coordinate and y-coordinate are whole numbers.

A point of a given line can be determined by substituting into its equation an arbitrary value of x or y. The resulting pair of x and y values correspond to the x and y coordinates of that point in the Cartesian coordinate system.

This is easy when your graph consists of just one line. But if you have several graphs and some of them, perhaps, consist of several lines, finding the points for all those lines can be an unenjoyable task.

When you want to construct several lines, this can be constructed more easily and quickly by calculating the points using spreadsheet formulas.

If you are new to spreadsheet formulas, you can just copy each formula from this web page and paste them directly into the spreadsheet cells.

These formulas are tested using Excel 2007. The formulas are very basic. I'm sure they will run in any version of Excel, or even in any spreadsheet program of your choice.

For simplicity, we will just use the constants of a linear equation in our spreadsheet calculations.

Given a linear equation in this form: Ax + By + C = 0,

A is the coefficient of x;
B is the coefficient of y;
C is the constant of the equation.

These are the headings to be entered in the first row:

cell	text
A1	A value
B1	B value
C1	C value
E1	x value
F1	y value

These are the values to be entered in cells E2 to E12:

cell	x value
E2	5
E3	4
E4	3
E5	2
E6	1
E7	0
E8	-1
E9	-2
E10	-3
E11	-4
E12	-5

These are the formulas to be entered in cells F2 to F12:

cell	formula
F2	=(-C2-E2*A2)/B2
F3	=(-C2-E3*A2)/B2
F4	=(-C2-E4*A2)/B2
F5	=(-C2-E5*A2)/B2
F6	=(-C2-E6*A2)/B2
F7	=(-C2-E7*A2)/B2
F8	=(-C2-E8*A2)/B2
F9	=(-C2-E9*A2)/B2
F10	=(-C2-E10*A2)/B2
F11	=(-C2-E11*A2)/B2
F12	=(-C2-E12*A2)/B2

Suppose we have the equation x - 2y = -5.

The transformed equation is x - 2y + 5 = 0.

The values we must enter for:

A is 1;
B is -2;
C is 5.

Here is how it might look like after entering the values.

And here is the graph of the line:

Make sure that the option for calculation of the sheet is set to automatic so that every time you enter new values each cell with formula will be recalculated.

If you want a longer line, simply replace any values in the column E with a higher value.

That's it. These simple spreadsheet formulas will calculate the points of any given linear equations just as quickly as you want.

Systems of Linear Inequalities And Their Graphs

2011-01-20T11:45:00.000-08:00

This set of problems is about the systems of:

two linear inequalities
three linear inequalities
four linear inequalities

The solid lines and the dashed lines are the loci of the points that satisfy the corresponding equalities.

The region where a given inequality holds true is indicated in black text.

The region where two or more inequalities hold true is indicated in yellow text.

Different color shades of the other regions simply indicate that different sets of corresponding inequalities apply to those regions.

Constructing the graphs of these inequalities requires at least two points for each of the corresponding equalities. After plotting, another two points on each side of the plotted line should be tested for the given inequality.

For those of you who want to draw the graphs the old fashioned way (that is, the paper-and-pencil method), finding the points can be a tedious task. This is where spreadsheet formulas can be a lot useful: you use spreadsheet formulas to calculate the points for you. So if you're unfamiliar with those spreadsheet formulas you might want to read my coming blog on simple spreadsheet formulas for calculating the x and y coordinates of points for a given linear equation.

Systems of Two Linear Inequalities

1. Draw the graph of the given system of inequalities:

3x - 2y > -2
x + 3y < 5

2. Sketch the graph of the following system:

3x - 4y < 2
5x - y < 9

3. Sketch the graph of the following system:

x + 2y > -4
3x + y < 3

Systems of Three Linear Inequalities

1. Draw the graph of the given system of inequalities:

2x + 3y > -2
x - 4y > -1
5x + 2y < 17

2. Draw the graph of the given system of inequalities:

x - 2y < -7
2x - y < 4
x + y < 2

3. Sketch the graph of the following system:

3x - 5y < 7
4x - y > -2
x + 4y > -9

System of Four Linear Inequalities

1. Draw the graph of the given system of inequalities:

3x - 2y > -13
2x + 3y < 6
x - 3y < 12
x + 6y > -21

Graphs of Inequalities With Two Variables

2010-12-16T12:22:00.000-08:00

In the following graphs of inequalities with 2 variables, the color-shaded regions indicate the regions containing the points that satisfy the given inequalities.

For any given inequality with two variables, there is a corresponding equality which is actually an equation of a straight line.

All points lying on the line satisfy the equality; and all points lying outside the line satisfy the corresponding inequalities.

Referring to the illustration below, the line x + 2y = - 2 is given.

All points along the line satisfy the statement x + 2y = - 2; any other points outside of the line will make either of the two corresponding inequality statements true, x + 2y > - 2 or x + 2y < - 2.

The two sets of points (the set of points above the line and the set of points below the line) define two inequalities of opposite directions as indicated in the figure.

The next graph below is for the line x - 3y = - 4 and its corresponding inequalities x - 3y > - 4 and x - 3y < - 4.

Observe that a change in the sign of slope of the line reverses the directions of the inequalities for the sets of points above and below the line.

Based on the above graphs, the following generalizations will be presented here: For any line whose sets of points can be differentiated into sets of points above and below it,

the set of points above the given line with a negative slope will always satisfy the corresponding inequality with greater than values;
the set of points below the given line with a positive slope will always satisfy the corresponding inequality with greater than values.

A quick way of determining the slope of a line is to convert its equation into a slope-intercept form, y = mx + b.

The shaded regions of the succeeding four graphs are bounded by dashed lines indicating that the points along the dashed lines do not make the inequality statements true.

Problem 1 Draw the graph of 3x + 2y > 0.

Problem 2 Draw the graph of 3x + 4y < 3.

Problem 3 Construct the graph describing the statement 2x - 5y < -10.

Problem 4 Make a graph of x - y > - 2.

These last four diagrams are about mixed expressions of inequality and equality. The shaded region contains the set of points satisfying the inequality part of the given mixed statement while the solid line border contains the points that satisfy the equality part of the same statement.

Problem 5 Make a graph of x - 4y ≤ 6.

Problem 6 Draw the graph of 3x - 5y ≥ - 3.

Problem 7 Sketch the graph that describes the expression 2x + 3y ≥ 6.

Problem 8 Sketch the graph that describes the expression 3x + 4y ≤ 6.

Graphs of Linear Inequalities

2010-11-05T12:48:00.001-07:00

In the following graphs, the shaded part of the number line (in red color) indicates the solution sets of the inequalities.

These problems, 20 in all, illustrate the following axioms (or postulates):

Axiom 1 A quantity added to both sides of an inequality does not change the direction of inequality.
Axiom 2 A quantity subtracted from both sides of an inequality does not change the direction of inequality.
Axiom 3 Multiplying both sides of an inequality by a positive number does not change the direction of inequality.
Axiom 4 Dividing both sides of an inequality by a positive number does not change the direction of inequality.
Axiom 5 Multiplying both sides of an inequality by a negative number changes the direction of inequality.
Axiom 6 Dividing both sides of an inequality by a negative number changes the direction of inequality.

Problems 4, 7, 8, 9, 12, 16 and 19 illustrate axioms 5 and 6.

The rest of the problems illustrate the remaining axioms.

1.	5x + 4 < 3x + 8
	Answer: { x \| x < 2 }

2.	x/3 - 3 < -1
	Answer: { x \| x < 6 }

3.	2x + 6 > 16
	Answer: { x \| x > 5 }

4.	x - 4 ≥ 9x - 60
	Answer: { x \| x ≤ 7 }

5.	8x + 5 ≥ 3x + 10
	Answer: { x \| x ≥ 1 }

6.	2x - 3 > 9
	Answer: { x \| x > 6 }

7.	x - 8 > 6x - 13
	Answer: { x \| x < 1 }

8.	4 - 5x > -6
	Answer: { x \| x < 2 }

9.	4 - 16x > 8
	Answer: { x \| x < - 1/4 }

10.	x/2 + 7 > 9
	Answer: { x \| x > 4 }

11.	9x + 5 > 4x + 40
	Answer: { x \| x > 7 }

12.	8 - 5x < 5
	Answer: { x \| x > 3/5 }

13.	3x/4 + 2 > 8
	Answer: { x \| x > 8 }

14.	2x - 5 ≤ x + 1
	Answer: { x \| x ≤ 6 }

15.	5x + 3 < 38
	Answer: { x \| x < 7 }

16.	7 - 8x < 3
	Answer: { x \| x > 1/2 }

17.	4x + 3 ≤ 3x + 10
	Answer: { x \| x ≤ 7 }

18.	5x + 7 < -8
	Answer: { x \| x < -3 }

19.	2 - 5x < -33
	Answer: { x \| x > 7 }

20.	2x - 5 > -21
	Answer: { x \| x > -8 }

Coordinate Graphs of Quadrilaterals With Their Diagonals And Medians

2010-10-15T12:34:00.000-07:00

The illustrated problems given below are graphs of the following quadrilaterals:

rectangles
squares
rhombus
trapezoids

Unlike my earlier post about quadrilaterals (see Coordinate Graphs of Quadrilaterals, dated July 2010), these problems focus on the properties of the diagonals and median of quadrilaterals.

These properties are as follows:

A diagonal divides a parallelogram into two congruent triangles (see Problems 1, 4 and 7).
Diagonals of a parallelogram bisect each other, that is, the point of intersection of the diagonals is their midpoint (see Problems 1, 4 and 7).
Midpoints of the sides of a parallelogram, when joined together, form another parallelogram (see Problems 2 and 3).
Diagonals of a rhombus bisect its angles and are perpendicular to each other (see Problem 7).
Diagonals of a rectangle are equal (see Problem 1 and 4).
Diagonals of an isosceles trapezoid are equal (see Problem 5 and 6).
The point of intersection of the diagonals of an isosceles trapezoid trisects the diagonals (see Problem 5 and 6).
The median of a trapezoid is parallel to its bases (see Problem 6).
The median of a trapezoid is equal to one half the sum of its bases (see Problem 6).
The median of a trapezoid intersects each of its diagonals at their midpoints (see Problem 6).

Concepts Illustrated By the Problems

distance between two points
properties of parallel and perpendicular lines
point of intersection of two straight lines
equation of a straight line
midpoint of a line segment
angle between two lines
angle bisector

Description of the Problems

Depending on the graph, a combination of the following information are provided:

equations of the line segments
slopes of the line segments
x-intercepts and y-intercepts of the lines
distances between the terminal points of the line segments
midpoints of the line segments

Using one or a combination of two or more of the above information, it is possible for a problem to be presented in several ways.

Line segments and their terminal points referred to in the problems are the sides and vertices, respectively, of the quadrilaterals.

Slope of a line segment with terminal points A and B is abbreviated as m_AB.

Distance between points A and B is abbreviated as distance_AB

Problem 1

Point of Intersection of the Diagonals: E (- 1/2, -1)

Distances Between Points

distance_AB = distance_CD = √ 68
distance_AD = distance_BC = √ 17
distance_AC = distance_BD = √ 85
distance_AE = distance_CE = distance_BE = distance_DE = √ 85/4

Equations of the Line Segments

line segment AB : x - 4y + 5 = 0 (x-intercept = -5; y-intercept = 5/4)
line segment BC : 4x + y - 14 = 0 (x-intercept = 14/4; y-intercept = 14)
line segment CD : x - 4y - 12 = 0 (x-intercept = 12; y-intercept = -3)
line segment AD : 4x + y + 20 = 0 (x-intercept = -5; y-intercept = -20)
line segment AC : 2x + 9y + 10 = 0 (x-intercept = -5; y-intercept = -10/9)
line segment BD : 6x - 7y - 4 = 0 (x-intercept = 2/3; y-intercept = - 4/7)

Slopes of the Line Segments

m_AB = m_CD = 1/4
m_AD = m_BC = -4
m_AC = - 2/9
m_BD = 6/7

Problem 2

Midpoints of Rectangle ABCD

E (-1, 9/2)
F (0, 0)
G (-3,- 7/2)
H (-4, 1)

Distances Between Points

distance_AB = distance_CD = √ 17
distance_AD = distance_BC = √ 68
distance_EF = distance_FG = distance_GH = distance_EH = √ 85/4
distance_AE = distance_BE = distance_CG = distance_DG = √ 17/4
distance_AH = distance_DH = distance_BF = distance_CF = √ 17

Equations of the Line Segments

line segment AB : x + 4y - 17 = 0 (x-intercept = 17; y-intercept = 17/4)
line segment BC : 4x - y = 0 (x-intercept = 0; y-intercept = 0)
line segment CD : x + 4y + 17 = 0 (x-intercept = -17; y-intercept = - 17/4)
line segment AD : 4x - y + 17 = 0 (x-intercept = - 17/4; y-intercept = 17)
line segment EF : 9x + 2y = 0 (x-intercept = 0; y-intercept = 0)
line segment FG : 7x - 6y = 0 (x-intercept = 0; y-intercept = 0)
line segment GH : 9x + 2y + 34 = 0 (x-intercept = -34/9; y-intercept = -17)
line segment EH : 7x - 6y + 34 = 0 (x-intercept = -34/7; y-intercept = 17/3)

Slopes of the Line Segments

m_AB = m_CD = - 1/4
m_AD = m_BC = 4
m_EF = m_GH = - 9/2
m_EH = m_FG = 7/6

Problem 3

Midpoints of Square ABCD

E (-1, 4)
F (4, 1)
G (1, -4)
H (-4, -1)

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √ 68
distance_EF = distance_FG = distance_GH = distance_EH = √ 34
distance_AE = distance_BE = distance_BF = distance_CF = distance_CG = distance_DG = distance_AH = distance_DH = √ 17

Equations of the Line Segments

line segment AB : x - 4y + 17 = 0 (x-intercept = -17; y-intercept = 17/4)
line segment BC : 4x + y - 17 = 0 (x-intercept = 17/4; y-intercept = 17)
line segment CD : x - 4y - 17 = 0 (x-intercept = 17; y-intercept = - 17/4)
line segment AD : 4x + y + 17 = 0 (x-intercept = - 17/4; y-intercept = -17)
line segment EF : 3x + 5y - 17 = 0 (x-intercept = 17/3; y-intercept = 17/5)
line segment FG : 5x - 3y - 17 = 0 (x-intercept = 17/5; y-intercept = - 17/3)
line segment GH : 3x + 5y + 17 = 0 (x-intercept = - 17/3; y-intercept = - 17/5)
line segment EH : 5x - 3y + 17 = 0 (x-intercept = - 17/5; y-intercept = 17/3)

Slopes of the Line Segments

m_AB = m_CD = 1/4
m_AD = m_BC = -4
m_EF = m_GH = - 3/5
m_FG = m_EH = 5/3

Problem 4

Point of Intersection of the Diagonals: E (0, 0)

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √ 52
distance_AC = distance_BD = √ 104
distance_AE = distance_CE = distance_BE = distance_DE = √ 26

Equations of the Line Segments

line segment AB : 2x + 3y - 13 = 0 (x-intercept = 13/2; y-intercept = 13/3)
line segment BC : 3x - 2y - 13 = 0 (x-intercept = 13/3; y-intercept = - 13/2)
line segment CD : 2x + 3y + 13 = 0 ( x-intercept = - 13/2; y-intercept = - 13/3)
line segment AD : 3x - 2y + 13 = 0 (x-intercept = - 13/3; y-intercept = 13/2)
line segment AC : 5x + y = 0 (x-intercept = 0; y-intercept = 0)
line segment BD : x - 5y = 0 (x-intercept = 0; y-intercept = 0)

Slopes of the Line Segments

m_AB = m_CD = - 2/3
m_BC = m_AD = 3/2
m_AC = -5
m_BD = 1/5

Problem 5

Point of Intersection of the Diagonals: E (-2, 1)

Distances Between Points

distance_AB = √ 20
distance_CD = √ 80
distance_AD = distance_BC = √ 50
distance_AC = distance_BD = √ 90
distance_AE = 1/3 (distance_AC) = (√ 90)/3 = √ 10
distance_BE = 1/3 (distance_BD) = (√ 90)/3 = √ 10

Equations of the Line Segments

line segment AB : x - 2y + 9 = 0 (x-intercept = -9; y-intercept = 9/2)
line segment BC : x + y - 3 = 0 (x-intercept = 3; y-intercept = 3)
line segment CD : x - 2y - 6 = 0 (x-intercept = 6; y-intercept = -3)
line segment AD : 7x + y + 33 = 0 (x-intercept = - 33/7; y-intercept = -33)
line segment AC : x + 3y - 1 = 0 (x-intercept = 1; y-intercept = 1/3)
line segment BD : 3x - y + 7 = 0 (x-intercept = - 7/3; y-intercept = 7)

Slopes of the Line Segments

m_AB = m_CD = 1/2
m_BC = -1
m_AD = -7
m_AC = - 1/3
m_BD = 3

Problem 6

Point of Intersection of the Diagonals: I (-1, 1)

Midpoints of

line segment BC: E (3/2, 7/2)
line segment AD: H (- 3/2, - 5/2)
diagonal AC: F (1/2, 3/2)
diagonal BD: G (- 1/2, - 1/2)

Distances Between Points

distance_AB = √ 20
distance_CD = √ 80
distance_AD = distance_BC = √ 50
distance_AC = distance_BD = √ 90
distance_EH = √ 45 = ½( distance_AB + distance_CD ) = ½(√ 20 + √ 80)
distance_AF = distance_CF = distance_BG = distance_DG = √ 45/2
distance_AI = 1/3 (distance_AC) = (√ 90)/3 = √ 10
distance_BI = 1/3 (distance_BD) = (√ 90)/3 = √ 10

Equations of the Line Segments

line segment AB : 2x - y + 8 = 0 (x-intercept = -4; y-intercept = 8)
line segment BC : x + 7y - 26 = 0 (x-intercept = 26; y-intercept = 26/7)
line segment CD : 2x - y - 7 = 0 (x-intercept = 7/2; y-intercept = -7)
line segment AD : x + y + 4 = 0 (x-intercept = -4; y-intercept = -4)
line segment AC : x - 3y + 4 = 0 (x-intercept = -4; y-intercept = 4/3)
line segment BD : 3x + y + 2 = 0 (x-intercept = - 2/3; y-intercept = -2)
line segment EH : 4x - 2y + 1 = 0 (x-intercept = - 1/4; y-intercept = 1/2)

Slopes of the Line Segments

m_AB = m_CD = m_EH = 2
m_BC = - 1/7
m_AD = -1
m_AC = 1/3
m_BD = -3

Problem 7

Point of Intersection of the Diagonals: E (1, 0)

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √ 40
distance_AC = √ 128
distance_BD = √ 32
distance_AE = distance_CE = √ 32
distance_BE = distance_DE = √ 8

Equations of the Line Segments

line segment AB : x + 3y - 9 = 0 (x-intercept = 9; y-intercept = 3)
line segment BC : 3x + y - 11 = 0 (x-intercept = 11/3; y-intercept = 11)
line segment CD : x + 3y + 7 = 0 (x-intercept = -7; y-intercept = - 7/3)
line segment AD : 3x + y + 5 = 0 (x-intercept = - 5/3; y-intercept = -5)
line segment AC : x + y - 1 = 0 (x-intercept = 1; y-intercept = 1)
line segment BD : x - y - 1 = 0 (x-intercept = 1; y-intercept = -1)

Slopes of the Line Segments

m_AB = m_CD = - 1/3
m_BC = m_AD = -3
m_AC = -1
m_BD = 1

Interior Angles of Rhombus ABCD

angle A = angle C = 53.13°
angle B = angle D = 126.87°
angle BAC = angle CAD = angle ACB = angle ACD = 26.57°
angle ABD = angle CBD = angle BDC = angle ADB = 63.43°

Systems of Linear Equations In Three Unknowns

2010-09-28T08:10:00.000-07:00

These problems are solved by converting the given system into a system of equations in two unknowns.

I kept the variables to minimum in order to avoid having to compute very large numbers that may distract the students from understanding the methods involved in solving this particular type of algebraic problem.

1.	3x - 5y + 4z = 23 and x - 2y + 2z = 9 and 5x - 6y - 3z = 14
	Answer: x = 7; y = 2; z = 3

2.	5x + 2y - 5z = 0 and 6x + y - 2z = 9 and 5x - 4y - 5z = -30
	Answer: x = 2; y = 5; z = 4

3.	5x - y - z = 14 and 5x - 2y - 3z = 7 and 3x - 2y + 3z = 5
	Answer: x = 4; y = 5; z = 1

4.	4x - 3y + 2z = -1 and 3x + y - 3z = 4 and 6x - y - 4z = -5
	Answer: x = 3; y = 7; z = 4

5.	3x + 6y + z = 11 and 2x - 5y - 5z = -13 and x + 4y + 3z = 11
	Answer: x = 1; y = 1; z = 2

6.	3x + y + 2z = 1 and 3x - y + 6z = -1 and 3x - 2y + 3z = 13
	Answer: x = 4; y = -5; z = -3

7.	5x - 2y - 2z = -8 and x + 6y - z = 0 and 5x + 4y + 2z = 30
	Answer: x = 2; y = 1; z = 8

8.	4x - 3y + z = -14 and 6x + y + 4z = 3 and 2x + 3y + 4z = 17
	Answer: x = -2; y = 3; z = 3

9.	3x - y - z = -9 and 2x + y + z = -1 and x - 5y - 3z = -9
	Answer: x = -2; y = -1; z = 4

10.	2x - y - 3z = -11 and x + 3y + z = 14 and x - 3y - 2z = -13
	Answer: x = 3; y = 2; z = 5

11.	3x + 4y + 4z = 34 and 2x - 3y + 2z = 15 and 2x + y - 3z = 4
	Answer: x = 6; y = 1; z = 3

12.	3x - y + 4z = 6 and x + 3y + z = 15 and 3x - 4y - 3z = -13
	Answer: x = 2; y = 4; z = 1

13.	3x + 3y + 2z = -8 and 3x - 2y - 5z = -8 and 5x + 4y + z = -18
	Answer: x = 1; y = -7; z = 5

14.	x - y + 4z = 4 and 5x + y + 3z = -11 and 5x + 3y + z = -3
	Answer: x = -7; y = 9; z = 5

15.	3x - 2y + 3z = -15 and 5x + 4y + z = -11 and x + 5y + 3z = 16
	Answer: x = -5; y = 3; z = 2

16.	3x + 5y - 2z = 14 and 5x + 3y - 4z = 16 and 3x - 4y + 3z = 20
	Answer: x = 5; y = 1; z = 3

17.	6x + y + 3z = -6 and 7x - 4y + 5z = -2 and x - 6y + 5z = -14
	Answer: x = 3; y = -3; z = -7

18.	6x - y - 6z = 23 and 5x - 4y + z = -20 and x - 6y - 5z = 22
	Answer: x = -2; y = 1; z = -6

19.	5x + 3y - 2z = 15 and 3x + 2y + 3z = 27 and 3x + 5y - 5z = 13
	Answer: x = 1; y = 6; z = 4

20.	4x - 3y + 2z = -2 and 3x - y + 3z = 8 and 6x - y + 4z = 14
	Answer: x = 1; y = 4; z = 3

21.	x - y - 4z = -22 and 5x - y - 3z = 11 and 5x - 3y - z = 3
	Answer: x = 7; y = 9; z = 5

22.	4x - 3y + 2z = 11 and 2x + 3y + z = 10 and 4x + 3y + 4z = 23
	Answer: x = 2; y = 1; z = 3

23.	4x + 3y + 2z = 27 and x + 2y + 3z = 13 and 2x + 3y + z = 18
	Answer: x = 4; y = 3; z = 1

24.	x - 3y - 4z = -33 and x + y - 5z = -38 and x - 5y - z = -8
	Answer: x = 6; y = 1; z = 9

25.	5x - 3y + 5z = -15 and 3x + y + 5z = -1 and 3x - y - 5z = 19
	Answer: x = 3; y = 5; z = -3

26.	x - y - 7z = -8 and 9x - 4y - 5z = 11 and 5x - y + 3z = 18
	Answer: x = 4; y = 5; z = 1

27.	2x - y - 5z = -13 and 2x - 3y - 5z = -17 and x - 2y - 2z = -8
	Answer: x = 2; y = 2; z = 3

28.	x - 3y + 3z = 3 and 2x + y + 2z = 9 and x - y + 2z = 4
	Answer: x = 3; y = 1; z = 1

29.	x - 3y + 6z = 25 and x - 4y + 9z = 38 and x - y + 5z = 24
	Answer: x = 1; y = 2; z = 5

30.	x - 3y + z = -10 and 3x + 3y - 4z = 21 and 2x - y - 3z = -5
	Answer: x = 5; y = 6; z = 3

2010-09-21T12:29:00.001-07:00

Coordinate Graphs of Triangles

The illustrated problems below are graphs of the following triangles:

isosceles triangles
scalene triangles
right triangles

Concepts Illustrated By the Problems

distance between two points
angle between two lines
point of intersection of two straight lines
equation of a straight line

Description of the Problems

For each problem, the following information are given:

equations of the line segments
slopes of the line segments
x-intercepts and y-intercepts of the lines
distances between the terminal points of the line segments
interior angles of the triangles

Using one or a combination of two or more of the above information, it is possible for a problem to be presented in several ways.

Line segments and their terminal points referred to in the problems are the sides and vertices, respectively, of the triangles.

Slope of a line segment with terminal points A and B is abbreviated as m_AB.

Distance between points A and B is abbreviated as distance_AB

Graphs of Isosceles Triangles

Show that points A(-5, 1), B(3, 5), C(2, -3) are vertices of an isosceles triangle.

Distance Between Two Points

distance_AB = √ 80
distance_BC = distance_AC = √ 65

Equations of the Line Segments

line segment AB: x - 2y + 7 = 0 (x-intercept = -7; y-intercept = 7/2)
line segment BC: 8x - y - 19 = 0 (x-intercept = 19/8, y-intercept = -19)
line segment AC: 4x + 7y + 13 = 0 (x-intercept = - 13/4, y-intercept = - 13/7)

Slopes of the Line Segments

m_AB = 1/2
m_BC = 8
m_AC = - 4/7

Interior Angles of the Triangle

angle A = angle B = 56.3°
angle C = 67.4°

Show that points A(-5, 4), B(3, 3), C(-1, -4) are vertices of an isosceles triangle.

Distance Between Two Points

distance_AB = distance_BC = √ 65
distance_AC = √ 80

Equations of the Line Segments

line segment AB: x + 8y - 27 = 0 ( x-intercept = 27, y-intercept = 27/8)
line segment BC: 7x - 4y - 9 = 0 (x-intercept = 9/7, y-intercept = - 9/4)
line segment AC: 2x + y + 6 = 0 (x-intercept = -3, y-intercept = -6)

Slopes of the Line Segments

m_AB = - 1/8
m_BC = 7/4
m_AC = -2

Interior Angles of the Triangle

angle A = angle C = 56.3°
angle B = 67.4°

Prove analytically that points A(-5, -5), B(-2, 6), C(5, -3) are vertices of an isosceles triangle.

Distance Between Two Points

distance_AB = distance_BC = √ 130
distance_AC = √ 104

Equations of the Line Segments

line segment AB: 11x - 3y + 40 = 0 (x-intercept = - 40/11, y-intercept = 40/3)
line segment BC: 9x + 7y - 24 = 0 (x-intercept = 24/9, y-intercept = 24/7)
line segment AC: x - 5y - 20 = 0 (x-intercept = 20, y-intercept = -4)

Slopes of the Line Segments

m_AB = 11/3
m_BC = - 9/7
m_AC = 1/5

Interior Angles of the Triangle

angle A = angle C = 63.4°
angle B = 53.1°

Graphs of Scalene Triangles

Show that points A(-4, 1), B(1, 0), C(-2, -5) are vertices of a scalene triangle.

Distance Between Two Points

distance_AB = √ 26
distance_BC = √ 34
distance_AC = √ 40

Equations of the Line Segments

line segment AB: x + 5y - 1 = 0 (x-intercept = 1, y-intercept = 1/5)
line segment BC: 5x - 3y - 5 = 0 (x-intercept = 1, y-intercept = - 5/3)
line segment AC: 3x + y + 11 = 0 (x-intercept = - 11/3, y-intercept = -11)

Slopes of the Line Segments

m_AB = - 1/5
m_BC = 5/3
m_AC = -3

Interior Angles of the Triangle

angle A = 60.3°
angle B = 70.3°
angle C = 49.4°

Show that the following points are vertices of a scalene triangle:

A (-5, -4)
B (3, 4)
C (4, -2)

Distance Between Two Points

distance_AB = √ 128
distance_BC = √ 37
distance_AC = √ 85

Equations of the Line Segments

line segment AB: x - y + 1 = 0 (x-intercept = -1, y-intercept = 1)
line segment BC: 6x + y - 22 = 0 (x-intercept = 11/3, y-intercept = 22)
line segment AC: 2x - 9y - 26 = 0 (x-intercept = 13, y-intercept = - 26/9)

Slopes of the Line Segments

m_AB = 1
m_BC = -6
m_AC = 2/9

Interior Angles of the Triangle

angle A = 32.5°
angle B = 54.5°
angle C = 93.0°

Show that the points A(-3, 5), B(5, -2), C(-1, -5) are vertices of a scalene triangle.

Distance Between Two Points

distance_AB = √ 113
distance_BC = √ 45
distance_AC = √ 104

Equations of the Line Segments

line segment AB: 7x + 8y - 19 = 0 (x-intercept = 19/7, y-intercept = 19/8)
line segment BC: x - 2y - 9 = 0 (x-intercept = 9, y-intercept = - 9/2)
line segment AC: 5x + y + 10 = 0 (x-intercept = -2, y-intercept = -10)

Slopes of the Line Segments

m_AB = - 7/8
m_BC = 1/2
m_AC = -5

Interior Angles of the Triangle

angle A = 37.5°
angle B = 67.8°
angle C = 74.7°

Graphs of Right Triangles

Prove that points A(-1, 5), B(5, -1), C(-4, 2) are vertices of a right triangle.

Distance Between Two Points

distance_AB = √ 72
distance_BC = √ 90
distance_AC = √ 18

Equations of the Line Segments

line segment AB: x + y - 4 = 0 (x-intercept = 4, y-intercept = 4)
line segment BC: x + 3y - 2 = 0 (x-intercept = 2, y-intercept = 2/3)
line segment AC: x - y + 6 = 0 (x-intercept = -6, y-intercept = 6)

Slopes of the Line Segments

m_AB = -1
m_BC = - 1/3
m_AC = 1

Interior Angles of the Triangle

angle A = 90.0°
angle B = 26.6°
angle C = 63.4°

Show that points A(-5, 3), B(2, -1), C(-1, -3) are vertices of a right triangle.

Distance Between Two Points

distance_AB = √ 65
distance_BC = √ 13
distance_AC = √ 52

Equations of the Line Segments

line segment AB: 4x + 7y - 1 = 0 (x-intercept = 1/4, y-intercept = 1/7)
line segment BC: 2x - 3y - 7 = 0 (x-intercept = 7/2, y-intercept = - 7/3)
line segment AC: 3x + 2y + 9 = 0 (x-intercept = -3, y-intercept = - 9/2)

Slopes of the Line Segments

m_AB = - 4/7
m_BC = 2/3
m_AC = - 3/2

Interior Angles of the Triangle

angle A = 26.6°
angle B = 63.4°
angle C = 90.0°

Prove that points A(-2, 5), B(4, -3), C(0, -6) are vertices of a right triangle.

Distance Between Two Points

distance_AB = √ 100
distance_BC = √ 25
distance_AC = √ 125

Equations of the Line Segments

line segment AB: 4x + 3y - 7 = 0 (x-intercept = 7/4, y-intercept = 7/3)
line segment BC: 3x - 4y - 24 = 0 (x-intercept = 8, y-intercept = -6)
line segment AC: 11x + 2y + 12 = 0 (x-intercept = - 12/11, y-intercept = -6)

Slopes of the Line Segments

m_AB = - 4/3
m_BC = 3/4
m_AC = - 11/2

Interior Angles of the Triangle

angle A = 26.6°
angle B = 90.0°
angle C = 63.4°

Graphs of Isosceles Right Triangles

Prove that the points A(-4, 1), B(1, 5), C(0, -4) are vertices of an isosceles right triangle.

Distance Between Two Points

distance_AB = √ 41
distance_BC = √ 82
distance_AC = √ 41

Equations of the Line Segments

line segment AB: 4x - 5y + 21 = 0 (x-intercept = - 21/4, y-intercept = 21/5)
line segment BC: 9x - y - 4 = 0 (x-intercept = 4/9, y-intercept = -4)
line segment AC: 5x + 4y + 16 = 0 (x-intercept = - 16/5, y-intercept = -4)

Slopes of the Line Segments

m_AB = 4/5
m_BC = 9
m_AC = - 5/4

Interior Angles of the Triangle

angle A = 90°
angle B = 45°
angle C = 45°

Show that points A(-5, 4), B(0, 2), C(-2, -3) are vertices of an isosceles right triangle.

Distance Between Two Points

distance_AB = √ 29
distance_BC = √ 29
distance_AC = √ 58

Equations of the Line Segments

line segment AB: 2x + 5y - 10 = 0 (x-intercept = 5, y-intercept = 2)
line segment BC: 5x - 2y + 4 = 0 (x-intercept = - 4/5, y-intercept = 2)
line segment AC: 7x + 3y + 23 = 0 (x-intercept = - 23/7, y-intercept = - 23/3)

Slopes of the Line Segments

m_AB = - 2/5
m_BC = 5/2
m_AC = - 7/3

Interior Angles of the Triangle

angle A = 45°
angle B = 90°
angle C = 45°

Prove that points A(-3, 4), B(5, 1), C(-6, -4) are vertices of an isosceles right triangle.

Distance Between Two Points

distance_AB = √ 73
distance_BC = √ 146
distance_AC = √ 73

Equations of the Line Segments

line segment AB: 3x + 8y - 23 = 0 (x-intercept = 23/3, y-intercept = 23/8)
line segment BC: 5x - 11y - 14 = 0 (x-intercept = 14/5, y-intercept = - 14/11)
line segment AC: 8x - 3y + 36 = 0 (x-intercept = - 9/2, y-intercept = 12)

Slopes of the Line Segments

m_AB = - 3/8
m_BC = 5/11
m_AC = 8/3

Interior Angles of the Triangle

angle A = 90°
angle B = 45°
angle C = 45°

2010-09-10T13:29:00.000-07:00

Starting tomorrow, Tutor Partner will be using a new design. Hence it will have a new appearance.

The content, though, will remain the same.

Thank you for your continuing support.

2010-09-02T10:56:00.000-07:00

Simultaneous Equations

1.	2x - 3y = -26 and 4x - 7y = -60
	Answer: x = -1; y = 8

2.	3x + 5y = 21 and 7x - 2y = -74
	Answer: x = -8; y = 9

3.	3x + 7y = -42 and 8x + 9y = -25
	Answer: x = 7; y = -9

4.	5x - 3y = 51 and 7x + 5y = 7
	Answer: x = 6; y = -7

5.	8x - 7y = 58 and 5x + 3y = 51
	Answer: x = 9; y = 2

6.	3x - 5y = 7 and 9x - 4y = -34
	Answer: x = -6; y = -5

7.	5x + 8y = -37 and 3x - 2y = 5
	Answer: x = -1; y = -4

8.	9x - 5y = -81 and 3x - 8y = -84
	Answer: x = -4; y = 9

9.	4x + 5y = 59 and 2x - 7y = -37
	Answer: x = 6; y = 7

10.	9x + 2y = 37 and x + 7y = 38
	Answer: x = 3; y = 5

11.	9x - 8y = -31 and 7x - 4y = -13
	Answer: x = 1; y = 5

12.	7x - 5y = -63 and 2x + 9y = 55
	Answer: x = -4; y = 7

13.	6x + 5y = 75 and 8x + 3y = 67
	Answer: x = 5; y = 9

14.	4x - 9y = 32 and 3x - 2y = 5
	Answer: x = -1; y = -4

15.	4x + 7y = -45 and 3x - 8y = 59
	Answer: x = 1; y = -7

16.	x + 8y = 19 and 5x + 2y = 19
	Answer: x = 3; y = 2

17.	5x + 3y = -15 and 6x + 5y = -11
	Answer: x = -6; y = 5

18.	3x - 4y = 19 and 9x + 5y = 91
	Answer: x = 9; y = 2

19.	2x - 3y = -14 and 4x - 7y = -36
	Answer: x = 5; y = 8

20.	5x + 7y = -69 and 3x + 2y = -26
	Answer: x = -4; y = -7

21.	8x + 7y = 79 and 2x - 9y = 9
	Answer: x = 9; y = 1

22.	9x - 7y = -12 and 9x - 5y = -6
	Answer: x = 1; y = 3

23.	7x - 6y = -7 and 5x + 4y = 53
	Answer: x = 5; y = 7

24.	7x - 2y = -37 and 6x - 7y = -37
	Answer: x = -5; y = 1

25.	7x - 8y = -10 and 3x + 2y = 12
	Answer: x = 2; y = 3

26.	6x + 5y = 53 and 2x - 7y = -43
	Answer: x = 3; y = 7

27.	3x - 8y = 61 and 6x + 5y = -46
	Answer: x = -1; y = -8

28.	4x - y = -3 and 5x + 4y = 33
	Answer: x = 1; y = 7

29.	9x + 7y = 47 and 5x + 2y = 28
	Answer: x = 6; y = -1

30.	9x + 8y = 49 and 7x + 4y = 27
	Answer: x = 1; y = 5

2010-07-17T13:08:00.000-07:00

Coordinate Graphs of Quadrilaterals

The illustrated problems given below are graphs of the following quadrilaterals:

rectangles
squares
rhombi
trapezoids

A quadrilateral is defined as a polygon with four sides.

A parallelogram is a quadrilateral that has two pairs of parallel opposite sides.

Description of Quadrilaterals

Rectangles and rhombi are both parallelograms. However, rhombi have no right interior angles while rectangles have right interior angles.

Generally, a rectangle has one of its two consecutive sides longer than the other one. But a square, which is also a rectangle, has two equal consecutive sides.

Like the square, a rhombus has two equal consecutive sides. But it has no right interior angles.

Lastly, a trapezoid is a quadrilateral with one pair of parallel opposite sides. Its non-parallel sides can be equal or unequal. If a trapezoid has equal non-parallel sides, then it's called an isosceles trapezoid.

Concepts Illustrated By the Problems

Each problem below illustrates the following concepts:

distance between 2 points
parallel and perpendicular lines
point of intersection of two straight lines
equation of a straight line

The equation of a straight line can be determined if any of the following are given:

2 points on the line
slope and a point of the line
slope and the x-intercept ( or y-intercept ) of the line
x-intercept and y-intercept of the line

Description of the Problems

For each problem (or graph), the following information are given:

equations of the line segments
slopes of the line segments
x-intercepts and y-intercepts of the lines
distances between the terminal points of the line segments
points on the extended line segments

Using one or a combination of two or more of the above information, it is possible for a problem to be presented in several ways.

Line segments and their terminal points referred to in the problems are the sides and vertices, respectively, of the quadrilaterals.

Slope of a line segment with terminal points A and B is abbreviated as m_AB.

Distance between points A and B is abbreviated as distance_AB

Points on a line refer to the points of the extended line of that line segment.

Graphs of Rectangles

Show that:

points A(-2, 4), B(-4, 0), C(0, 3), and D(-2, -1) are vertices of a rectangle;
points A(-2, 4), B(-4, 0), E(4, 1), and F(2, -3) are vertices of a rectangle;
points C(0, 3), D(-2, -1), E(4, 1), and F(2, -3) are vertices of a square.

Equations of the Line Segments

line segment AB (2x - y + 8 = 0): x-intercept = -4; y-intercept = 8
line segment CD (2x - y + 3 = 0): x-intercept = -3/2; y-intercept = 3
line segment EF (2x - y - 7 = 0): x-intercept = 7/2; y-intercept = -7
line segment AC or CE or AE (x + 2y - 6 = 0): x-intercept = 6; y-intercept = 3
line segment BD or DF or BF (x + 2y + 4 = 0): x-intercept = -4; y-intercept = -2

Slopes of the Line Segments

m_AB = m_CD = m_EF = 2
m_AC = m_CE = m_AE = - 1/2
m_BD = m_DF = m_BF = - 1/2

Distances Between Points

distance_AB = distance_CD = distance_EF = √ 20
distance_AC = distance_BD = √ 5
distance_CE = distance_DF = √ 20
distance_AE = distance_BF = √ 45

Points On The Lines

line 2x - y + 8 = 0 (line segment AB): (-5, -2), (-4, 0), (-3, 2), (-2, 4), (-1, 6), (1, 10),(2, 12), (3, 14), (4, 16), (5, 18)
line 2x - y + 3 = 0 (line segment CD): (-5, -7), (-4, -5), (-3, -3), (-2, -1), (-1, 1), (1, 5), (2, 7), (3, 9), (4, 11), (5, 13)
line 2x - y - 7 = 0 (line segment EF): (-5, -17), (-4, -15), (-3, -13), (-2, -11), (-1, -9), (1, -5), (2, -3), (3, -1), (4, 1), (5, 3)
line x + 2y - 6 = 0 (line segment AC or CE or AE): (-5, 11/2), (-4, 5), (-3, 9/2), (-2, 4), (-1, 7/2), (1, 5/2), (2, 2), (3, 3/2), (4, 1), (5, 1/2)
line x + 2y + 4 = 0 (line segment BD or DF or BF): (-5, 1/2), (-4, 0), (-3, -1/2), (-2, -1), (-1, -3/2), (1, -5/2), (2, -3), (3, -7/2), (4, -4), (5, -9/2)

Show that the quadrilateral bounded by the lines 3x - 2y + 8 = 0 and 3x - 2y - 5 = 0 and line segments AB and CD is a rectangle.

Equations of the Line Segments

line 3x - 2y + 8 = 0: x-intercept = - 8/3; y-intercept = 4
line 3x - 2y - 5 = 0: x-intercept = 5/3; y-intercept = - 5/2
line segment AB (2x + 3y - 12 = 0): x-intercept = 6; y-intercept = 4
line segment CD (2x + 3y + 14 = 0): x-intercept = -7; y-intercept = -14/3

Slopes of the Line Segments

m_AB = m_CD = - 2/3
m_AD = m_BC = 3/2

Distances Between Points

distance_AD = distance_BC = √ 52
distance_AB = distance_CD = √ 13

Points On The Lines

line 3x - 2y + 8 = 0 (line segment AD): (-5, -7/2), (-4, -2), (-3, -1/2), (-2, 1), (-1, 5/2), (1, 11/2), (2, 7), (3, 17/2), (4, 10), (5, 23/2)
line 3x - 2y - 5 = 0 (line segment BC): (-5, -10), (-4, -17/2), (-3, -7), (-2, -11/2), (-1, -4), (1, -1), (2, 1/2), (3, 2), (4, 7/2), (5, 5)
line 2x + 3y - 12 = 0 (line segment AB): (-5, 22/3), (-4, 20/3), (-3, 6), (-2, 16/3), (-1, 14/3), (1, 10/3), (2, 8/3), (3, 2), (4, 4/3), (5, 2/3)
line 2x + 3y + 14 = 0 (line segment CD): (-5, -4/3), (-4, -2), (-3, -8/3), (-2, -10/3), (-1, -4), (1, -16/3), (2, -6), (3, -20/3), (4, -22/3), (5, -8)

Prove analytically that points A(-5, 2), B(3, 4), C(4, 0) and D(-4, -2) are vertices of a rectangle.

Equations of the Line Segments

line segment AB : x - 4y + 13 = 0; x-intercept = -13; y-intercept = 13/4
line segment BC : 4x + y - 16 = 0; x-intercept = 4; y-intercept = 16
line segment CD : x - 4y - 4 = 0; x-intercept = 4; y-intercept = - 1
line segment AD : 4x + y + 18 = 0; x-intercept = - 9/2; y-intercept = -18

Slopes of the Line Segments

m_AB = m_CD = 1/4
m_AD = m_BC = - 4

Distances Between Points

distance_AB = distance_CD = √ 68
distance_BC = distance_AD = √ 17

Points On The Lines

line x - 4y + 13 = 0 (line segment AB): (-5, 2), (-4, 9/4), (-3, 5/2), (-2, 11/4), (-1, 3), (1, 7/2), (2,15/4 ), (3, 4), (4, 17/4), (5, 9/2)
line 4x + y - 16 = 0 (line segment BC): (-5, 36), (-4, 32), (-3, 28), (-2, 24), (-1, 20), (1, 12), (2, 8), (3, 4), (4, 0), (5, -4)
line x - 4y - 4 = 0 (line segment CD): (-5, -9/4), (-4, -2), (-3, -7/4), (-2, -3/2), (-1, -5/4), (1, -3/4), (2, -1/2), (3, -1/4), (4, 0), (5, 1/4)
line 4x + y + 18 = 0 (line segment AD): (-5, 2), (-4, -2), (-3, -6), (-2, -10), (-1, -14), (1, -22), (2, -26), (3, -30), (4, -34), (5, -38)

Graphs of Squares

Show that points A(-1, 3), B(3, 2), C(2, -2) and D(-2, -1) are vertices of a square.

Equations of the Line Segments

line segment AB : x + 4y - 11 = 0; x-intercept = 11; y-intercept = 11/4
line segment BC : 4x - y - 10 = 0; x-intercept = 5/2; y-intercept = -10
line segment CD : x + 4y + 6 = 0; x-intercept = -6; y-intercept = -3/2
line segment AD : 4x - y + 7 = 0; x-intercept = -7/4; y-intercept = 7

Slopes of the Line Segments

m_AB = m_CD = - 1/4
m_BC = m_AD = 4

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √17

Points On The Lines

line x + 4y - 11 = 0 (line segment AB): (-5, 4), (-4, 15/4), (-3, 7/2), (-2, 13/4), (-1, 3), (1, 5/2), (2, 9/4), (3, 2), (4, 7/4), (5, 3/2)
line 4x - y - 10 = 0 (line segment BC): (-5, -30), (-4, -26), (-3, -22), (-2, -18), (-1, -14), (1, -6), (2, -2), (3, 2), (4, 6), (5, 10)
line x + 4y + 6 = 0 (line segment CD): (-5, -1/4), (-4, -1/2), (-3, -3/4), (-2,-1 ), (-1, -5/4), (1, -7/4), (2, -2), (3, -9/4), (4, -5/2), (5, -11/4)
line 4x - y + 7 = 0 (line segment AD): (-5, -13), (-4, -9), (-3, -5), (-2,-1 ), (-1, 3), (1, 11), (2, 15), (3, 19), (4, 23), (5, 27)

Show that the points of intersection of the following lines are vertices of a square:

x + y - 4 = 0
x - y - 4 = 0
x + y + 4 = 0
x - y + 4 = 0

Equations of the Line Segments

line segment AB : x + y - 4 = 0; x-intercept = 4; y-intercept = 4
line segment BC : x - y - 4 = 0; x-intercept = 4; y-intercept = -4
line segment CD : x + y + 4 = 0; x-intercept = -4; y-intercept = -4
line segment AD : x - y + 4 = 0; x-intercept = -4; y-intercept = 4

Slopes of the Line Segments

m_AB = m_CD = - 1
m_BC = m_AD = 1

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √ 32

Points On The Lines

line x + y - 4 = 0 (line segment AB): (-5, 9), (-4, 8), (-3, 7), (-2, 6), (-1, 5), (1, 3), (2, 2), (3, 1), (4, 0), (5, -1)
line x - y - 4 = 0 (line segment BC): (-5, -9), (-4, -8), (-3, -7), (-2, -6), (-1, -5), (1, -3), (2, -2), (3, -1), (4, 0), (5, 1)
line x + y + 4 = 0 (line segment CD): (-5, 1), (-4, 0), (-3, -1), (-2, -2), (-1, -3), (1, -5), (2, -6), (3, -7), (4, -8), (5, -9)
line x - y + 4 = 0 (line segment AD): (-5, -1), (-4, 0), (-3, 1), (-2, 2), (-1, 3), (1, 5), (2, 6), (3, 7), (4, 8), (5, 9)

Show that the points A(1, 5), B(3, 2), C(0, 0), and D(-2, 3) are vertices of a square.

Equations of the Line Segments

line segment AB : 3x + 2y - 13 = 0; x-intercept = 13/3; y-intercept = 13/2
line segment BC : 2x - 3y = 0; x-intercept = 0; y-intercept = 0
line segment CD : 3x + 2y = 0; x-intercept = 0; y-intercept = 0
line segment AD : 2x - 3y + 13 = 0; x-intercept = -13/2; y-intercept = 13/3

Slopes of the Line Segments

m_AB = m_CD = - 3/2
m_BC = m_AD = 2/3

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √ 13

Points On The Lines

line 3x + 2y - 13 = 0 (line segment AB): (-5, 14), (-4, 25/2), (-3, 11), (-2, 19/2), (-1, 8), (1, 5), (2, 7/2), (3, 2), (4, 1/2), (5, -1)
line 2x - 3y = 0 (line segment BC): (-5, -10/3), (-4, -8/3), (-3, -2), (-2, -4/3), (-1, -2/3), (1, 2/3), (2, 4/3), (3, 2), (4, 8/3), (5, 10/3)
line 3x + 2y = 0 (line segment CD): (-5, 15/2), (-4, 6), (-3, 9/2), (-2, 3), (-1, 3/2), (1, -3/2), (2, -3), (3, -9/2), (4, -6), (5, -15/2)
line 2x - 3y + 13 = 0 (line segment AD): (-5, 1), (-4, -5/3), (-3, 7/3), (-2, 3), (-1, 11/3), (1, 5), (2, 17/3), (3, 19/3), (4, 7), (5, 23/3)

Graphs of Rhombi

Prove that points A(1, 5), B(4, 1), C(1, -3) and D(-2, 1) are vertices of a rhombus.

Equations of the Line Segments

line segment AB : 4x + 3y - 19 = 0; x-intercept = 19/4; y-intercept = 19/3
line segment BC : 4x - 3y - 13 = 0; x-intercept = 13/4; y-intercept = - 13/3
line segment CD : 4x + 3y + 5 = 0; x-intercept = - 5/4; y-intercept = - 5/3
line segment AD : 4x - 3y + 11 = 0; x-intercept = - 11/4; y-intercept = 11/3

Slopes of the Line Segments

m_AB = m_CD = - 4/3
m_BC = m_AD = 4/3

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √ 25

Points On The Lines

line 4x + 3y - 19 = 0 (line segment AB): (-5, 13), (-4, 35/3), (-3, 31/3), (-2, 9), (-1, 23/3), (1, 5), (2, 11/3), (3, 7/3), (4, 1), (5, -1/3)
line 4x - 3y - 13 = 0 (line segment BC): (-5, -11), (-4, -29/3), (-3, -25/3), (-2, -7), (-1, -17/3), (1, -3), (2, -5/3), (3, -1/3), (4, 1), (5, 7/3)
line 4x + 3y + 5 = 0 (line segment CD): (-5, 5), (-4, 11/3), (-3, 7/3), (-2, 1), (-1, -1/3), (1, -3), (2, -13/3), (3, -17/3), (4, -7), (5, -25/3)
line 4x - 3y + 11 = 0 (line segment AD): (-5, -3), (-4, -5/3), (-3, -1/3), (-2, 1), (-1, 7/3), (1, 5), (2, 19/3), (3, 23/3), (4, 9), (5, 31/3)

Show that points A(-2, 3), B(3, 4), C(2, -1), and D(-3, -2) are vertices of a rhombus.

Equations of the Line Segments

line segment AB : x - 5y + 17 = 0; x-intercept = -17; y-intercept = 17/5
line segment BC : 5x - y - 11 = 0; x-intercept = 11/5; y-intercept = -11
line segment CD : x - 5y - 7 = 0; x-intercept = 7; y-intercept = - 7/5
line segment AD : 5x - y + 13 = 0; x-intercept = - 13/5; y-intercept = 13

Slopes of the Line Segments

m_AB = m_CD = 1/5
m_BC = m_AD = 5

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √ 26

Points On The Lines

line x - 5y + 17 = 0 (line segment AB): (-5, 12/5), (-4, 13/5), (-3, 14/5), (-2, 3), (-1, 16/5), (1, 18/5), (2, 19/5), (3, 4), (4, 21/5), (5, 22/5)
line 5x - y - 11 = 0 (line segment BC): (-5, -36), (-4, -31), (-3, -26), (-2, -21), (-1, -16), (1, -6), (2, -1), (3, 4), (4, 9), (5, 14)
line x - 5y - 7 = 0 (line segment CD): (-5, -12/5), (-4, -11/5), (-3, -2), (-2, -9/5), (-1, 8/5), (1, -6/5), (2, -1), (3, -4/5), (4, -3/5), (5, -2/5)
line 5x - y + 13 = 0 (line segment AD): (-5, -12), (-4, -7), (-3, -2), (-2, 3), (-1, 8), (1, 18), (2, 23), (3, 28), (4, 33), (5, 38)

Prove that points A(-5, 2), B(3/5, 4), C(5, 0), and D(-3/5, -2) are vertices of a rhombus.

Equations of the Line Segments

line segment AB : 5x - 14y + 53 = 0; x-intercept = - 53/5; y-intercept = 53/14
line segment BC : 10x + 11y - 50 = 0; x-intercept = 5; y-intercept = 50/11
line segment CD : 5x - 14y - 25 = 0; x-intercept = 5; y-intercept = - 25/14
line segment AD : 10x + 11y + 28 = 0; x-intercept = - 14/5; y-intercept = - 28/11

Slopes of the Line Segments

m_AB = m_CD = 5/14
m_BC = m_AD = - 10/11

Distances Between Points

distance_AB = distance_BC = distance_CD = distance_AD = √ 884/25

Points On The Lines

line 5x - 14y + 53 = 0 (line segment AB): (-5, 2), (-4, 33/14), (-3, 19/7), (-2, 43/14), (-1, 24/7), (1, 29/7), (2, 9/2), (3, 34/7), (4, 73/14), (5, 39/7)
line 10x + 11y - 50 = 0 (line segment BC): (-5, 100/11), (-4, 90/11), (-3, 80/11), (-2, 70/11), (-1, 60/11), (0, 50/11), (1, 40/11), (2, 30/11), (3, 20/11), (4, 10/11), (5, 0)
line 5x - 14y - 25 = 0 (line segment CD): (-5, 25/7), (-4, -45/14), (-3, -20/7), (-2, -5/2), (-1, -15/7), (1, -10/7), (2, -15/14), (3, -5/7), (4, -5/14), (5, 0)
line 10x + 11y + 28 = 0 (line segment AD): (-5, 2), (-4, 12/11), (-3, 2/11), (-2, -8/11), (-1, -18/11), (1, -38/11), (2, -48/11), (3, -58/11), (4, -68/11), (5, -78/11)

Graphs of Trapezoids

Prove that the points A(-1, 4), B(4, 5), C(-1, -5) and D(-4, -2) are vertices of a trapezoid.

Equations of the Line Segments

line segment AB : x - 5y + 21 = 0; x-intercept = - 21; y-intercept = 21/5
line segment BC : 2x - y - 3 = 0; x-intercept = 3/2; y-intercept = - 3
line segment CD : x + y + 6 = 0; x-intercept = - 6; y-intercept = - 6
line segment AD : 2x - y + 6 = 0; x-intercept = -3; y-intercept = 6

Slopes of the Line Segments

m_AB = 1/5
m_CD = -1
m_BC = m_AD = 2

Distances Between Points

distance_AB: √ 26
distance_BC: √ 125
distance_CD: √ 18
distance_AD: √ 45

Points On The Lines

line x - 5y + 21 = 0 (line segment AB): (-5, 16/5), (-4, 17/5), (-3, 18/5), (-2, 19/5), (-1, 4), (1, 22/5), (2, 23/5), (3, 24/5), (4, 5), (5, 26/5)
line 2x - y - 3 = 0 (line segment BC): (-5, -13), (-4, -11), (-3, -9), (-2, -7), (-1, -5), (1, -1), (2, 1), (3, 3), (4, 5), (5, 7)
line x + y + 6 = 0 (line segment CD): (-5, -1), (-4, -2), (-3, -3), (-2, -4), (-1, -5), (1, -7), (2, -8), (3, -9), (4, -10), (5, -11)
line 2x - y + 6 = 0 (line segment AD): (-5, -4), (-4, -2), (-3, 0), (-2, 2), (-1, 4), (1, 8), (2, 10), (3, 12), (4, 14), (5, 16)

Show that points A(- 80/17, 71/17), B(-1, 5), C(2, 0), and D(- 8/17, - 49/17) are vertices of an isosceles trapezoid.

Equations of the Line Segments

line segment AB : 2x - 9y + 47 = 0; x-intercept = - 47/2; y-intercept = 47/9
line segment BC : 5x + 3y - 10 = 0; x-intercept = 2; y-intercept = 10/3
line segment CD : 7x - 6y - 14 = 0; x-intercept = 2; y-intercept = - 7/3
line segment AD : 5x + 3y + 11 = 0; x-intercept = - 11/5; y-intercept = - 11/3

Slopes of the Line Segments

m_AB = 2/9
m_CD = 7/6
m_BC = m_AD = - 5/3

Distances Between Points

distance_AB = distance_CD = √ 245/17
distance_BC = √ 34
distance_AD = √ 1152/17

Points On The Lines

line 2x - 9y + 47 = 0 (line segment AB): (-5, 37/9), (-4, 13/3), (-3, 41/9), (-2, 43/9), (-1, 5), (1, 49/9), (2, 17/3), (3, 53/9), (4, 55/9), (5, 19/3)
line 5x + 3y - 10 = 0 (line segment BC): (-5, 35/3), (-4, 10), (-3, 25/3), (-2, 20/3), (-1, 5), (1, 5/3), (2, 0), (3, -5/3), (4, -10/3), (5, -5)
line 7x - 6y - 14 = 0 (line segment CD): (-5, - 49/6), (-4, -7), (-3, - 35/6), (-2, - 14/3), (-1, - 7/2), (1, - 7/6), (2, 0), (3, 7/6), (4, 7/3), (5, 7/2)
line 5x + 3y + 11 = 0 (line segment AD): (-5, 14/3), (-4, 3), (-3, 4/3), (-2, -1/3), (-1, -2), (1, -16/3), (2, -7), (3, -26/3), (4, -31/3), (5, -12)

Prove that points A(-4, 1), B(-1, 4), C(5, 1) and D(-1, -5) are vertices of an isosceles trapezoid.

Equations of the Line Segments

line segment AB : x - y + 5 = 0; x-intercept = - 5; y-intercept = 5
line segment BC : x + 2y - 7 = 0; x-intercept = 7; y-intercept = 7/2
line segment CD : x - y - 4 = 0; x-intercept = 4; y-intercept = - 4
line segment AD : 2x + y + 7 = 0; x-intercept = - 7/2; y-intercept = - 7

Slopes of the Line Segments

m_AB = m_CD = 1
m_BC = - 1/2
m_AD = -2

Distances Between Points

distance_AB = √ 18
distance_CD = √ 72
distance_BC = distance_AD = √ 45

Points On The Lines

line x - y + 5 = 0 (line segment AB): (-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)
line x + 2y - 7 = 0 (line segment BC): (-5, 6), (-4, 11/2), (-3, 5), (-2, 9/2), (-1, 4), (1, 3), (2, 5/2), (3, 2), (4, 3/2), (5, 1)
line x - y - 4 = 0 (line segment CD): (-5, -9), (-4, -8), (-3, -7), (-2, -6), (-1, -5), (1, -3), (2, -2), (3, -1), (4, 0), (5, 1)
line 2x + y + 7 = 0 (line segment AD): (-5, 3), (-4, 1), (-3, -1), (-2, -3), (-1, -5), (1, -9), (2, -11), (3, -13), (4, -15), (5, -17)

2010-06-18T13:32:00.000-07:00

Algebra: Root of a Linear Equation In One Unknown (Fractional)

2010-06-03T08:23:00.000-07:00

Analytic Geometry: Graphs of Intersecting And Parallel Lines

The following illustrated problems are about:

intersecting lines (perpendicular and non-perpendicular)
parallel lines

All the calculated values are given so as to allow flexibility in presenting the problems in different ways.

1-1 Graphs of Perpendicular Bisector of a Line Segment

Find the point at which the line x - 2y + 3 = 0 bisects perpendicularly the line segment formed by the x-intercept and y-intercept of the line 2x + y - 4 = 0.

point of intersection (midpoint of line segment AB): (1,2)
slope of the line x - 2y + 3 = 0: m = 1/2
angle of inclination of line x - 2y + 3 = 0: 26.6°
points on the line x - 2y + 3 = 0: {(-5, -1), (-4, -1/2), (-2, 1/2), (-1, 1), (0, 3/2), (-3, 0), (1, 2), (2, 5/2), (3, 3), (4, 7/2), (5, 4)}
slope of the line 2x + y - 4 = 0: m = -2
angle of inclination of line 2x + y - 4 = 0: 116.6°
points on the line 2x + y - 4 = 0: {(-5, 14), (-4, 12), (-3, 10), (-2, 8), (-1, 6), (0, 4), (2, 0), (3, -2), (4, -4), (5, -6)}

Find the point at which the line x + 5y + 12 = 0 bisects perpendicularly the line segment formed by the x-intercept and y-intercept of the line 5x - y - 5 = 0.

point of intersection (midpoint of line segment AB): (1/2, -5/2)
slope of the line x + 5y + 12 = 0: m = -1/5
angle of inclination of line x + 5y + 12 = 0: 168.7°
points on the line x + 5y + 12 = 0: {(-5, -7/5), (-4, -8/5), (-3, -9/5), (-2, -2), (-1, -11/5), (0, -12/5), (-12, 0), (1, -13/5), (2, -14/5), (3, -3), (4, -16/5), (5, -17/5)}
slope of the line 5x - y - 5 = 0: m = 5
angle of inclination of line 5x - y - 5 = 0: 78.7°
points on the line 5x - y - 5 = 0: {(-5, -30), (-4, -25), (-3, -20), (-2, -15), (-1, -10), (0, -5), (1, 0), (2, 5), (3, 10), (4, 15), (5, 20)}

Find the point at which the line 5x + 3y = 0 bisects perpendicularly the line segment AB.

point of intersection (midpoint of line segment AB): (-3/2, 5/2)
slope of the line 5x + 3y = 0: m = -5/3
angle of inclination of line 5x + 3y = 0: 120.96°
points on the line 5x + 3y = 0: {(-5, 25/3), (-4, 20/3), (-3, 5), (-2, 10/3), (-1, 5/3), (0, 0), (1, -5/3), (2, -10/3), (3, -5), (4, -20/3), (5, -25/3)}
slope of the line segment AB: m = 3/5
equation of line segment AB: 3x - 5y + 17 = 0

1-2 Graphs of Perpendicular Lines

Find the point of intersection of lines 8x + 5y + 13 = 0 and 5x - 8y - 3 = 0.

point of intersection: (-1, -1)
slope of the line 8x + 5y + 13 = 0: m = -8/5
angle of inclination of line 8x + 5y + 13 = 0: 122.0°
points on the line 8x + 5y + 13 = 0: {(-5, 27/5), (-4, 19/5), (-3, 11/5), (-2, 3/5), (0, -13/5), (-13/8, 0), (1, -21/5) (2, -29/5), (3, -37/5), (4, -9), (5, -53/5)}
slope of the line 5x - 8y - 3 = 0: m = 5/8
angle of inclination of line 5x - 8y - 3 = 0: 32.0°
points on the line 5x - 8y - 3 = 0: {(-5, -7/2), (-4, -23/8), (-3, -9/4), (-2, -13/8), (0, -3/8), (3/5, 0), (1, 1/4), (2, 7/8), (3, 3/2), (4, 17/8), (5, 11/4))}

Find the point of intersection of lines x + 5y - 5 = 0 and 5x - y - 2 = 0.

point of intersection: (15/26, 23/26)
slope of the line x + 5y - 5 = 0: m = -1/5
angle of inclination of line x + 5y - 5 = 0: 168.7°
points on the line x + 5y - 5 = 0: {(-5, 2), (-4, 9/5), (-3, 8/5), (-2, 7/5), (-1, 6/5), (0, 1), (5, 0), (1, 4/5), (2, 3/5), (3, 2/5), (4, 1/5)}
slope of the line 5x - y - 2 = 0: m = 5
angle of inclination of line 5x - y - 2 = 0: 78.7°
points on the line 5x - y - 2 = 0: {(-5, -27), (-4, -22), (-3, -17), (-2, -12), (-1, -7), (0, -2), (2/5, 0), (1, 3), (2, 8), (3, 13), (4, 18), (5, 23)}

1-3 Graphs of Intersecting Non-Perpendicular Lines

Find the point of intersection of lines 5x + 6y - 8 = 0 and 2x + y + 1 = 0.

point of intersection: (-2, 3)
slope of the line 5x + 6y - 8 = 0: m = -5/6
angle of inclination of line 5x + 6y - 8 = 0: 140.2°
points on the line 5x + 6y - 8 = 0: {(-5, 11/2), (-4, 14/3), (-3, 23/6), (-1, 13/6), (0, 4/3), (8/5, 0), (1, 1/2), (2, -1/3), (3, -7/6), (4, -2), (5, -17/6)}
slope of the line 2x + y + 1 = 0: m = -2
angle of inclination of line 2x + y + 1 = 0: 116.6°
points on the line 2x + y + 1 = 0: {(-5, 9), (-4, 7), (-3, 5), (-1, 1), (0, -1), (-1/2, 0), (1, -3), (2, -5), (3, -7), (4, -9), (5, -11)}

Find the point of intersection of lines 5x + y + 16 = 0 and 3x - y + 8 = 0.

point of intersection: (-3, -1)
slope of the line 5x + y + 16 = 0: m = -5
angle of inclination of line 5x + y + 16 = 0: 101.3°
points on the line 5x + y + 16 = 0: {(-5, 9), (-4, 4), (-2, -6), (-1, -11), (0, -16), (-16/5, 0), (1, -21), (2, -26), (3, -31), (4, -36), (5, -41)}
slope of the line 3x - y + 8 = 0: m = 3
angle of inclination of line 3x - y + 8 = 0: 71.6°
points on the line 3x - y + 8 = 0: {(-5, -7), (-4, -4), (-2, 2), (-1, 5), (0, 8), (-8/3, 0), (1, 11), (2, 14), (3, 17), (4, 20), (5, 23)}

2 Graphs of Parallel Lines

Show that lines 3x + 5y - 4 = 0 and 3x + 5y + 15 = 0 are parallel.

points on the line 3x + 5y - 4 = 0: {(-5, 19/5), (-4, 16/5), (-3, 13/5), (-2, 2), (-1, 7/5), (0, 4/5), (4/3, 0), (1, 1/5), (2, -2/5), (3, -1), (4, -8/5), (5, -11/5)}
slope of the lines: m = -3/5
angle of inclination of the lines: 149.0°
points on the line 3x + 5y + 15 = 0: {(-5, 0), (-4, -3/5), (-3, -6/5), (-2, -9/5), (-1, -12/5), (0, -3), (1, -18/5), (2, -21/5), (3, -24/5), (4, -27/5), (5, -6)}

Show that lines 4x - 3y + 1 = 0 and 4x - 3y - 3 = 0 are parallel.

points on the line 4x - 3y + 1 = 0: {(-5, -19/3), (-4, -5), (-3, -11/3), (-2, -7/3), (-1, -1), (0, 1/3), (-1/4, 0), (1, 5/3), (2, 3), (3, 13/3), (4, 17/3), (5, 7)}
slope of the lines: m = 4/3
angle of inclination of the lines: 53.1°
points on the line 4x - 3y - 3 = 0: {(-5, -23/3), (-4, -19/3), (-3, -5), (-2, -11/3), (-1, -7/3), (0, -1), (3/4, 0), (1, 1/3), (2, 5/3), (3, 3), (4, 13/3), (5, 17/3)}

2010-05-25T08:38:00.000-07:00

Algebra: Root of a Linear Equation With One Unknown (Non-Fractional

1.	4x + 5 = 7x - 10
	Answer: x = 5

2.	3x - 4 = 8x + 41
	Answer: x = -9

3.	11x + 7 = 122 - 12x
	Answer: x = 5

4.	9x + 7 = 187 - 9x
	Answer: x = 10

5.	10x + 7 = 7x - 26
	Answer: x = -11

6.	x + 3 = 7x + 21
	Answer: x = -3

7.	8x + 7 = 37 - 2x
	Answer: x = 3

8.	3x + 11 = 9x - 13
	Answer: x = 4

9.	2x + 3 = 58 - 3x
	Answer: x = 11

10.	11x + 5 = 29 + 7x
	Answer: x = 6

11.	4x - 6 = 7 - 13x
	Answer: x = 13/17

12.	x - 4 = 5 - 5x
	Answer: x = 3/2

13.	14x + 5 = 5x + 9
	Answer: x = 4/9

14.	11x - 2 = 8x - 4
	Answer: x = -2/3

15.	6x + 2 = 7 - 14x
	Answer: x = 1/4

16.	4x - 13 = 14 - 11x
	Answer: x = 9/5

17.	4x - 3 = 4 - 10x
	Answer: x = 1/2

18.	3x + 10 = 15 - 12x
	Answer: x = 1/3

19.	5x + 7 = 10 - 10x
	Answer: x = 1/5

20.	11x - 1 = 2 - 13x
	Answer: x = 1/8

2010-05-09T14:03:00.000-07:00

Algebra: Graphing of Linear Equation Problems

2010-04-22T08:01:00.000-07:00

Algebra: The Polynomial Arithmetic

The following sets of problems are about:

Addition of Polynomials
Subtraction of Polynomials
Multiplication of Polynomials
Division of Polynomials

Addition of Polynomials

1.	9x - y - 2z and 7x + 5y - 8z
	Answer: 16x + 4y - 10z

2.	8m - 7n + 7 and 5m + 3n - 2
	Answer: 13m - 4n + 5

3.	5a + 3b + 9 and 2a - 9b + 7
	Answer: 7a - 6b + 16

4.	9a + 4b + 9c and 3a + 3b + 2c
	Answer: 12a + 7b + 11c

5.	6x - 7y - 6z, 9x + 8y - 7z and 9x - 5y - 7z
	Answer: 24x - 4y - 20z

6.	2x - 9y + 2z, 6x - 5y + z and 7x + 6y + 9
	Answer: 15x - 8y + 3z + 9

7.	5a + 6b - 3, 1a + b - 6c and 3a - 7b + 5
	Answer: 9a - 6c + 2

8.	6x - 7y - 6z, 9x + 8y - 7z and 9x - 5y - 7z
	Answer: 24x - 4y - 20z

9.	2x - 9y + 2z, 6x - 5y + z and 7x + 6y + 9
	Answer: 15x - 8y + 3z + 9

10.	3a + 5b + 9, 8a - 9b - 7c and 9a + b - 3
	Answer: 20a - 3b - 7c + 6

11.	6a² - 6ab - 12a and 7a² - 28ab - 7a
	Answer: 13a² - 34ab - 19a

12.	14a²b + 56ab² + 21ab and 42a²b - 7ab² - 7ab
	Answer: 56a²b + 49ab² + 14ab

13.	3a⁵ + 5a⁴b + 8a⁴ and 20a⁵ + 35a⁴b - 10a⁴
	Answer: 23a⁵ + 40a⁴b - 2a⁴

14.	35x² + 28xy + 21x and 20x² + 4xy - 28x
	Answer: 55x² + 32xy - 7x

15.	10x²y + 5xy² + 40xy and 49x²y + 14xy² + 7xy
	Answer: 59x²y + 19xy² + 47xy

16.	49x⁵ - 35x⁴y + 49x⁴ and 8x⁵ + 14x⁴y + 14x⁴
	Answer: 57x⁵ - 21x⁴y + 63x⁴

17.	4x⁴ + x³ + 9 and 2x⁴ + 7x³ + 3
	Answer: 6x⁴ + 8x³ + 12

18.	8m³ + m² - 6 and 7m³ + 8m² - 7
	Answer: 15m³ + 9m² - 13

19.	4b² + 9b + 5 and 2b² + 5b + 2
	Answer: 6b² + 14b + 7

20.	63x² - 9xy - 63x and 30x² - 40xy - 45x
	Answer: 93x² - 49xy - 108x

Subtraction of Polynomials

1.	9x + 2y + 3z and 7x + 6y + 8z
	Answer: 2x - 4y - 5z

2.	2m + 8n + 2 and m - 7n - 8
	Answer: m + 15n + 10

3.	8a + 2b - 1 and 7a - b - 3
	Answer: a + 3b + 2

4.	6a + 5b + 8c and 4a - 7b + 7c
	Answer: 2a + 12b + c

5.	6x - 3y + 8z, x - 3y - 5z and 2x + 7y - 2z
	Answer: 3x - 7y + 15z

6.	9x + 4y + 9z, 4x - 7y + z and x + 5y - 2
	Answer: 4x + 6y + 8z + 2

7.	9a + 3b - 5, 8a + 7b + 4c and a - b + 4
	Answer: - 3b - 4c - 9

8.	8x - 4y + 5z, 3x + 4y - 7z and 5x + 4y - z
	Answer: -12y + 13z

9.	9x - 5y - 9z, 2x + 2y - 3z and 5x + 6y - 5
	Answer: 2x - 13y - 6z + 5

10.	9a + 3b - 8, a + 5b + 2c and 7a - 8b + 3
	Answer: a + 6b - 2c - 11

11.	4a² - 20ab - 4a and a² - 8ab + 3a
	Answer: 3a² - 12ab - 7a

12.	63a²b + 36ab² - 45ab and 9a²b - ab² - 5ab
	Answer: 54a²b + 37ab² - 40ab

13.	24a⁵ - 21a⁴b - 24a⁴ and 20a⁵ + 35a⁴b - 45a⁴
	Answer: 4a⁵ - 56a⁴b + 21a⁴

14.	24x² + 24xy + 9x and 12x² + 24xy - 4x
	Answer: 12x² + 13x

15.	21x²y + 42xy² + 7xy and 9x²y + 9xy² + 63xy
	Answer: 12x²y + 33xy² - 56xy

16.	18x³ - 63x²y - 81x² and 6x³ - 5x²y - 8x²
	Answer: 12x³ - 58x²y - 73x²

17.	9x² + 2x + 7 and 4x² + 3x + 8
	Answer: 5x² - x - 1

18.	9m² - 7m + 6 and 8m² - 2m + 1
	Answer: m² - 5m + 5

19.	8b² - 3b + 1 and 7b² - b + 5
	Answer: b² - 2b - 4

20.	5x² - 7x + 8 and 4x² - 5x + 7
	Answer: x² - 2x + 1

Multiplication of Polynomials

1.	x - 3y and x - 7y
	Answer: x² - 10xy + 21y²

2.	4a - 7b and 4a - 7b
	Answer: 16a² - 56ab + 49b²

3.	6a + b and 6a + b
	Answer: 36a² + 12ab + b²

4.	8x + y and 9x + 4y
	Answer: 72x² + 41xy + 4y²

5.	2x - 7y and 2x + 7y
	Answer: 4x² - 49y²

6.	3a - 4b and 9a + 2b
	Answer: 27a² - 30ab - 8b²

7.	6mx + ny and 8mx - ny
	Answer: 48m²x² + 2mxny - n²y²

8.	7ax + by and 8ax + 7by
	Answer: 56a²x² + 57axby + 7b²y²

9.	(x - 5y - 6)²
	Answer: x² - 10xy + 25y² - 12x + 60y + 36

10.	(8p - 7q + 1)²
	Answer: 64p² - 112pq + 49q² + 16p - 14q + 1

11.	(8p - 7q + 4)²
	Answer: 64p² - 112pq + 49q² + 64p - 56q + 16

12.	(a - 5b - 6y)²
	Answer: a² - 10ab + 25b² - 12a + 60b + 36y²

13.	(5x⁴)(x + y)
	Answer: 5x⁵ + 5x⁴y

14.	(4x)(7x⁴ - 8y²)
	Answer: 28x⁵ - 32xy²

15.	(20xy)(3x⁴ - 8)
	Answer: 60x⁵y - 160xy

16.	(3ab⁴)(3a⁴ + 7)
	Answer: 9a⁵b⁴ + 21ab⁴

17.	(14s³t²)(3s³ - 7u² + 7)
	Answer: 42s⁶t² - 98s³t²u² + 98s³t²

18.	(s + t)(u³ + v⁴)
	Answer: su³ + sv⁴ + tu³ + tv⁴

19.	(w + 4x)(5y + z)
	Answer: 5wy + wz + 20xy + 4xz

20.	(5s + 7t)(7u - 4v)
	Answer: 35su - 20sv + 49tu - 28tv

21.	(3w³ + 8x⁴)(2y² + z³)
	Answer: 6w³y² + 3w³z³ + 16x⁴y² + 8x⁴z³

22.	(3w + 4x²)(4y² - 3z⁴)
	Answer: 12wy² - 9wz⁴ + 16x²y² - 12x²z⁴

23.	(7a + 9b)(2b + 3)
	Answer: 14ab + 21a + 18b² + 27b

24.	(x + 9y)(7y - 4)
	Answer: 7xy - 4x + 63y² - 36y

25.	(8s³ + t)(4t⁴ - 7)
	Answer: 32s³t⁴ - 56s³ + 4t⁵ - 7t

26.	(s - t)(u + 8v)
	Answer: su + 8sv - tu - 8tv

27.	(x + 6)(x + 7)
	Answer: x² + 13x + 42

28.	(x - 7)(x + 4)
	Answer: x² - 3x - 28

29.	(x - 9)(x - 1)
	Answer: x² - 10x + 9

30.	(8x + 1)(3x + 2)
	Answer: 24x² + 19x + 2

31.	(x - 5)(4x + 3)
	Answer: 4x² - 17x - 15

32.	(9x - 1)(5x - 2)
	Answer: 45x² - 23x + 2

33.	(x + 9)(x + 9)
	Answer: x² + 18x + 81

34.	(3x + 5)(3x + 5)
	Answer: 9x² + 30x + 25

35.	(x - 1)(x - 1)
	Answer: x² - 2x + 1

36.	(9x - 1)(9x - 1)
	Answer: 81x² - 18x + 1

37.	[ 9x + 8(y - z) ] [ (9x - 8(y - z) ]
	Answer: 81x² - 64(y - z)²

38.	(x + 2y) (x - 2y) (x² + 4y²)
	Answer: x⁴ - 16y⁴

39.	(x - 8) (x² + 8x + 64)
	Answer: x³ - 512

40.	(x² - 7) (x⁴ + 7x² + 49)
	Answer: x⁶ - 343

41.	(7y - 8) (49y² + 56y + 64)
	Answer: 343y³ - 512

42.	(2y² - 9) (4y⁴ + 18y² + 81)
	Answer: 8y⁶ - 729

43.	(3a - 4b) (9a² + 12ab + 16b²)
	Answer: 27a³ - 64b³

44.	(3a² - 7b) (9a⁴ + 21a²b + 49b²)
	Answer: 27a⁶ - 343b³

45.	(9xy² - 4z) (81x²y⁴ + 36xy²z + 16z²)
	Answer: 729x³y⁶ - 64z³

46.	x (2s + 9t)(4s² - 18st + 81t²)
	Answer: 8s³x + 729t³x

47.	x (8s² + 2t²)(64s⁴ - 16s²t² + 4t⁴)
	Answer: 512s⁶x + 8t⁶x

48.	(8x + 5y) (64x² - 40xy + 25y²)
	Answer: 512x³ + 125y³

49.	(7x² + 9y²) (49x⁴ - 63x²y² + 81y⁴)
	Answer: 343x⁶ + 729y⁶

50.	4x⁹ (y + 7z) (y - 7z)
	Answer: 4x⁹y² - 196x⁹z²

Division of Polynomials

1.	63a⁴b⁷x⁹ ÷ 9a³b⁵x⁴
	Answer: 7ab²x⁵

2.	9x⁷y⁶z⁴ ÷ x²y³z³
	Answer: 9x⁵y³z

3.	-14m⁴n⁷ ÷ 2m³n⁴
	Answer: -7mn³

4.	72x²y² ÷ 8xy
	Answer: 9xy

5.	24a⁶b⁵c⁷ ÷ 3a²b⁴c³
	Answer: 8a⁴bc⁴

6.	28x⁵y²z + 63x³y³z + 49x²y³z ÷ 7x²y²z
	Answer: 4x³ + 9xy + 7y

7.	x² - 8x - 209 ÷ x - 19
	Answer: x + 11

8.	x² - 29x + 204 ÷ x - 9
	Answer: x - 20; remainder = 24

9.	x² - 22x + 88 ÷ x - 4
	Answer: x - 18; remainder = 16

10.	x⁴ + 3x² - 460 ÷ x² + 23
	Answer: x² - 20

11.	x⁴ - 21x² + 130 ÷ x² - 10
	Answer: x² - 11; remainder = 20

12.	x⁴ - 6x³y - 25x²y² + 52xy³ + 14y⁴ ÷ x² + 2xy - 7y²
	Answer: x² - 8xy - 2y²

13.	x⁴ - 8x³y + 7x²y² + 34xy³ + 15y⁴ ÷ x² - 5xy - 3y²
	Answer: x² - 3xy - 5y²

14.	8x³ - 57x² + 48x + 42 ÷ x - 6
	Answer: 8x² - 9x - 6; remainder = 6

15.	56x⁵ - 111x⁴ + 94x³ - 70x² + 27x - 47 ÷ 7x³ - 6x² - 2x - 5
	Answer: 8x² - 9x + 8; remainder = -2x - 7

16.	6a⁵x⁴y⁹ ÷ a³x⁶y⁶
	Answer: 6a²x^-2y³

17.	56p⁹q²r⁴ ÷ 8p³q⁶r⁵
	Answer: 7p⁶q^-4r^-1

18.	63x⁵ + 88x⁴ + 41x³ - 63x - 63x² + 9 ÷ 7x³ + 9x² + 9x - 1
	Answer: 9x² + x - 7; remainder = x + 2

19.	32x³ + 44x² + 25x + 21 ÷ 8x + 9
	Answer: 4x² + x + 2; remainder = 3

20.	x⁴ + 11x³y + 25x²y² - 7xy³ - 6y⁴ ÷ x² + 8xy + 3y²
	Answer: x² + 3xy - 2y²

2010-03-19T06:51:00.000-07:00

Algebra: Factoring of Fractional Polynomials

The following problems consist of three sets as described below.

Set 1 problems are reduction of fractions to their lowest terms.
Set 2 problems are multiplication of fractions.
Set 3 problems are division of fractions.

Most of the expressions are either products of a monomial and a binomial or products of two binomials.

Set 1

Reduction of fractional polynomials to their lowest terms.

Set 2

Multiplication of fractional polynomials.

Set 3

Division of fractional polynomials.

2010-02-10T07:20:00.000-08:00

Algebra: Factoring of Polynomials

For convenience and quick browsing, the problems are grouped into polynomials that are:

product of a monomial and a polynomial
grouped into terms containing a common factor
difference of two squares
difference of two cubes
sum of two cubes
product of two binomials

The last group of polynomials given above consists of polynomials that are:

product of two binomials having a common factor
product of any two binomials
square of a binomial sum
square of a binomial difference

Some of these grouped polynomials contain factors which are a combination of a monomial, a binomial, and a trinomial. Polynomials that are:

difference of two squares can be the products of two binomials, or products of two trinomials, or products of a monomial and two binomials.
sum of two cubes can be products of a binomial and a trinomial, or products of a monomial, a binomial, and a trinomial

Polynomials With Monomial As One of Its Factors

1.	15x⁴ + 15x³y
	Answer: (3x³)(5x + 5y)

2.	4x³ - 4x²y⁴
	Answer: (4x²)(x - y⁴)

3.	5x⁵ - 5x³y²
	Answer: (x³)(5x² - 5y²)

4.	63s⁶t³ - 42s³t³u + 126s³t³
	Answer: (21s³t³)(3s³ - 2u + 6)

5.	14x⁵ + 6x²y
	Answer: (2x²)(7x³ + 3y)

6.	18x⁶y³ - 15x²y³z² + 3x²y³
	Answer: (3x²y³)(6x⁴ - 5z² + 1)

7.	14a⁴b⁴ + 56a³b⁴
	Answer: (14a³b⁴)(a + 4)

8.	7x⁵ + xy²
	Answer: (x)(7x⁴ + y²)

9.	7x⁴ + 21x³y³
	Answer: (7x³)(x + 3y³)

10.	24x⁶y⁴ - 192x³y⁴
	Answer: (24x³y⁴)(x³ - 8)

11.	35x³ - 49xy²
	Answer: (7x)(5x² - 7y²)

12.	48a⁴b² + 12a³b²
	Answer: (12a³b²)(4a + 1)

13.	56x⁴ - 40xy⁴
	Answer: (8x)(7x³ - 5y⁴)

14.	20a³b² + 60ab²
	Answer: (20ab²)(a² + 3)

15.	120s⁵t³ - 75s²t³u² + 30s²t³
	Answer: (15s²t³)(8s³ - 5u² + 2)

16.	24x⁴y - 6x³y
	Answer: (6x³y)(4x - 1)

17.	105x⁵y - 35x³y
	Answer: (35x³y)(3x² - 1)

18.	30x⁵ - 15x²y³
	Answer: (5x²)(6x³ - 3y³)

19.	6x⁴ + 8x³y⁴
	Answer: (x³)(6x + 8y⁴)

20.	20x⁴y³ - 40xy³
	Answer: (20xy³)(x³ - 2)

Polynomials That Can Be Grouped Into Terms Having A Common Factor

1.	su⁴ + sv + tu⁴ + tv
	Answer: (s + t)(u⁴ + v)

2.	wy + 3wz + xy + 3xz
	Answer: (w + x)(y + 3z)

3.	24su - 42sv + 20tu - 35tv
	Answer: (6s + 5t)(4u - 7v)

4.	20w²y⁴ + 35w²z⁴ + 8xy⁴ + 14xz⁴
	Answer: (5w² + 2x)(4y⁴ + 7z⁴)

5.	64w⁴y³ - 40w⁴z² + 8xy³ - 5xz²
	Answer: (8w⁴ + x)(8y³ - 5z²)

6.	3ab + 3a + b² + b
	Answer: (3a + b)(b + 1)

7.	5xy - 20x + 2y² - 8y
	Answer: (5x + 2y)(y - 4)

8.	28s³t - 35s³ + 24t² - 30t
	Answer: (7s³ + 6t)(4t - 5)

9.	7su + sv - 7tu - tv
	Answer: (s - t)(7u + v)

10.	su³ + sv⁴ + tu³ + tv⁴
	Answer: (s + t)(u³ + v⁴)

11.	8wy + 32wz + 3xy + 12xz
	Answer: (8w + 3x)(y + 4z)

12.	6su - 16sv + 3tu - 8tv
	Answer: (2s + t)(3u - 8v)

13.	12w²y + 15w²z² + 8x³y + 10x³z²
	Answer: (3w² + 2x³)(4y + 5z²)

14.	3w³y⁴ - 8w³z³ + 15x⁴y⁴ - 40x⁴z³
	Answer: (w³ + 5x⁴)(3y⁴ - 8z³)

15.	4ab + 2a + 10b² + 5b
	Answer: (2a + 5b)(2b + 1)

16.	28xy - 4x + 7y² - y
	Answer: (4x + y)(7y - 1)

17.	24st² - 15s + 64t⁴ - 40t²
	Answer: (3s + 8t²)(8t² - 5)

18.	5su + 2sv - 5tu - 2tv
	Answer: (s - t)(5u + 2v)

19.	12ab + 42a + 14b² + 49b
	Answer: (6a + 7b)(2b + 7)

20.	6xy - 10x + 9y² - 15y
	Answer: (2x + 3y)(3y - 5)

Polynomials That Are The Difference of Two Squares

You will find that some of the problems given below can be factored twice, that is, after factoring an expression one of the resulting factors is still factorable.

Also, in some problems a group of terms is treated as a single term.

1.	x² - 25
	Answer: (x - 5)(x + 5)

2.	36x² - 25
	Answer: (6x - 5)(6x + 5)

3.	25x² - 49
	Answer: (5x - 7)(5x + 7)

4.	x² - 9
	Answer: (x - 3)(x + 3)

5.	16x² - 25
	Answer: (4x - 5)(4x + 5)

6.	x² - (y - z)²
	Answer: [ x + (y - z) ] [ (x - (y - z) ]

7.	(x + 8)² - 9
	Answer: [ (x + 8) + 3 ] [ (x + 8) - 3 ]

8.	(x + 3)² - 1
	Answer: [ (x + 3) + 1 ] [ (x + 3) - 1 ]

9.	49x² - 9(y - z)²
	Answer: [ 7x + 3(y - z) ] [ (7x - 3(y - z) ]

10.	(x + 5)² - 9
	Answer: [ (x + 5) + 3 ] [ (x + 5) - 3 ]

11.	(x + 1)² - 36
	Answer: [ (x + 1) + 6 ] [ (x + 1) - 6 ]

12.	(x + 8)² - 25
	Answer: [ (x + 8) + 5 ] [ (x + 8) - 5 ]

13.	16x² - 49(y - z)²
	Answer: [ 4x + 7(y - z) ] [ (4x - 7(y - z) ]

14.	49x⁴y² - 49x⁴z²
	Answer: 49x⁴ (y + z) (y - z)

15.	49x²y² - 441x²z²
	Answer: 49x² (y + 3z) (y - 3z)

16.	9x²y² - 36x²z²
	Answer: 9x² (y + 2z) (y - 2z)

17.	16x²y² - 144x²z²
	Answer: 16x² (y + 3z) (y - 3z)

18.	4x²y² - 100x²z²
	Answer: 4x² (y + 5z) (y - 5z)

19.	x⁴ - 256y⁴
	Answer: (x + 4y) (x - 4y) (x² + 16y²)

20.	x⁴ - 81y⁴
	Answer: (x + 3y) (x - 3y) (x² + 9y²)

Polynomials That Are The Difference of Two Cubes

1.	27x³y⁶ - 125z³
	Answer: (3xy² - 5z) (9x²y⁴ + 15xy²z + 25z²)

2.	343y⁶ - 512
	Answer: (7y² - 8) (49y⁴ + 56y² + 64)

3.	8a⁶ - 125b³
	Answer: (2a² - 5b) (4a⁴ + 10a²b + 25b²)

4.	x³ - 8
	Answer: (x - 2) (x² + 2x + 4)

5.	8a³ - 512b³
	Answer: (2a - 8b) (4a² + 16ab + 64b²)

6.	343x³y⁶ - 64z³
	Answer: (7xy² - 4z) (49x²y⁴ + 28xy²z + 16z²)

7.	512y³ - 125
	Answer: (8y - 5) (64y² + 40y + 25)

8.	27a⁶ - 125b³
	Answer: (3a² - 5b) (9a⁴ + 15a²b + 25b²)

9.	8a³ - 125b³
	Answer: (2a - 5b) (4a² + 10ab + 25b²)

10.	125y⁶ - 216
	Answer: (5y² - 6) (25y⁴ + 30y² + 36)

11.	343y³ - 216
	Answer: (7y - 6) (49y² + 42y + 36)

12.	x⁶ - 8
	Answer: (x² - 2) (x⁴ + 2x² + 4)

13.	8y³ - 125
	Answer: (2y - 5) (4y² + 10y + 25)

14.	8y³ - 216
	Answer: (2y - 6) (4y² + 12y + 36)

15.	343y⁶ - 27
	Answer: (7y² - 3) (49y⁴ + 21y² + 9)

16.	x⁶ - 64
	Answer: (x² - 4) (x⁴ + 4x² + 16)

17.	x⁶ - 27
	Answer: (x² - 3) (x⁴ + 3x² + 9)

18.	x³ - 27
	Answer: (x - 3) (x² + 3x + 9)

19.	8y⁶ - 343
	Answer: (2y² - 7) (4y⁴ + 14y² + 49)

20.	27y⁶ - 27
	Answer: (3y² - 3) (9y⁴ + 9y² + 9)

21.	512a³ - 8b³
	Answer: (8a - 2b) (64a² + 16ab + 4b²)

22.	64a⁶ - 343b³
	Answer: (4a² - 7b) (16a⁴ + 28a²b + 49b²)

23.	8a³ - 8b³
	Answer: (2a - 2b) (4a² + 4ab + 4b²)

24.	x³ - 125
	Answer: (x - 5) (x² + 5x + 25)

25.	512a⁶ - 8b³
	Answer: (8a² - 2b) (64a⁴ + 16a²b + 4b²)

Polynomials That Are The Sum of Two Cubes

1.	512x⁶ + 27y⁶
	Answer: (8x² + 3y²) (64x⁴ - 24x²y² + 9y⁴)

2.	8x⁶ + 27y⁶
	Answer: (2x² + 3y²) (4x⁴ - 6x²y² + 9y⁴)

3.	8a⁶ + 125b⁶
	Answer: (2a² + 5b²) (4a⁴ - 10a²b² + 25b⁴)

4.	512x³ + 8y³
	Answer: (8x + 2y) (64x² - 16xy + 4y²)

5.	8x⁶ + 64y⁶
	Answer: (2x² + 4y²) (4x⁴ - 8x²y² + 16y⁴)

6.	8m³ + 343n³
	Answer: (2m + 7n) (4m² - 14mn + 49n²)

7.	64s³x + 8t³x
	Answer: x (4s + 2t)(16s² - 8st + 4t²)

8.	8x³ + 216y³
	Answer: (2x + 6y) (4x² - 12xy + 36y²)

9.	125x⁶y + 343y
	Answer: y (5x² + 7)(25x⁴ - 35x² + 49)

10.	125x³ + 8y³
	Answer: (5x + 2y) (25x² - 10xy + 4y²)

11.	512x⁶ + 343y⁶
	Answer: (8x² + 7y²) (64x⁴ - 56x²y² + 49y⁴)

12.	8x³ + 125y³
	Answer: (2x + 5y) (4x² - 10xy + 25y²)

13.	125a⁶x + 64b⁶x
	Answer: x (5a² + 4b²)(25a⁴ - 20a²b² + 16b⁴)

14.	64m³y + 27n³y
	Answer: y (4m + 3n)(16m² - 12mn + 9n²)

15.	64s³x + 125t³x
	Answer: x (4s + 5t)(16s² - 20st + 25t²)

16.	216x³z + 343y³z
	Answer: z (6x + 7y)(36x² - 42xy + 49y²)

17.	8s⁶x + 125t⁶x
	Answer: x (2s² + 5t²)(4s⁴ - 10s²t² + 25t⁴)

18.	343a³c + 64b³c
	Answer: c (7a + 4b)(49a² - 28ab + 16b²)

19.	512s⁶x + 343t⁶x
	Answer: x (8s² + 7t²)(64s⁴ - 56s²t² + 49t⁴)

20.	8s⁶u + 27t⁶u
	Answer: u (2s² + 3t²)(4s⁴ - 6s²t² + 9t⁴)

The following problems are about polynomials which are the products of two binomials.

Polynomials That Are The Products of Two Binomials Having a Common Factor

1.	x² + 11x + 30
	Answer: (x + 5)(x + 6)

2.	x² - 11x + 24
	Answer: (x - 3)(x - 8)

3.	x² + 8x + 15
	Answer: (x + 3)(x + 5)

4.	49x² - 35x + 4
	Answer: (7x - 1)(7x - 4)

5.	x² - 9x + 14
	Answer: (x - 2)(x - 7)

6.	x² + 5x + 4
	Answer: (x + 1)(x + 4)

7.	x² - x - 30
	Answer: (x - 6)(x + 5)

8.	x² + 7x + 10
	Answer: (x + 5)(x + 2)

9.	25x² + 65x + 42
	Answer: (5x + 6)(5x + 7)

10.	x² + x - 56
	Answer: (x - 7)(x + 8)

11.	x² - 9x + 8
	Answer: (x - 1)(x - 8)

12.	x² + 2x - 15
	Answer: (x - 3)(x + 5)

13.	x² + 13x + 42
	Answer: (x + 6)(x + 7)

14.	x² + 11x + 24
	Answer: (x + 8)(x + 3)

15.	x² - 11x + 30
	Answer: (x - 5)(x - 6)

16.	x² - 13x + 40
	Answer: (x - 5)(x - 8)

17.	x² + x - 30
	Answer: (x - 5)(x + 6)

18.	x² + 3x + 2
	Answer: (x + 2)(x + 1)

19.	x² - 2x - 3
	Answer: (x - 3)(x + 1)

20.	x² - 8x + 15
	Answer: (x - 5)(x - 3)

Polynomials That Are The Products of Any Two Binomials

1.	24x² + 35x - 49
	Answer: (8x - 7)(3x + 7)

2.	2x² + 7x - 4
	Answer: (2x - 1)(x + 4)

3.	24x² + 34x + 7
	Answer: (6x + 7)(4x + 1)

4.	32x² + 28x + 3
	Answer: (8x + 1)(4x + 3)

5.	33x² + 14x + 1
	Answer: (3x + 1)(11x + 1)

6.	12x² + 11x - 5
	Answer: (3x - 1)(4x + 5)

7.	5x² + 33x - 56
	Answer: (5x - 7)(x + 8)

8.	8x² - 9x + 1
	Answer: (x - 1)(8x - 1)

9.	3x² + 19x + 6
	Answer: (x + 6)(3x + 1)

10.	6x² - 25x + 14
	Answer: (3x - 2)(2x - 7)

11.	35x² + 17x - 30
	Answer: (7x - 5)(5x + 6)

12.	7x² + 38x + 15
	Answer: (7x + 3)(x + 5)

13.	3x² + 32x + 64
	Answer: (x + 8)(3x + 8)

14.	3x² - 13x + 12
	Answer: (x - 3)(3x - 4)

15.	10x² + 21x + 9
	Answer: (2x + 3)(5x + 3)

16.	6x² - 23x + 20
	Answer: (3x - 4)(2x - 5)

17.	10x² - 47x + 42
	Answer: (2x - 7)(5x - 6)

18.	6x² - 43x + 42
	Answer: (6x - 7)(x - 6)

19.	15x² + 4x - 32
	Answer: (3x - 4)(5x + 8)

20.	3x² - 14x - 5
	Answer: (x - 5)(3x + 1)

Polynomials That Are The Square of a Binomial Sum

1.	169x² + 364x + 196
	Answer: (13x + 14)(13x + 14)

2.	64x² + 80x + 25
	Answer: (8x + 5)(8x + 5)

3.	25x² + 40x + 16
	Answer: (5x + 4)(5x + 4)

4.	x² + 10x + 25
	Answer: (x + 5)(x + 5)

5.	64x² + 48x + 9
	Answer: (8x + 3)(8x + 3)

6.	25x² + 60x + 36
	Answer: (5x + 6)(5x + 6)

7.	16x² + 8x + 1
	Answer: (4x + 1)(4x + 1)

8.	x² + 22x + 121
	Answer: (x + 11)(x + 11)

9.	169x² + 78x + 9
	Answer: (13x + 3)(13x + 3)

10.	x² + 4x + 4
	Answer: (x + 2)(x + 2)

11.	49x² + 56x + 16
	Answer: (7x + 4)(7x + 4)

12.	x² + 2x + 1
	Answer: (x + 1)(x + 1)

13.	x² + 14x + 49
	Answer: (x + 7)(x + 7)

14.	169x² + 156x + 36
	Answer: (13x + 6)(13x + 6)

15.	x² + 16x + 64
	Answer: (x + 8)(x + 8)

16.	x² + 26x + 169
	Answer: (x + 13)(x + 13)

17.	64x² + 16x + 1
	Answer: (8x + 1)(8x + 1)

18.	x² + 8x + 16
	Answer: (x + 4)(x + 4)

19.	25x² + 110x + 121
	Answer: (5x + 11)(5x + 11)

20.	49x² + 42x + 9
	Answer: (7x + 3)(7x + 3)

Polynomials That Are The Square of a Binomial Difference

1.	9x² - 30x + 25
	Answer: (3x - 5)(3x - 5)

2.	169x² - 364x + 196
	Answer: (13x - 14)(13x - 14)

3.	25x² - 40x + 16
	Answer: (5x - 4)(5x - 4)

4.	49x² - 42x + 9
	Answer: (7x - 3)(7x - 3)

5.	9x² - 60x + 100
	Answer: (3x - 10)(3x - 10)

6.	64x² - 80x + 25
	Answer: (8x - 5)(8x - 5)

7.	36x² - 84x + 49
	Answer: (6x - 7)(6x - 7)

8.	x² - 26x + 169
	Answer: (x - 13)(x - 13)

9.	9x² - 12x + 4
	Answer: (3x - 2)(3x - 2)

10.	x² - 2x + 1
	Answer: (x - 1)(x - 1)

11.	x² - 14x + 49
	Answer: (x - 7)(x - 7)

12.	x² - 4x + 4
	Answer: (x - 2)(x - 2)

13.	4x² - 20x + 25
	Answer: (2x - 5)(2x - 5)

14.	x² - 8x + 16
	Answer: (x - 4)(x - 4)

15.	25x² - 80x + 64
	Answer: (5x - 8)(5x - 8)

16.	x² - 16x + 64
	Answer: (x - 8)(x - 8)

17.	x² - 6x + 9
	Answer: (x - 3)(x - 3)

18.	64x² - 16x + 1
	Answer: (8x - 1)(8x - 1)

19.	x² - 10x + 25
	Answer: (x - 5)(x - 5)

20.	49x² - 70x + 25
	Answer: (7x - 5)(7x - 5)

2009-12-11T00:23:00.000-08:00

Dissociation of Weak Acid and Base Problems

Some problems are solved using both approximation and the quadratic equation. Values in the parenthesis are obtained using the quadratic equation.

Approximation is made by assuming that the amount of dissociated weak acid or base is negligible such that the given amount of undissociated acid or base remains practically the same.

When approximation introduces large % error, the quadratic equation alone is used as indicated in the Answer part of the problem.

For convenience, problems are grouped into aqueous solutions containing:

a weak acid or a weak base

a strong acid or a strong base

a salt of a weak acid or a salt of a weak base

a weak acid and its salt (weak acid/conjugate base composition) or a weak base and its salt (weak base/conjugate acid composition)

For a more detailed discussion of this topic and other chemistry topics, go to www.chemistrypartner.blogspot.com.

Aqueous Solution of a Weak Acid or a Weak Base

The pH of an aqueous solution containing a weak acid or a weak base is calculated from their dissociation constant equations shown below:

1.	If a solution of HCOOH has a concentration of 0.49 M, what is its pH? K_a for formic acid is 1.8 x 10^-4.
	Answer: pH = 2.03 ( 2.03 )

2.	What is the pH of an aqueous solution of CH₃(CH₂)₂CH₂NH₂ having a concentration of 0.04 M? The dissociation constant of 1-butylamine is 4 x 10^-4.
	Answer: pH = 11.58 (Quadratic equation is used.)

3.	What is the pH of a 0.85 M HNO₂ solution. ( K_a = 4.5 x 10^-4 )
	Answer: pH = 1.71; ( 1.71)

4.	What is the pH of an aqueous solution of (CH₃)₂NH having a concentration of 0.08 M? The dissociation constant of dimethylamine is 5.9 x 10^-4.
	Answer: pH = 11.84 ( 11.82)

5.	What is the pH of a 0.13 M HC₂H₂ClO₂ solution. ( K_a = 1.36 x 10^-3 )
	Answer: pH = 1.9 (Quadratic equation is used.)

Aqueous Solution of a Strong Acid or a Strong Base

The [H₃O⁺] or pH of an aqueous solution containing a strong acid or base is calculated using the ion-product constant of water, K_w.

K_w = [H₃O⁺] [OH^-]

1 x 10^-14 = [H₃O⁺] [OH^-]

6.	Find the [H₃O⁺] and [OH^-] of a 0.48 M of HCN solution. K_a = 6.2 x 10^-10.
	Answer: [H₃O⁺] = 1.73 x 10^-5 M; [OH^-] = 5.8 x 10^-10 M ( 1.73 x 10^-5; 5.8 x 10^-10 )

7.	A one liter aqueous solution contains 3.37 g KOH. What is the pH of this solution? The dissociation constant for water, K_w, is 1 x 10^-14. ( KOH = 56.1056 g )
	Answer: pH = 12.78

8.	A one liter aqueous solution contains 6 g NaOH. What is the pH of this solution? The dissociation constant for water, K_w, is 1 x 10^-14. ( NaOH = 39.99707 g )
	Answer: pH = 13.18

Aqueous Solution of a Salt of a Weak Acid or a Salt of a Weak Base

The general relationship of the conjugate acid/base pairs is used to solve problems 9 to 14.

This relationship is expressed as follows:

K_aK_b = K_w

K_aK_b = 1 x 10^-14

9.	Calculate the pH of a 0.71 M NaC₆H₅COO solution. (K_a for benzoic acid is 6.3 x 10^-5; K_w = 1 x 10^-14)
	Answer: pH = 9.03 ( 9.03 )

10.	What is the pH of a solution of 0.36 M NaF? Calculate the percent hydrolysis. (K_a = 6.7 x 10^-4; K_w = 1 x 10^-14)
	Answer: pH = 8.37; 6.44 x 10^-4 % ( 8.37; 6.44 x 10^-4 % )

11.	Calculate the dissociation constant of the conjugate base of HC₂H₂ClO₂. K_a = 1.36 x 10^-3.
	Answer: 7.35 x 10^-12

12.	What is the pH of an aqueous solution of 0.1 M CH₃CH₂NH₃Cl? ( Given: K_b = 4.28 x 10^-4; K_w = 1 x 10^-14)
	Answer: pH = 5.82 ( 5.82 )

13.	What is the dissociation constant of the conjugate acid of NH₃. K_b = 1.8 x 10^-5.
	Answer: 5.56 x 10^-10

14.	Find the pH of a 0.49 M NaCH₃COO solution. (K_a for acetic acid is 1.8 x 10^-5; K_w = 1 x 10^-14)
	Answer: pH = 9.22 ( 9.22 )

Aqueous Solution of a Weak Acid And Its Salt or a Weak Base And Its Salt

Calculation of [H₃O⁺] or pH of a buffer solution is accomplished using the dissociation constant equation of a weak acid or a weak base.

For an aqueous solution containing a weak acid and its salt:

Inverting the last term of the equation gives us: The last equation is the general form of the Henderson-Hasselbach equation.

15.	270 mL of 0.03 M NaOH solution is added to 270 mL of 0.04 M HF solution. Find the pH of the resulting solution. Given: K for hydrofluoric acid is 6.7 x 10^-4.
	Answer: pH = 3.65 ( 3.68)

16.	If 3 L of 0.02 M CH₃NH₂ solution has a pH of 10.41, find the weight of CH₃NH₃Cl dissolved in the solution. ( K_b for methylamine is 4.8 x 10^-4; mol. wt._methylamine = 67.5181 g/mol )
	Answer: 7.57 g

17.	What is the pH of an aqueous solution having a concentration of 0.03 M HF and 0.02 M NaF? ( K_a = 6.7 x 10^-4)
	Answer: pH = 3.00 ( 3.03 )

18.	An aqueous solution of 0.11 M HNO₂ and 0.11 M NaNO₂ has a volume of 530 mL. If 1.2 g of NaOH is added to this solution, find the change in pH. Assume no change in the volume of solution. ( K_a = 4.5 x 10^-4; NaOH = 39.99707 g )
	Answer: pH = 3.84 ( 3.84)

19.	An aqueous solution has a strength of 0.07 M HF and 0.04 M NaF. Calculate the pH of this solution. ( K_a = 6.7 x 10^-4)
	Answer: pH = 2.93 ( 2.95 )

20.	In a buffer solution of 0.07 M C₆H₅COOH and 0.07 M NaC₆H₅COOH, HCl is added to it such that its calculated concentration in the solution is 0.06 M. Assuming a constant volume of solution, find the change in pH of the solution. ( K_a = 6.3 x 10^-5)
	Answer: pH = 3.09 ( 3.12 )

a	b	c
3	4	5
6	8	10
5	12	13
9	12	15
8	15	17
12	16	20
7	24	25
15	20	25
24	7	25
10	24	26
20	21	29
18	24	30
16	30	34
21	28	35
12	35	37
15	36	39
24	32	40
9	40	41
27	36	45
14	48	50
30	40	50
24	45	51
20	48	52
28	45	53
33	44	55
40	42	58
36	48	60
60	11	61
56	33	65
60	25	65
60	32	68
56	42	70
55	48	73
60	45	75

a	b	c
3	4	5
6	8	10
5	12	13
9	12	15
8	15	17
12	16	20
7	24	25
15	20	25
24	7	25
10	24	26
20	21	29
18	24	30
16	30	34
21	28	35
12	35	37
15	36	39
24	32	40
9	40	41
27	36	45
14	48	50
30	40	50
24	45	51
20	48	52
28	45	53
33	44	55
40	42	58
36	48	60
60	11	61
56	33	65
60	25	65
60	32	68
56	42	70
55	48	73
60	45	75

a	b	c
3	4	5
6	8	10
5	12	13
9	12	15
8	15	17
12	16	20
7	24	25
15	20	25
24	7	25
10	24	26
20	21	29
18	24	30
16	30	34
21	28	35
12	35	37
15	36	39
24	32	40
9	40	41
27	36	45
14	48	50
30	40	50
24	45	51
20	48	52
28	45	53
33	44	55
40	42	58
36	48	60
60	11	61
56	33	65
60	25	65
60	32	68
56	42	70
55	48	73
60	45	75