Pythagorean Theorem and Right Triangles

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:

• Coordinate Graphs of Quadrilaterals With Their Diagonals And Median
• Coordinate Graphs of Quadrilaterals

The Exterior Angle Bisectors and the Excenters of a Triangle

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

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

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

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

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

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

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

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