principles of mathematics 11 section 4, introduction...

Module 3, Section 4

Analytic Geometry II

IntroductionAs the lesson titles show, this section extends what you havelearned about Analytic Geometry to several related topics, suchas applications and the graphing of systems of “inequations”(inequalities). The last lesson shows how Analytic Geometry hasa lot in common with classical geometry and it’s Theorems.

Section 4 — Outline

Lesson 1 Applications of Systems

Lesson 2 Non-Linear Systems

Lesson 3 Graphing Linear Inequalities

Lesson 4 Quadratic, Rational, and Absolute Value Inequalities in Two Variables

Lesson 5 Verifying and Proving Assertions in Coordinate Geometry

Review

Principles of Mathematics 11 Section 4, Introduction 201

Module 3

Notes

202 Section 4, Introduction Principles of Mathematics 11

Module 3

Module 3

Lesson 1

Applications of Systems

Outcome

When you complete this lesson, you will be able to

• solve systems of linear equations and apply them in problem-solving situations

Overview

We will be using the techniques of solving systems of equationsand applying them to problem-solving situations.

Example 1An airplane took 4 hours to fly 1920 km when it had a tail wind.Flying back against the same wind, and travelling at the sameair speed, the plane took 1 hour longer. Find the speed of thewind and the plane’s air speed.

Solution

Let x = air speed of the airplaney = speed of the wind

distance = rate x time

4(x + y) = 1920 (1)

5(x – y) = 1920 (2)

Note: Flying with the wind increases the speed and flyingagainst the wind decreases the speed.

Equation (1) ÷ 4: x + y = 480 (3)

Equation (2) ÷ 5: x – y = 384 (4)

Equation (1) + Equation (2): 2x = 864

x = 432 km/hr

Substitute x = 432 into Equation 3: 432 + y = 480

y = 48 km/hr

The air speed of the airplane is 432 km/hr.

The speed of the wind is 48 km/hr.

Principles of Mathematics 11 Section 4, Lesson 1 203

Module 3

Example 2The sum of three numbers is 18. The third number is five timesthe sum of the first two numbers. The sum of the third number,three times the first number, and twice the second number is17. Write the three equations and find the numbers.

Solution

Let the three numbers be x, y, and z.

Rearrange equations:

Equation (1) + Equation (2):Equation (1) – Equation (3):

Equation (4) 6:

Equation (5) + Equation (6):

Substitute x = –4 into Equation (6):

Substitute x = –4, y = 7 into Equation (1):

The three numbers are –4, 7, and 15 or written as an orderedtriple (–4, 7, 15).

− + + ==

4 7 1815

z

z

− + ==

4 3

7

y

y

(4)(5)(6)

6 6 18

2 1

34

4

x y

x y

x yx

x

+ =− − =

+ =− =

= −

(1)(2)(3)

x y z

x y zx y z

+ + =+ − =+ + =

18

5 5 03 2 17

x y z

z x y

z x y

+ + == +

+ + =

18

5

3 2 17

204 Section 4, Lesson 1 Principles of Mathematics 11

( )

Example 3For breakfast, Jean had one cup of oatmeal porridge, one cup ofskim milk, and one cup of orange juice for a total of 375 calories.John had two cups of oatmeal porridge, two cups of skim milk,and one cup of orange juice for a total of 640 calories. Joan hadone cup of oatmeal porridge, three-quarter cup of skim milk,and one-half cup of orange juice for a total of 290 calories. Howmany calories are in one cup of oatmeal porridge, one cup ofskim milk, and one cup of orange juice?

Solution


Module 3

Jean:John:

Joan:

(1)(2)

(3)

(1)(2)(3)

(4)(5)

Let x = number of calories in a cup of oatmeal

y = number of calories in a cup of skim milk

z = number of calories in a cup of orange juice

x y zx y z

x y z

+ + =+ + =

+ + =

3752 2 640

34

12

290

x y zx y z

x y z

+ + =+ + =

+ + =

375

2 2 6404 3 2 1160 Equation (3) x 4

Substitute y = 120 in Eqn (4):

Substitute y = 120, x = 145 in Eqn (1):

Eqn (1) – Eqn (2)Eqn (2) x 2Eqn (3) – Eqn (5)

− − = −+ + =

− = −=

x yx y z

yy

2654 4 2 1280

120

120

− − = −− = −

=

xxx

120 265

145145145 120 375

110+ + =

=zz

Ordered triple: (145, 120, 110).∴ x = 145 calories in a cup of oatmeal

y = 120 calories in a cup of skim milkz = 110 calories in a cup of orange juice

Self-Marking Activity

1. The length of each of the congruent sides of an isoscelestriangle is one and one-half times the length of the base. Theperimeter of the triangle is 60 cm. Find the length of eachside of the triangle.

2. The sum of the digits of a three-digit number is 13. If thetens and hundreds digits are interchanged, the new numberis 90 less than the original, and if the units and hundredsdigits are interchanged, the resulting number is 99 less thanthe original. Find the original number.

Hint:

3. Melissa has loonies and $5 and $10 bills in her purse thatare worth $96. If she had one more loonie, she would havejust as many loonies as $5 and $10 bills combined. If she has23 bills and loonies in all, how many of each denominationdoes she have?

Hint:

4. A cauliflower, one bunch of celery, and one bag of radishescost $3.67. Two bunches of celery, a cauliflower, and fourbags of radishes cost $8.53. Three bunches of celery, twocauliflowers, and one bag of radishes cost $7.04. Find thecost of each type of vegetable.

You use variables to represent coin-valuein much the same way as you did forplace-values in the previous question.

Let x = hundreds digit

y = tens digit

z = units digit

Note: Remember a number like 576 means100 • 5 + 7• 10 + 6 when you interpret its placevalue.

Accordingly, the number in this question is100x + 10y + z.


Module 3

5. The sum of the length, width, and height of a rectangularbox is 75 cm. The length is twice the sum of the width andheight, and twice the width exceeds the height by 5 cm. Findthe dimensions of the box.

6. Twenty-five coins whose value is $2.75 are made up ofnickels, dimes, and quarters. If the nickels were dimes, thedimes were quarters, and the quarters were nickels, the totalvalue would be $3.75. How many coins of each type arethere?

7. In a triangle whose perimeter is 52 cm, the length of thelongest side is 2 cm less than the sum of the lengths of theother two sides. The length of the longest side is 3 cm morethan twice the length of the shortest side. How long is eachside of the triangle?

Check your answers in the Module 3 Answer Key.


Module 3


Module 3

Notes

Module 3

Lesson 2

Non-Linear Systems

Outcomes


• solve quadratic-linear systems of equations graphically andalgebraically

• solve quadratic-quadratic systems graphically andalgebraically

OverviewIn the previous lessons, you investigated systems of linearequations in two variables graphically and algebraically. In thislesson, you will investigate systems containing one linearequation and one quadratic equation (second-degree equation).

The graph of a second-degree equation is referred to as a conic.You will study these in Principles of Mathematics 12 but atpresent you should be familiar with the different types ofgraphs that could result.

The following diagrams are given as a guide to the variousgraphs that could result from graphing a quadratic equation.

Parabolas

a) b)

c) d)

x

y

x

y

x

y

x

y


Module 3

Circles

Ellipses

a) b)

Hyperbolas

a) b)

c)

x

y

x

y

x

y

x

y

x

y


x

y

A quadratic-linear system of equations includes an equation ofone of the conics and a linear equation. The solution of twoequations in two variables still corresponds to a point ofintersection of their graphs.

Draw several different sketches showing intersections of a lineand a parabola that result in different numbers of intersectionpoints. How many points of intersection are possible for a lineand a parabola?

Repeat this process for a line and a circle, a line and an ellipse,and a line and a hyperbola.

Example 1Identify the number of solutions in each system.

Solution

a) 2 solutions because there are two points of intersection

b) 1 solution because there is one point of intersection

c) no solution because the graphs do not intersect

A quadratic-linear system has 0, 1, or 2 real number solutions.As in the case of systems of linear equations, you can findapproximate solutions by graphing or exact solutionsalgebraically.


Module 3

x

y

0 x

y

0

l

x

y

0

a) b) c)

Example 2

Solve the following system of equations by graphing:

Solution

The first equation represents a line with a slope of –2 and a y-intercept of 2. The graph ofthe second equation is a circlewith centre (0, 0) and radius5. The intersection pointsappear to be near

and (3, –4).

Example 3Solve the following system of equations algebraically.

Solution

Solve the linear equation for y.

Substitute the expression for y in the first equation.

To find the corresponding y-values, substitute the x-values intothe linear equation and solve for y.

y x x

x x x

x x

x x

x x

= − +

− = − +

= − +

= − −= =

2

2

2

6 9

5 6 9

0 5 4

0 4 1

4 1or

x y

y x

+ == −

55

y x x

x y

= − += +

2 6 95

−112

5,

x

y

0

5

5

y x

x y

= − +

+ =

2 2

252 2


Module 3

( ) ( )

The solutions are (4, 1) and (1, 4), meaning there are two pointsof intersection on the graphs. Check your answers algebraicallyand graphically.

A system of two or more quadratic relations or conics is aquadratic-quadratic system. A quadratic-quadratic systemmay have 0, 1, 2, 3, or 4 real number solutions as illustrated inthe following diagrams.

No Solution 1 Solution

2 Solutions 3 Solutions

4 Solutions

They may be solved graphically or algebraically rememberingthat the graphs might be approximate solutions but thealgebraic solutions are exact.

ll

l l

l

l

l

ll

l

x

x yy

y

=+ =+ =

=

1

51 5

4

x

x yy

y

=+ =+ =

=

4

54 5

1


Module 3

Example 4Solve by (a) graphing, and (b) algebraically.

Solution

You can use a table of values or a graphing calculator to get the accompanying graph.

x –4 –3 3 4±2 0 0 ±2

±3.2 ≈±3.2

x –6 –3 –2 –1.5 –1 0 1 1.5 2 3 61 2 3 4 6 undefined –6 –4 –3 –2 –1

a) The points of intersection are at (–3, 2) and (3, –2).

x

y

0

= −6

yx

± 11± 11= ± −2 5y x

5− 5

x yxy

2 2 5

6 0

− =+ =


Module 3

b) Algebraically:

Solve the second equation for y:

Substitute for y in the first equation.

Solve.

Because x2 + 4 has no real roots, then

Substitute the x-values into

Solution is at (3 –2) and (–3, 2).

Notice that this corresponds to the graph.

If x

y

y

= −

= −−

=

363

2

If x

y

y

=

= −

= −

3632

yx

=−6

.

x

x x

x

2 9 0

3 3 0

3 3

− =− + =

= + −or

x x

x x

4 2

2 2

5 36 0

9 4 0

− − =

− + =

− =

− − =

− =

2 2

22

222

5

65

365 Multiply by and rearrange.

x y

xx

xxx

−6x

xy

yx

=

= −6

6

x y

xy

2 2 5

6 0

− =+ =


Module 3

( ) ( )

( ) ( )

Example 5Solve:

This example can be done algebraically by substitution or byaddition-subtraction.

Solution 1: SubstitutionSolve the first equation for x2: x2 = 49 – 4y2

Substitute 49 – 4y2 for x2 in the second equation and solve.

Substitute for y in the first equation to find thex-values.

The solution is made up of

3 10 3 10 3 10 3 10, , , , , , , .− − − −

x

x

x

x

2 2

2

2

4 10 49

40 49

9

3

+ =

+ =

== ±

10 10or −

2 49 4 3 12

98 8 3 12

11 110

10

10 10

2 2

2 2

2

2

− − = −

− − = −

− = −

=

= −

y y

y y

y

y

y or

x y

x y

2 2

2 2

4 49

2 3 12

+ =− = −


Module 3

( )

)(

)( )( )( )(

Solution 2: Addition Subtraction

Multiply first equation x2

+ 4y2

= 49 by 2 and subtract 2x

2– 3y

2= –12

The next steps are a repeat of Solution 1.

This system is graphed below.

x

y

0

2 8 98

2 3 12

11 110

10

10

2 2

2 2

2

2

x y

x y

y

y

y

+ =

− = −

=

=

= ±


Module 3

Example 6Solve:

Solution

y x

x y

2 4

4

=− = −


Module 3

Graph Algebraic solution:

Solve Eqn (2) for x in termsof y and substitute thatexpression into Eqn (1).

This quadratic equation does not factorso use the discriminant to determinewhether it has solutions.

Because b2 – 4ac < 0, there is nosolution. This means the graphs of thetwo equations do not intersect.

(1)

(2)

y = 4 x 2

x y = 4x

y

y x

x y

2 44

=− = −

x y

y x

y y

y y

y y

= −== −

= −− + =

4

4

4 4

4 16

4 16 0

2

2

2

2

a b c

b ac

= = − =

− = − − = −

1 4 16

4 4 4 1 16 482 2

( )

( ) ( )( )

Self-Marking Activity 1. Solve each system both graphically and algebraically.

a) b) c)

d) e)

2. The line y = x intersects the parabola y = 4x – x2 at the originand the point Q. Let P be the vertex of the parabola.

a) Find the coordinates of Q.

b) Find the coordinates of P.

c) Find the length of OQ.d) Find the distance from P to OQ.

e) Find the area of OQP.

3. Solve algebraically.a) b) c)

d) e) f) x y

y x

2 2

2 2

16

2 32 2 0

+ =− + =

x y

x y

2 2

2 2

9

4

+ =+ =

2 4

2 4

2 2

2 2

x y

x y

+ =− =

x xy y

xy

2 2 90

+ + ==

x y

xy

2 22 22 0

− =+ =

x y

y x

2 2

2

18

3

+ ==

P

Q

O

x y

y x

2 2 16

2 10

+ == −

x y

y x

2 2 18+ ==

xy

x y

=+ =

1

2x y

x y

− + − =+ + =

2 3 4

3 0

2 2y x

y x

==

48

2


Module 3

( ) ( )

4. Find the dimensions of a rectangle if its diagonal is 17 cmand the perimeter is 46 cm.

5. The perimeter of a rectangle is 26 cm. Its area is 12 cm2.Find the dimensions of the rectangle.

6. The sum of two numbers is 27. Their product is 126. Whatare the two numbers?

7. The sum of two numbers is 3. The sum of the squares ofthese numbers is 17. What are the two numbers?

8. The sum of the squares of two positive numbers is 65. Thedifference of their squares is 33. What are the two numbers?

9. The area of a rectangle is 60 cm2. The measure of each of itsdiagonals is 13 cm. Determine the length and width of therectangle.



Module 3

Module 3

Lesson 3

Graphing Linear Inequalities In Two Variables

Outcomes


• graph the solution of a linear inequality in two variables

• find the solution to a system of inequalities

Overview

You have learned several ways of solving systems of linearequations. You will now investigate the idea of a linearinequality because much of the mathematics that you use indaily life does not come in the form of an equation. Statementssuch as, “I want to find a job that pays more than the minimumwage.” or “I have to spend less than $10 this week.” areexamples of statements of inequalities.

The graph of a line separates the plane into three distinctregions: two half planes and the line itself. The line itself isthe boundary line of each half plane. It divides the coordinateplane into two half planes.

A linear inequality has a boundary line that can be expressedin the form y = mx + b. The solution to a linear inequality is theset of all ordered pairs that make the inequality true.

When the inequality is ≤ (read “less than or equal to”) or ≥(read “greater than or equal to”), the solution includes thepoints on the boundary line and the graph has a solid boundaryline.

x

y

B o u n d a r y

L i n e

H a l f -

P l a n e

H a l f -

P l a n e


Module 3

When the inequality is < or > the solution does not include thepoints on the boundary line and the graph will have a dashedboundary line.

Example 1Graph 2x + y < 4.

Solution

The boundary line is 2x + y = 4. You can graph the line bytransferring it into y = –2x + 4. You can use the slope interceptform, with b = 4 and the slope – 2.

Use a dashed line to show that the points on the line are notsolutions to 2x + y < 4.

x

y

x

y

x

y

x

y


To determine which half plane is the solution, select a pointfrom each half plane and see which ordered pair makes theinequality true. A good test point is the origin (0, 0) as long asthe boundary line does not go through that point.

(0, 0) on the left side (4, 0) on the right sidey < –2x + 4 y < –2x + 4

0 < –2(0) + 4 0 < –2(4) + 4

0 < 4 True 0 < –4 False

Because the test point (0, 0) makes the inequality true, shadethe half plane in which this point is located.

Note: Some graphing calculators, like the TI-83Plus, can graphinequalities and shade one side for you. You may use thatfeature if you have it on assignments and tests in this course,but show the working for a test point if the question asks for it.

88 4 x

y

8

8

4

4

4

x

y

84

4

8

8

4

48


Module 3

Extensions of this would be:

a) 2x + y ≤ 4

becomes y ≤ –2x + 4The boundary line would be solid and shaded below the line.

b) 2x + y > 4

becomes y > –2x + 4

The boundary line would be dashed to show that the pointson the line are not solutions of y > –2x + 4. Use the same testpoints as in the initial questions.

(0, 0) on the left side (4, 0) on the right sidey > –2x + 4 y > –2x + 4

0 > –2(0) + 4 0 > –2(4) + 4

0 > 4 False 0 > –4 True

The coordinates of (4, 0) make the inequality true.

88 4 x

y

8

8

4

4

4

88 4 x

y

8

8

4

4

4


Module 3

c) y ≥ –2x + 4

The boundary line is solid and the shading is then the sameas in b).

A linear inequality in x and y is an inequality that can bewritten in one of the following forms: ax + by < c, ax + by ≤ c,ax + by > c, or ax + by ≥ c.

An ordered pair (x, y) is a solution of a linear inequality in xand y if the inequality is true when a and b are substitutedfor x and y, respectively.

For instance (4, 0) is a solution to y > –2x + 4 because 0 isgreater than –2(4) + 4 = –4.

To sketch the graph of a linear inequality in two variables,first sketch the line given by the corresponding equation.Use a dashed line for inequalities with < or >. This lineseparates the coordinate plane into two half planes.

• In one of the half planes are points that are solutions ofthe inequality.

• In the other half plane no point is a solution.

You can decide whether the points in an entire half planesatisfy the inequality by testing one point in the half plane.

88 4 x

y

8

8

4

4

4


Module 3

Example 2Sketch the graph of the inequality y ≥ 3.

Graph the solid horizontal liney = 3.

Next, test a point (0, 0).

y ≥ 3

0 ≥ 3 False

The origin is not a solution andit is below the line. Therefore,the graph is all points on orabove the line y = 3.

Example 3Sketch the graph of the inequality x < 2.

SolutionSketch the dashed vertical line x < 2.

Next, choose a test point (0, 0).

x < 2

0 < 2 True

This is a solution whichrepresents all points to the leftof 2.

Two or more linear inequalities on the same coordinate planemake up a system of linear inequalities.

The following process may help you sketch the graph of asystem of linear inequalities.

1. Sketch the line that corresponds to each inequality. It isuseful to put the line in the form y = mx + b. Use a dashedline for inequalities with < or > and a solid line forinequalities with ≤ or ≥.

123 x

y

1

1

2

2 3

3

1

2

3

x1123

1

2

2 3

3

4

y


Module 3

2. Lightly shade the half plane that is the graph of each linearinequality. Using a different-coloured pencil or differentshading to shade each inequality may help you distinguishthe different half planes.

3. The graph of the system is the intersection of each of the halfplanes. If you used coloured pencils or different shading it isthe region that has been shaded with each colour.

Example 4Given: x + y ≥ 2

2x – y < 1

SolutionGraph x + y ≥ 2, or y ≥ 2 – x, showing the boundary as a solidline.

In the same coordinate plane, graph 2x – y < 1, or y > 2x – 1,showing the boundary as a dashed line.

The graph of the solution set of the system is the doubly shadedregion above and on the graph of x + y = 2 and also above thegraph of 2x – y = 1.

x + y ≥ 2 and 2x – y < 2

The word “and” means the intersection of the two sets of points.The solution is the dark shaded region.

Algebraically it can be written as

( ){ } ( ){ }+ ≥ − <∩, | 2 , |2 1x y x y x y x y

x

y

2x

y =

1

x + y = 2


Module 3

Note: 2x – y < 1–y < –2x + 1y > 2x – 1

when you divide by(–1) the sign of theinequality changes.

x + y 2 or 2x – y 1

The word “or” means the union of the two sets of points. Thesolution is all the shaded area in the graph.

Algebraically it can be written as


1. Sketch the graph of the system of linear inequalities.

2. Match the system of inequalities with its graph.a) b)

c) d)

i) ii)

iii) iv)

x

y

22

4

42

2

4

4x

y

22

4

42

2

4

4

x

y

22

4

4 62

2

4

x1

y

1

2

3

1

123

x

y

y x

≤<≥ − +

3

11

y x

y

x

<> −≤

30

y

x

y x

> −≥ −≤ − +

1

31

yy

≤> −

42

y x

y

x

≤ +

≥≤

12

2

0

0

( ){ } ( ){ }+ ≥ − <, | 2 , |2 1x y x y x y x y


Module 3

3. Write a system of linear inequalities for the region shownbelow.

4. Sketch the following systems defined by

a) b)

c) d)

5. Graph the system x + y > 6 and 2x + 3y < 5.

6. Graph the system x – 2y < 5 or 2x + y > 3.

7. Sketch the region defined by 5x – 6y ≥ 6, 3x + y ≤ 4, and y ≥ –3.

8. a) Draw the region defined by x ≥ 0, y ≥ 0, 3x + y ≤ 4, andy – 2x > – 1.

b) Name the coordinates of the vertices in part (a).

9. Find the area of the region determined by the graph of thesolutions of y ≤ x, x ≤ 6, and y ≥ 0.


2 554

3

x y

y x

+ <

≥ −

x yx y

+ >− + <

3

4 4

2 2

5 2

x yx y

− ≥− ≥

2 2

2 1

x yy x

− >− <

y

x2 4 6 8

2

4

6

8

24

4

2

l l

ll

A (0, 3) B (9, 3)

C (5, 1)D ( 4 , 1 )


Module 3


Module 3

Notes

Lesson 4

Quadratic, Rational, and Absolute Value Inequalities

Outcomes


• sketch the graph of a quadratic inequality in two variables

• solve a quadratic inequality in one variable

• solve a rational inequality in one variable• solve an absolute value inequality in one variable

Overview

The graphs of linear and quadratic inequalities in two variablesare closely related to the graphs of lines and parabolas. If a linehas the equation y = mx + b, any point (x, y) above the linesatisfies the inequality y > mx + b. Any point (x, y) below theline y = mx + b satisfies the inequality y < mx + b.

If the quadratic has the equation y = ax2 + bx + c, any point (x, y)above the parabola satisfies the inequality y > ax2 + bx + c. Anypoint (x, y) below the parabola y = ax2 + bx + c satisfies theinequality y < ax2 + bx + c.

x

y

y = ax + bx + c2

y < ax + bx + c2

y > ax + bx + c2

x

y

y = m x + b

y > mx + b

y < m x + b

Module 3


Module 3

In this lesson, you will investigate four types of quadraticinequalities:

y < ax2

+ bx + c y ax2

+ bx + cy > ax

2+ bx + c y ax

2+ bx + c

The steps used to sketch the graph of a quadratic inequality areas follows:

1. Sketch the graph of the parabola y = ax2 + bx + c. Use adashed parabola for inequalities with < or > and a solidparabola for inequalities with or .

2. Test a point inside and one outside the parabola.

3. Only one of the test points will be a solution. Shade theregion that contains the test point.

Example 1Sketch the graph of y x2 – 2x – 8.

Solution

The parabola has its vertex at

Substitute x = 1 for

Vertex: (1, –9)

Zeros:

Draw the sketch using a solid line for the parabola.

x

y

y = x 2x 82

42

2

8

x x

x x

x x

2 2 8 0

4 2 0

4 2

− − =− + =

= = −or

y x x= − −

= − −

= − −= −

2

2

2 8

1 2 1 8

1 2 8

9

xba

=−

= −−

=2

22 1

1.


( )

( )

( )

( )( )

Choose one point inside region and one outside region andcheck.

(0, 0) (5, 0)

y ≥ x2– 2x – 8 y ≥ x2

– 2x – 80 ≥ 0 – 2(0) – 8 0 ≥ 52 – 2(5) – 8

0 ≥ –8 true 0 ≥ 7 false

The region containing (0, 0) must be shaded as the graph of theinequality.

At this stage you have graphed quadratic inequalities in twovariables. Because there were two variables, the solution setwas represented graphically by a two-dimensional area whichyou shaded.To solve a quadratic inequality which has only one variable, youcan use a number line or a sign graph of the function. The stepsare as follows:

1. Determine the zeros of the function and plot them on anumber line.

2. Use open circles for inequalities with < or >.

3. Use solid circles for inequalities with ≤ or ≥.

4. Test one value of x from each one of the intervals determinedby the zeros to determine the sign of each factor.

5. Determine the intervals for which the values are true.

x

y

y = x 2x 82

42

2

8


Module 3

Example 2Solve x

2– 2x – 8 0 using a sign graph.

Solution

Find and plot the zeros:

The zeros are –2 and 4.

Determine the signs by testing one value of x in each interval.

Interval: x –2 or (– , –2]. Test x = –3.

Interval: –2 x 4 or [–2, 4]. Test x = 0.

Interval: x 4 or [4, ). Test x = 5.

The quadratic (x + 2)(x – 4) is positive or zero in the interval(– , –2] and in the interval [4, ). Therefore, the solution is allreal numbers in the interval (– , –2] or [4, ).

This can also be written as x –2 or x 4.

0 42

+ +

f x x x

f

= + −

= + −

+ + = +

2 4

5 5 2 5 4

signs of factors signs of product

0 42

+

f x x x

f

= + −

= + −

+ − = −

2 4

0 0 2 0 4

signs of factors signs of product

0 42

+

f x x x

f

= + −

− = − + − −

− − = +

2 4

3 3 2 3 4

signs of factorssigns of product

0 42

x x

x x

2 2 8 0

4 2 0

− − =− + =


Module 3

( )

( )

( )

( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )

( )

( )

( )

( ) ( )

( ) ( )

( ) ( )

( )( )

When f(x) > 0 the graph is above the x-axis.

When f(x) < 0 the graph is below the x-axis.

The above method uses the fact that a polynomial can changesign only at its zeros (that is, the x-values that make thepolynomial zero). Between consecutive zeros a polynomial mustbe entirely positive or negative. When the real zeros of apolynomial are put in order, they divide the number line intointervals in which the polynomial has no sign changes. Thesezeros are referred to as the critical numbers of the inequalityand the resulting intervals are the test intervals for theinequality.

Not all polynomials change sign at a zero.

Example 3Solve (x – 1)2 0.

Solution

Find zeros and plot.

Critical number: x – 1 = 0, x = 1

Determine signs of factors in the interval.

Interval: [1, ), x = 2

Interval: [– ,1], x = 0

The signs are positive on each side of the zero. A polynomial willnot change sign at a zero if a corresponds to the squared factor(x – a)2.

x x− − ≥

− − = +

1 1 0

signs of factors sign of product

x x− − ≥

+ + = +

1 1 0

signs of factors sign of product

0 1 2

+ +


Module 3

( ) ( )

( ) ( )

( )( )

( )( )

(x – 1)2 ≥ 0 is always true, but if the inequality is given as (x – 1)2 > 0 it is true for all values except x = 1.

The concept of critical numbers and test intervals can beextended to inequalities involving rational expressions. Thisuses the fact that the value of a rational expression can changesign only at its zeros (x-value for which its numerator is zero)and its undefined values (x-values at which the denominator iszero). These are the two types of numbers that make up thecritical numbers of a rational inequality.

Example 4

Solution

Determine critical numbers and plot.

The critical numbers are x – 1 = 0, x = 1 (the numerator is zerowhen x = 1) and (x – 2)(x + 3) = 0, x = 2 and –3 (the denominatoris zero when x = 2 and –3.

Critical numbers are –3, 1, 2.

Open dots are used because f(x) is < 0.

Interval (–∞, –3) Test point: –4

Note: In the original equation, two of these factors were in thedenominator. But the sign is what we want, and it is the same iffactors are all multiplied together or if some are divided. So wesimply multiply them—that’s easier to do—and take the sign ofthe product.

Interval (–3, 1) Test point: 0x x x+( ) −( ) −( )

+( ) −( ) −( )+( ) −( ) −( )

3 1 2

0 3 0 1 0 2

signs of faactors sign of product1 244444444 344444444 1 244 34= + 44

0 1 2

+ +

3123

x x x+( ) −( ) −( )− +( ) − −( ) − −( )−( ) −( ) −( )

3 1 2

4 3 4 1 4 2

signs oof factors sign of prod1 244444444444 344444444444 = −

uuct1 244 344

xx x

−− +

<12 3

0( )( )

Solve:


Module 3

Interval (1, 2)

Interval (2, ∞) Test point: x = 3

The solution is the intervals for which the sign of the quotientis < 0 (negative), (–∞, –3), or (1, 2).

Example 5

Solution

Zeros are at x + 1 = 0, x – 3 = 0, and x – 4 = 0.

x = –1, 3, 4

Critical numbers are –1, 3, 4

Because f(x) can equal 0, use solid dots for –1 and 3 and an opendot for 4 as the denominator can never be zero.

Interval (–∞, –1]. Test point: x = –2.

0 1 2

+ +

3123 4 5ll

x x x x+( ) −( ) −( ) −( )−( ) −( ) −( ) −( )

1 3 3 4

signs of factors1 22444444444 3444444444 1 244 344= +

sign of product

( )( )x xx

− +−

≤1 3

40

2Solve:

x x x+( ) −( ) −( )+( ) −( ) −( )

+( ) +( ) +( )

3 1 2

3 3 3 1 3 2

signs of faactors sign of product1 2444444 3444444 1 244 344= +

x+ 3( ) x−1( ) x−2( )32

+ 3

32−1

32−2

+( ) +( ) −( )signs of factors

! "###### $###### = −sign of product! "## $##

Test point: x =32


Module 3

Interval [–1, 3]. Test point: x = 0.

Interval [3, 4)


The solution interval is the intervals where the quotient isnegative or zero [–1, 4) or, –1 ≤ x < 4.

Example 6

Solution

For x2 + x + 3, the discriminant b2 – 4ac = 12 – 4(1)(3) = –11 whichmeans that the numerator has no real zeros.

∴ x – 3 = 0 and x + 2 = 0

x = +3 and x = –2 are the critical numbers

xx xx x x

x x

x x xx

−−

+>

+( )− −( )−( ) +( )

>

+ − +−

31

20

2 33 2

0

2 33

2

(( ) +( )>

+ +−( ) +( )

>

x

x xx x

20

33 2

02

xx x−

>+31

2Solve:

x x x x+( ) −( ) −( ) −( )+( ) +( ) +( ) +( )

1 3 3 4

signs of factors1 22444444444 3444444444 1 244 344= +

sign of product

x x x x+( ) −( ) −( ) −( )+( ) +( ) +( ) −( )

1 3 3 4

signs of factors1 22444444444 3444444444 1 244 344= −

sign of product

Test point:72

x x x x+( ) −( ) −( ) −( )+( ) −( ) −( ) −( )

1 3 3 4

signs of factors1 22444444444 3444444444 1 244 344= −

sign of product


Module 3

Interval (–∞, –2) Test point: x = –4

Interval (–2, 3) Test point: x = 0


The solution intervals where the quotient is > 0 is (–∞, –2) or(3, ∞).

The next inequality to investigate is the absolute valueinequality. The absolute value of a number, x, denoted by |x|can be interpreted geometrically as the distance from x to zeroin either direction on the number line. Examine the followingchart involving this interpretation.

* Remember –3 < x < 3 means –3 < x and x < 3.

x x x x2 3 2 3+ +( ) +( ) −( )

+( ) +( ) +( )signs of factors

1 2444444444444 344444444444 1 244 344= +sign of product

x x x x2 3 2 3+ +( ) +( ) −( )

+( ) +( ) −( )signs of factors

1 2444444444444 344444444444 1 244 344= −sign of product

02 3

+ +x x x x2 3 2 3+ +( ) +( ) −( )

+( ) −( ) −( )signs of factors

1 2444444444444 344444444444 1 244 344= +sign of product


Module 3

Statement

|x| = 3

|x| < 3

|x| > 3

GeometricInterpretation

Values of x that lie3 units from 0

Values of x that lieless than 3 unitsfrom 0

Values of x that liemore than 3 unitsfrom 0

Graph Solution

x = –3 and x = 3

–3 < x < 3*(–3, 3)

x < –3 or x > 3(–∞, –3) (3, ∞)

3 3

–3

3

3

3

0 3• •

–3 3

–3–2–1 0 1 2 3

–3

3 2 1

3

0 1 2 30

–3 3

–3

–3

3

3

0 3

The following represents the meaning and the solution ofabsolute value equations and inequalities involving |x| and c,where c represents a distance with c ≥ 0.

These statements are valid if ≤ is used instead of < and ≥replaces >.

Equations and inequalities involving |x – a|, where a is aconstant, can be solved by interpreting x – a as the distancefrom x to a specific point a on the number line.

The next two tables explain |x – a|. The first table uses aspecific value of 1 for a, and the second table explains the“universal” case for any value of a.


Module 3

Statement

|x| = c

|x| < c

|x| > c


The distance from xto 0 is exactly cunits.

The distance from xto 0 is less than cunits.

The distance from xto 0 is more than cunits.

Graph Solution

x = c or x = –c

–c < x < c or(–c, c)

x < –c or x > c(–∞, –c) (c, ∞)

0 ccll

0 cc

0 cc

Statement

|x – 1| = 3

|x – 1|< 3

|x – 1|> 3


The distance from xto 1 is 3.

The distance from xto 1 is less than 3.

The distance from xto 1 is greater than3.

Graph Solution

x = –2 or x = 4

–2 < x < 4

x < –2 or x > 4

–3 3

–2

–3

2

3

1 4• •

3 3

–2

3

2

3

1 4

–2 1 4

Example 7Solve |x – 1| > 7 (a) geometrically and (b) algebraically.

Solution

a) Geometrically

|x – 1| > 7 means the distance from x to 1 is greater than 7units

b) Algebraically

Translated it means that x – 1 < –7 or x – 1 > 7.

i.e., x <–6 or x > 8.

1 86

7 7


Module 3

Statement

|x – a| = c

|x – a|< c

|x – a|> c


The distance from xto a is c.

The distance from xto a is less than c.

The distance from xto a is greater than c.

Graph Solution

x = a – c or x = a + c

–c < x – a < cie. a – c < x < a < c

x – a > c or x – a < –c

ie. x > a + c orx < a – c

l l lll llll llll

Example 8Solve |2x + 1| ≤ |–7| (a) geometrically and (b) algebraically.

Solution

a) Geometrically

Make use of the fact that |ab| = |a| • |b|.

Thus, –4 ≤ x ≤ 3 or [– 4, 3].

b) Algebraically

|2x + 1| ≤ 7

This means

RewrittenSubtract 1 from each partDivide by 2

− ≤ + ≤− ≤ ≤− ≤ ≤

7 2 1 7

8 2 6

4 3

x

x

x

or equal to72

.

x to is less than−12

This means that the distance from

Remember x+12

= x− −12

2 ⋅ x+12

≤7

2 x+12

≤7

x+12

≤72

x− −12

≤

72

2x+1 ≤ −7

2 x+12

≤ −7


Module 3

Self-Marking Activity 1. Determine whether the ordered pair is a solution of the

inequality.

a) y > –x2

+ x + 4 (1, 6)b) y ≤ 2x2 – x + 1 (–1, 5)

2. Match the inequality with its graph.

a) y > –x2

– 6x – 2

b) y ≥ 2x2

+ 3x – 5c) y ≤ –x2 + 3x + 5

d) y > x2 – 2x – 6

3. Sketch the graph of the inequalities:

a)

b)

c)

y x x

y x x

y x x

≤ +

> + +

< − −

2

2

2

4

12

3 2

3 6

x

y

2

6

4

iv)

8x

y

2 64

4

6

ii i)

2 4 2

x

y

2 64

4

6

x

y

2 442

4

6

i) ii)


Module 3

4. Find the solution set for the following. Express your answerin set notation and interval notation.

5. Solve the following inequalities. Check your solutions.

6. Solve the following inequalities geometrically andalgebraically.


a) d)

b) e)

c)

5 8 2 4 8 4

2 1 5 3 4 10

3 4 9

x x

x x

x

− ≤ + > −

+ ≤ − + >

− >

a) d)

b) e)

c) f)

xx

xx

xx

xx x

x xx

xx

x xx x

++

<++

≤+−

+− −

≥− −

−≥

+−

≤− −+ −

≤

31

02 5

111

12

012

250

24

02 5 33 5 2

0

2

2

2

2

2

2

2

b g

a)

b)

c)

d)

x x

x x

x x

x x

2

2

2

2

3 4 0

2 3 5 0

20 0

3 18

+ − ≤

+ − >

− − <

+ ≥


Module 3

Module 3

Lesson 5

Verifying and Proving Assertions inCoordinate Geometry

OutcomeWhen you complete this lesson, you will be able to

• use the properties of slope, midpoint and distance to verifyand prove assertions in coordinate geometry

OverviewWe will be using the slope, midpoint, distance formulas to verifyor prove assertions in coordinate geometry, pertaining totriangles, quadrilaterals, and circles.

You may find the following formulas useful in doing your proofsor verifications.

Also included is a review of the properties of quadrilaterals.

Parallelograms

• Opposite sides are parallel.

• Opposite sides are equal.

>

>

>>

Slope m( ) =y2 − y1x2 − x1

Midpoint =x1 + x2

2,y1 + y2

2

Distance d( ) = x1 − x2( )2 + y1 − y2( )2

Distance d( ) =Ax2 + By1 +C

A2 + B2


If two lines are parallel their slopes are equal.If l1 ⊥ l2 thenm1 ˘m2 = -1

Module 3

• Opposite angles are equal.

• Adjacent angles are supplementary.

• Diagonals bisect each other.

Rectangle

• All angles are right angles.

• Diagonals are equal.

AD = BC

A B

DC

x + y = 180°

x

x

y

y


Rhombus

• All sides are equal; opposite sides are parallel.

• Diagonals are perpendicular to each other and bisect eachother.

Square (rectangle and rhombus)

• All angles are right angles.

• All sides are equal.

• Diagonals are equal and perpendicular to each other.

AD = BC

A B

DC


Module 3

Example 1Show that A(2, –1), B(1, 3), C(6, 5), and D(7, 1) are the verticesof a parallelogram.

Solution

There are several ways to solve this problem.

1. You could show that the opposite sides have the same slopeand, therefore, are parallel.

Using the slope formula:

sides are parallel and the given ordered pairs are vertices ofa parallelogram.

2. You could show that the opposite sides are equal in length bydetermining their lengths.

Because AB = CD = and BC = AD = , the figure isa parallelogram.

2917

d = x2 − x1( )2 + y2 − y1( )2

AB = 2−1( )2 + −1−3( )2 = 1+16 = 17

BC = 1−6( )2 + 3−5( )2 = 25+ 4 = 29

CD = 7−6( )2 + 1−5( )2 = 1+16 = 17

AD = 7−2( )2 + 1− −1( )( )2 = 25+ 4 = 29

Because and , oppositem m m mAB CD BC DA= = − = =425

my yx x

m

m

m

AB

BC

CD

=−−

=− −( )

−= −

=−−

=

=−

2 1

2 1

3 11 2

4

3 51 6

25

5 116 7

4

1 12 7

25

−= −

=− −

−=mDA

l

l

l

lx

y

B(1, 3)

C(6, 5)

D(7, 1)

A(2, 1)


Module 3

Example 2Decide whether ABCD is a rectangle, rhombus, or a squaregiven these vertices: A(3, 9), B(1, –2), C(–9, –7), and D(–7, 4).

Solution

In order to be a rectangle or square, the figure must have fourright angles.

If DC ⊥ CB, then ∠C would be 90°.

DC ⊥ CB if mDC • mBC = –1.

When l1 ⊥ l2, then m1m2 = –1.

∴ quadrilateral ABCD is not a rectangle or a square

Because is notm m CDC BC⋅ = ⋅

≠ − ∠ °

112

510

1 90, .

m

m

DC

BC

=− −( )

− − −( )= =

=− − −( )

− −=

−−

=

4 77 9

112

112

7 29 1

510

5510

l

l

l

l

x

y

B(1, 2)

C( 9, 7)

D( 7, 4)

A(3, 9)


Module 3

If it is a rhombus, then all four sides must be congruent.

Because all the lengths are equal, the figure must be arhombus.

The following proof involves using coordinate geometry to provea theorem in plane geometry. The connection between these twogeometries is a very important one in mathematics.

Example 3Prove that the segment determined by the midpoint of two sidesof a triangle is:

• parallel to the third side

• half the length of the third side

Solution

Let B be at the origin and BC lie along the x-axis so that C hascoordinates (2, 0). Arbitrarily choose A to have coordinates (2a,2b).

x

y

(2, 0)(0, 0)B C

A(2a, 2b)

D(a + 1, b)E(a, b)

d x x y y

AB

= −( ) −( )

= −( ) + − −( )( ) = + =

2 12

2 12

2 23 1 9 2 4 121 125 ==

= − −( )( ) + − − −( )( ) = + = =

= − −

5 5

1 9 2 7 100 25 125 5 5

9

2 2BC

CD −−( )( ) + − −( ) = + = =

= − −( ) + −( ) =

7 7 4 4 121 125 5 5

7 3 4 9

2 2

2 2AD 1100 25 125 5 5+ = =


Module 3

Now, find the midpoints of AB and AC at E and D respectively.

Midpoint of AB Midpoint of AC

Determine the slope of DE and BC

Because mDE = mBC, then DE || BC.

∴ The line segment determined by the midpoints of two sides ofa triangle is parallel to the third side and half its length.

Note: The proof could also be done by choosing A to be (a,b) andC to be (c,0).

Length of

Length of

DE a a b b

BC

= + −( ) + −( ) = =

= −( )

1 1 1

2 0

2 2

2 ++ −( ) = =

∴

0 0 4 2

12

2

DE BC

mb ba a

m

DE

BC

=−

+ −= =

=−−

= =

101

0

0 02 0

02

0

Ex x y y

Dx x y y1 2 1 2 1 2 1 2

2 2 2 2+ +

+ +

, ,

+ +

+ +

E

a bD

a b

E

0 22

0 22

2 22

0 22

, ,

aa b D a b, ,( ) +( )1


Module 3


1. Three vertices of a rectangle ABCD are A(–9, 0), B(5, 4), andC(7, –3).

a) Find the coordinates of the 4th vertex of the rectangle.b) Find the perimeter of the rectangle.c) Find the area of the rectangle.

2. A triangle has vertices at A(–4, –2), B(2, –8), and C(4, 6). Isthis a right triangle? Verify your answer.

3. Show that the quadrilateral with vertices A(–5, –2),B(1, –1), C(4, 4), and D(–2, 3) is a parallelogram.

4. Line l1 contains the points (x, 3) and (–2, 1). Line l1 ⊥ line l2.Line l2 contains the points (5, –2) and (1, 4). Find the value ofx. Explain your rationale.

5. Line l3 contains the points (r, 3) and (–2, 1). Line l3 is parallelto line l4 which contains the points (5, –2) and (1, 4). Find thevalue of r. Describe the procedure used.

6. Prove that the quadrilateral defined by the following lines isa square: l1: 4y = 3x – 6, l2: 4x + 3y = 33, l3: 4y = 3x + 19, and l4: 4x + 3y – 8 = 0.

7. A triangle is given by the vertices P(–5, 4), Q(1, 8), and R(–1, –2). Is the median from P to RQ perpendicular to RQ?Explain your answer.

8. A quadrilateral PQRS has vertices at P(5, –6), Q(3, 0), R(–1, 2), and S(–5, –4). Verify that the midpoints of the sidesof this quadrilateral form the vertices of a parallelogram.


Review Section 4 before attempting the review questionsbeginning on the next page. These questions should help youconsolidate your knowledge as you prepare for the SectionAssignment 3.4.


Module 3

Review1. Solve each system for x and y.

a)

b)

c)

d)

2. Graph the following systems of inequalities. State thesolution algebraically.

3. Graph: y < x2

– x – 20.

4. Solve the following inequalities:

5. A triangle is given by the vertices P(–5, 4), Q(1, 8), andR(–1, –2). Is the median from P to QR perpendicular to QR?Explain your answer.


Now do the Section Assignment which follow this section.

a)

b)

c)

d)

x xx

x xx

x

2

2

3 4 01

20

2 7

3 1 8

+ − ≤+

− −≥

+ >

+ ≤

a) and

b) or

2 4 2 6

2 6 2 4

x y x yx y x y

− > + <+ ≤ − ≥

+ = −+ =

2 3 63 2 25

x yx y

+= −

+= −

33

33

1

x y

xy

x y

x y

2 2

2 2

25

7

+ =

− =

x yx y

2 2 102

+ =+ =

Principles of Mathematics 11 Section 4, Review 253

Module 3

Notes

254 Section 4, Review Principles of Mathematics 11

Module 3

PRINCIPLES OF MATHEMATICS 11

Section Assignment 3.4

Module 3

Principles of Mathematics 11 Section Assignment 3.4 255

256 Section Assignment 3.4 Principles of Mathematics 11

Module 3

Module 3

Section Assignment 3.4

Analytical Geometry II

1. Find the points where the line y = x + 3 intersects the

parabola y = x2 + 3x.

2. Graph the inequalities.

a) y ≥ 3x – 2


Total Value: 40 marks(Mark values in brackets)

Scientific and/or graphing calculator is permitted.(4)

(2)

Module 3

b) 2x + y < 8

3. Graph the systems of inequalities. Clearly mark the solutionset.

a) x + y > –24x – y < 4


(3)

(4)

b) 2x + 3y ≤ 6y ≥ –1x ≤ 1

4. Graph the following inequality. Clearly mark the solutionset.

y > x2 + 4x + 3


Module 3

(4)

(3)

5. Solve the system algebraically.

6. Give the solution set for the inequality, geometrically andalgebraically.

+ <6 4x

− =

+ =

2 2

2 2

11

4 16

x y

x y


Module 3

(4)

(2)

7. State the solution set for each of the following inequalities.

a)

b) ( )( ) ( )

+<

+ −1

02 3x

x x

+ >3 2 4x


Module 3

(3)

(4)

8. Give the solution set for the following. Express your answerin set notation and interval notation.

9. Given right triangleABC with D themidpoint of BA.. Showthat D is equidistantfrom B and C.A (a,0)

B (0,b)

C

D

x x2 3 10 0− − ≥


Module 3

(3)

(4)

(Total: 40)

principles of mathematics 11 section 4, introduction...

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