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Module 3, Section 3

Analytic Geometry I

IntroductionThis section begins by developing the equation of a circle.Besides finding the distance between two points, you will learnto find the distance from a point to a line. You will solvesystems of two linear equations in two unknowns and systemsof two non-linear equations in two unknowns, graphically andalgebraically. Systems of three linear equations will also besolved.

Section 3 — Outline

Lesson 1 Circle

Lesson 2 Distance Between Points and Lines

Lesson 3 Systems of Linear Equations

Lesson 4 Solving Systems of Equations Algebraically

Lesson 5 Systems of Equations Containing Three Variables

Principles of Mathematics 11 Section 3, Introduction 135

Module 3

Notes

136 Section 3, Introduction Principles of Mathematics 11

Module 3

Module 3

Lesson 1

Circle

Outcome

When you complete this lesson, you will be able to

• relate the properties of a circle such as its centre and radius withthe equation of a circle

Overview

In previous modules, you gained experience with straight lineequations and their graphs, and quadratic equations and theirgraphs.

In dealing with sets of points, there are three importantdefinitions that must be understood.

Definitions

1. A locus (or curve) is the set of points and only those pointswhose coordinates satisfy certain conditions.

2. The equation of a locus (or curve) is the condition which thecoordinates of each of the points of that locus and only thosepoints, must satisfy.

3. The graph of a locus (or curve) is the geometricrepresentation of the set of points of the locus.

Example 1Graph and state the equation of a straight line. What is theline’s locus?

Solution

If you use the basic straight line y = x, the locus is that set ofordered pairs whose coordinates satisfy certain conditions.Representative samples of the locus could be (–1, –1), (0, 0), (1, 1).

The equation of the locus is y = x.

Principles of Mathematics 11 Section 3, Lesson 1 137

Module 3

The graph is the geometric representation of the set of points.

Definition

A circle is the set of all points (x, y) that are equidistant from afixed point called the centre of the circle. The distance, r,between the centre of the circle and any point (x, y) on the circleis the radius.

If the centre of the circle is located at (0, 0), the origin, and r isthe radius, allow P(x, y) to be any point in this circle. Thedistance from (x, y) to (0, 0) is equal to the radius (r).

We use the formula for the distance between two points toarrive at the standard form of an equation with radius, r, andcentre (0, 0).

( ) ( )= − + −2 2

1 2 2 1d x x y y

5

x

y

50r

P(x, y)l

x

y

y = x

138 Section 3, Lesson 1 Principles of Mathematics 11

The standard form of the equation of a circle with radius, r, andcentre (0, 0) is x

2+ y

2= r

2.

Example 2Sketch the circle whose equation is x2 + y2 = 36.

Identify four points on the graph that are on the circumferenceof the circle. Verify your solution on the graphing calculator.

Solution

Comparing it to x2 + y2 = r2

x2

+ y2

= 36 has C(0, 0) and radius = 6

Four of the points on the circle are: (6, 0), (0, 6), (–6, 0), (0, –6)

Note: The equation x2 + y2 = 1 is the equation of a circle withcentre (0, 0) and r = 1. This is referred to as a unit circle whichwill be used extensively in Principles of Mathematics 12.

You can use your graphing calculator to graph the circle. Acircle is not a function but it is made up of two semi-circles eachof which can be described as a function. To do this, solve for yfirst:

x y

y x

y x

2 2

2 2

2

36

36

36

+ =

= −

= ± −

x

y

100

10

Distance formula

Simplify

Square both sides

x y r

x y r

x y r

− + − =

+ =

+ =

0 02 2

2 2

2 2 2


Module 3

( ) ( )

Press and enter:

The graph on your screen may appear to be oval even though itis actually a circle. This is because the spacing between theunits on the axes is not consistent. Your graph can be adjusted

by using 5: ZSquare.

Suppose you have a circle with its centre at (h, k) and a radiusof r. Then, the distance from C(h, k) to any point P(x, y) on thecircle is

Standard Form of a Circle

The standard form of a circle is (x – h)2 + (y – k)2 = r2, where (h, k) is the centre and r is the radius.

Remember that x2 + y2 = r2 and (x – h)2 + (y – k)2 = r2 are thesame size and (x – h)2 + (y – k)2 = r2 is x2 + y2 = r2 shifted h unitshorizontally and k units vertically.

Example 3Write the equation in standard form for the circle in thediagram.

Solution

Centre: C(3, –2); r = 4

Standard form of circle:

(x – h)2 + (y – k)2 = r2

h = 3, k = –2, r = 4(x – 3)2 + (y – (–2))2 = 42

(x – 3)2 + (y + 2)2 = 16

x

y

0

4

P(x, y)

C(3, 2)

x

y

Radius : r

C e n t r e : (h, k )

P(x , y ) is a point onthe circ le

x h y k r

x h y k r

− + − =

− + − =

2 2

2 2 2

or

ZOOM

(use the (–) key for the firstnegative sign)

y x

y x

12

22

36

36

= −

= − −

Y=


Module 3

( ) ( )

( )( )

When the equation of a circle is given in the formx2 + y2 + dx + ey + f = 0, you can find the centre and radius of thecircle by completing the square.

Example 4Find the centre and radius of the circlex2 + y2 – 8x + 12y + 35 = 0 and provide a sketch.

Solution

x

y

C (4, 6)

( )− =4, 6 17C r

x x y y

x x y y

x y

2 2

22

22

2 2

2 2

8 12 35

882

12122

35 4 6

4 6 17

− + + = −

− + + + + = − + +

− + + =


Module 3

Rearrange termsComplete the square

Simplify( ) ( )

)()(

Example 5

Write the equation of the circle with centre on the line x = –3, andwhich is tangent to the y-axis at T(0,5).

Solution

The centre of the circle is on the line x = –3.

Radius OT is perpendicular to they-axis at T (the point of tangency), so OT (a radius) = 3.

Coordinates of centre O are (–3,5)The equation of the given circle

is (x + 3)2 + (y – 5)2 = 9.

Example 6

Find the centre and radius of the circle

Solution

The circle is not defined because r2 = –1 is impossible.

x x y y

x y

x y

2 2 2 2 2 2

2 2

2 2

4 2 2 1 6 2 1

2 1 6 4 1

2 1 1

− + − + + + = − + − +

− + + = − + +

− + + = −

+ − + + =2 2 4 2 6 0x y x y


Module 3

( ) ( )

( ) ( )

( ) ( )

)( ( )

x

y

T (0,5)

–3

O(–3,5)

x = –3

Example 7A two-lane highway goes through a semi-circular tunnel that is 5 metres high at the top. If each traffic lane is 4 metres wide, howhigh is the tunnel at the edge of each lane.

Solution

Self-Marking Activity

1. Write the equation for each circle in standard form.

2. Determine the centre and radius of each circle whoseequation is given by:

3. Write the equation for each circle in standard form.

a) b)5

x

y

50

x

y

200

20

a)

b)

c)

x y

x y

x y

+ + − =

+ + =

+ + + =

2 1 12

119

1 1 64

2 2

2 2

2 2

a) centre

b) centre

c) centre

( , )

( , )

( , )

0 0 3

2 112

2 3 5

r

r

r

=

− =

− =


Module 3

(5, 0)

(4, 3)

5 m

4 4

x y2 2 25+ =

Substitute x = 4.

4 25

25 16

93

2 2

2

2

+ =

= −

==

y

y

yy m

( ) ( )

( ) ( )

( )

4. If the given equation has a circle as its graph, find the centreand radius.

5. Sketch a graph of the following equations:

6. Write equations for each of the following circles with thegiven properties. Leave answers in standard form.a) Centre (0, 2) passes through originb) Centre (5, 0) has a diameter of 10c) Centre (4, 3) passing through (1, 2)d) Centre on the line y = 2 tangent to the x-axis at (3, 0)e) Centre in first quadrant, radius 3; tangent to y-axis at

(0, 1)f) Centre (0, 0) and area 12 π square unitsg) Centre (2, –1) and circumference 20 π units

7. Some irrigation systems rotate around a pivot in a circularformation. Write an equation to model the circular boundaryof the field if the line of sprinklers is 400 m long.

Graph this circle. If you are on the edge of the circle and youry-coordinate is 400, what would be the x-coordinate?

8. The circle with equation (x – 1)2 + (y + 4)2 = 16 is translated 3units to the left and 2 units downward. Find the equation ofthe resulting circle.

( ) ( )2 2

2 2

a) 2 4 9

b) 6 2 6 0

x y

x y x y

+ + − =

+ − + + =

2 2

2 2

2 2

2 2

a) 4 2 4 0

b) 4 0

c) 6 12 0

d) 10 4 0

x y x y

x y

x y y

x y x y

+ + − − =

+ + =

+ + − =

+ − − =


Module 3

9. Find the equation of the illustrated circle.

10. The equation of the large circle is (x – 6)2 + y2 = 16. Find theequation of the small circle.

Check your answers in the Module 3 Answer Key.

x

y

l

x

y

lC (3, 3)


Module 3


Module 3

Notes

Module 3

Lesson 2

Distance Between Points and Lines

Outcome


• find the distance between points and lines.

Overview

In Principles of Mathematics 10 you learned how to determinethe midpoint of a line segment joining two points, P1(x1,y1) andP2(x2,y2).

Formula for the midpoint:

Example 1

Find the midpoint of the line segment joining points P(-3,5) andQ(2,-1).

Formula for the midpoint:

Solution

In Lesson 1, you used to find the

distance between two points P1(x1, y1) and P2(x2, y2).

One other interesting relation that you will examine is thedistance from a point to a line. This is understood to be theshortest distance from the point to the line.

To show you how the formula can be derived, you will firstexamine a situation in geometry and then apply that to thecurrent situation in analytical geometry.

( ) ( )= − + −2 2

2 1 2 1d x x y y

( ) ++=

=

5 –1–3 2,

2 2

1– ,2

2

M

+ + =

1 2 1 2,

2 2x x y y

M

+ + =

1 2 1 2,

2 2x x y y

M


Module 3

In ABC with sides oflength a and b, AB is thehypotenuse and its length is

Area of ABC can be listed in the following two ways:

Because the areas are equal regardless of which lengths areused as the base and altitude, the two expressions for the areaequal each other.

In a right triangle with sides of lengths a and b, the altitude tothe hypotenuse is

To find the distance from a point to a line, the previousderivation must be applied

The diagram showsthe point P(x1, y1),where d is therequired distance tothe line Ax + By + C= 0.

hab

a b=

+2 2.

12

12

2 2

2 2

2 2

ab a b h

ab a b h

ab

a bh

= + ⋅

= + ⋅

+=

Area Area= =12

ab h

a b2 2+ .AB

C

bha


R

x

y

d

Q

− −By CA

y11,

xAx C

B11,

− −

P(x1, y1)

(Using a as the baseand b as the height)

(Using the hypotenuseas the base)

Construct horizontal and vertical line segments PQ and PRwhere Q and R are points on the line Ax + By + C = 0. The y-coordinate of Q is y1 because it is the same vertical distance asP above the x-axis.

Substituting into Ax + By + C = 0 gives the x-coordinate of Q.

Coordinates of Q =

Similarly, R has the same x-coordinate as P because P and R areboth the same distance from the y-axis. The x-coordinate of Rwill be x1.

Substituting into Ax + By + C = 0 gives the y-coordinate:

The coordinates of R are

Because PQ is a horizontal line segment, its length is

Because PR is a vertical line segment, its length is

For convenience let M = Ax1 + By1 + C, and find an expressionfor d using

hab

a b=

+2 2

PRA C

BA B C

B= − − − = + +

yx x y

11 1 1

PQB C

AA B C

A= − − − = + +

xy x y

11 1 1

xx

11,

− −A CB

A B CB A C

A CB

x y

y x

yx

1

1

1

0+ + == − −

= − −

− −B CAy

y11,

A B CA B C

B CA

x y

x y

xy

+ + == − −

= − −

1

1

1

0


Module 3

Expression for d:

The distance (perpendicular) from a point P(x1, y1) to the line Ax + By + C = 0 is

You do not have to know the derivation of the formula.

dx y

=+ +

+

A B C

A B1 1

2 2

dx y

=+ +

+

A B C

A B1 1

2 2

=+

M

A B2 2

=+

M MAB

M A BAB

2 2

=+

M

AB

MA B

AB

2

2 2

2

=+

M

AB

M1

A1B

2

22 2

=+

MA

MB

MA

MB

2

2

2

2

d =+

PQ PR

PQ PR2 2


Module 3

( )( )

( )

( )( )

Example 2

Find the perpendicular distance from P(–3, 6) to the line 3x – 2y – 6 = 0.

Solution

The perpendicular distance from a point to a line is given by

where x1 = –3, y1 = 6, A = 3, B = –2, C = –6

Example 3

For the lines x + 3y = 6 and x + 3y = 3, find the

a) horizontal distance between the linesb) vertical distance between the lines

c) shortest distance between the lines

Solution

x

y

x + 3 y = 6

x + 3y = 3

2

1

1 3 6

d =− − −

+ −

=− − −

+

=−

=

=

3 3 2 6 6

3 2

9 12 6

9 427

132713

27 1313

2 2

Rationalize your answer

dx y

=+ +

+

A B C

A B1 1

2 2


Module 3

( ) ( )

( ) ( )

a) Horizontal distance is the distance between the x-intercepts= | 6 – 3 | = 3. Remember, to find the x-intercept, let y = 0.

b) Vertical distance is the distance between the y-intercepts = | 2 – 1 | = 1. Remember, to find the y-intercept, let x = 0.

c) Choose a point on x + 3y = 3. A convenient point is one of theintercepts.

Therefore, (0, 1) can represent P(x1, y1).

The other line, x + 3y – 6 = 0, represents Ax1 + By1 + C = 0.

A = 1, B = 3, C = –6

You can check your answer by taking a point on the linex + 3y = 6 and using x + 3y – 3 = 0 as the line Ax + By + C = 0.

A point that represents the x-intercept is (6, 0), where

A = 1, B = 3, C = –3

dx y

=+ +

+

=+ −

+

= = =

A B C

A B1 1

2 2

2 2

1 6 3 0 3

1 33

10310

3 1010

dx y

=+ +

+

=+ −

+

=−

= =

A B C

A B1 1

2 2

2 2

1 0 3 1 6

1 33

10310

3 1010


Module 3

( ) ( )

( ) ( )

Example 4A line is parallel to 12x – 5y = 6. Find its equation if it is 2 units awayfrom the point P(–3,–1).

Solution

The equation of the desired line will be 12x – 5y + C = 0 since it is parallel to the line 12x – 5y = 6. We need todetermine the value of C.

For substitution into the formula: A = 12, B = –5, x1 = –3, and y1 = –1

The equation of the line could be 12x – 5y + 57 = 0 or 12x – 5y + 5 = 0

( )( ) ( )( )( ) ( )

+ +=

+

+=

= +

= + = += =

∴ = =

2 2

12 –3 –5 –1 C2

12 –5

–31 C2

1326 –31 C

26 –31 C or –26 –31 C57 C 5 C

C 5 7 o r C 5


Module 3

x

y

10

12x

– 5y

+ 5

7 =

0

12x

– 5y

+ 5

= 0

} }

22

Example 5Find the equation of a circle with centre (–2, 4) and tangent tothe line x + y – 10 = 0

Solution

C(–2, 4) and tangent to x + y – 10 = 0.

The perpendicular distance from the centre to the tangentline is the radius.

Equation of circle:


Module 3

x y1 12 4 10= − = = = = −, , A 1, B 1, C

radiusA B C

A B=

+ +

+

=− + + −

+

=− + −

=−

=

x y1 1

2 2

2 2

1 2 1 4 10

1 12 4 10

28

282

x h y k r

x y

x y

− + − =

− − + − =

+ + − =

2 2 2

2 22

2 2

2 482

2 4 32

( ) ( )( ) ( ) ( )

( ) ( )

( ) ( )

( )( )

)(


1. Calculate the distance between the following pairs of points:a) A(4, 6) and B(6, 5)

b) C(–4, –2) and D(2, 2)

2. Find the distance between the point and the line:

Point Linea) (2, 3) 4x + 3y = 10

b) (–2, –1) y = x – 2

c) (6, 2) x + 1 = 0

d) (0, 8) 6x + y = 0

3. Find the midpoint between A(3, –4) and B(–15, 2).

4. The lines y = 3x + 1 and y = 3x – 9 are parallel. Determinethe vertical distance between the two lines, the horizontaldistance between the two lines, and the shortest distancebetween the two lines.

5. Find the distance between: (If the distance to be found is notspecified, it is assumed to be the shortest distance.)

a) 2x – 3y = 6 and 2x – 3y = 9

b) 3x – 4y – 12 = 0 and 3x – 4y – 6 = 0

6. In a triangle whose vertices are A(–1, 2), B(2, 5), and C(3, –1), find

a) the length of the altitude from A to BC

b) area of the triangle

7. Find an equation of the line that is parallel to 3x – 4y = 6 but5 units away.

8. A line is parallel to x – 4y = 7. Find its equation if it is3 units away from (4, 1).


Module 3

9. Given ∆ ABC with vertices at A(5, 4), B(7 –2), and C(–3, 4):

a) Find the distance between the midpoints of sides AC andBC.

b) Find the length of AB. What is the relationship betweenthe length of the line segment between the midpoints andAB.

c) Find the length of the median drawn from C.(Remember, a median is a line drawn from a point to themidpoint of a line segment.)



Module 3

Module 3

Lesson 3

Systems of Linear Equations

Outcomes


• find the solution of a pair of linear equations in two variablesgraphically

• model situations using systems of equations

Overview

A system of linear equations is a set of two or more linearequations with the same variables. The solution of the system isthe set of all ordered pairs that make all equations true.

One way to find the solution to, or “solve,” a system of equations isto look for the points where all of the equations intersect. Thecoordinates of the intersection points solve the system. Thismethod called solving by graphing is shown for a system ofequations:

Example 1

Solve the system of equations y = 3x – 2 and y = –4x + 5.

Find the point(s) of intersection of the graphs of the following bygraphing.

SolutionIf you do not have a graphing calculator, you'll have to do this byhand.

Step 1: Graph y = 3x – 2.

Step 2: Graph y = –4x + 5. y-intercept is 5 so plot the point(0,5). Slope is –4/1 from (0,5)move 1 unit to the right and 4units down to the point (1,1).Draw the line joining (0,5) and(1,1).

y-intercept is –2 so plot thepoint (0,–2). Slope is 3/1 so from(0,–2) move 1 unit to the rightand 3 units upwards to thepoint (1,1). Draw the linejoining (0,–2) and (1,1).


Module 3

Clearly, there isone point ofintersection,namely (1,1). So,the system ofequations hasexactly onesolution: x = 1 andy = 1.

Now if you do have a graphing calculator…

To graph a function you use the blue key located on theleft below the screen. You will see a screen with

Plot 1 Plot 2 Plot 3\y1=\y1=\y3=

etc.

The cursor will be beside Y1=

Note: If the Plot1 or other plots are darkened then thestatistical plots are on and you won’t be able to graph. Scroll upto the darkened plot using the arrow keys, and press ENTER .Then scroll back down to Y1=. This should turn off the plot

and the plots will no longer appear darkened.

If you want to graph a function type in its equation:

Example 1: Graph y = 3x – 2

Enter at the cursor beside y1= 3 x,t,θ,n – 2 ENTER

To set the standard viewing window press ZOOM 6

The graph will appear on the screen.

Y=


5

–2

y

x

Point of Intersection(1,1)

y = 3x – 2

y = –4x + 5

On the same screen we will also graph y = –4x + 5. Enter thefunction at y2=. Be careful to use the (-) key for the negative.After entering and pressing ENTER press GRAPH (far right ofthe top line).

To find where the two lines intersect, press 2nd TRACE for theCALC option and choose option 5: intersect by scrolling down toit or typing in 5. The graphs are shown and the display showy1 = 3x – 2 at the top and the words first curve? at the bottom.Press ENTER to indicate that this is the first curve you wish toconsider. It then shows the second equation and asks forconfirmation — press ENTER . It asks you for a guess. If thecursor is not close to the point of intersection move it until it isand press ENTER . The intersection is shown at the bottom ofthe screen: x = 1 y = 1.

Find the point of intersection for the following equations usingyour graphing calculator.

y1 = 3x – 2

y2 = – 4x + 5

Solution

Press and enter the equations y1 and y2 into the function

register.

Press (CALC) and choose 5: Intersect.

The point of intersection to this system is (1, 1).

The following exploration involves three types of systems oflinear equations. For each type of system, graph each pair oflinear equations in the same coordinate plane.

Example 2

Find the point(s) of intersection of the graphs of the following:

− + =

= −

6 10 18 0 Equation 15

3 Equation 23

y x

y x

2nd

Y=


Module 3

Solution

First we put Equation 1 into slope y-intercept form (y = mx + b)

We can see that this is the same line as defined by Equation 2.Hence, there would be an infinite number of points ofintersection. This system of equations, therefore, is said to havean infinite number of solutions.

This line is the graphicalrepresentation of:

Example 3Solve the following system graphically:

+ − =+ + =

3 4 12 0 Equation 16 8 16 0 Equation 2

x yx y

− + =

= −

6 10 18 0

and5

33

y x

y x

− + == −

−=

= −

6 10 18 06 10 18

10 186

53

3

y x

y x

xy

y x


Module 3

y

x

(3,2)

Solution

In y = mx + b form we have:

When we graph these lines on the same grid we can see thatthey are parallel and, therefore, have no points of intersection.So this system has no solution.

The previous three examples show that a system of two linearequations in two variables can have exactly one solution,infinitely many solutions, or no solutions at all. Systems areclassified based on their number of solutions. Theseclassifications (Independent, Dependent, Inconsistent) will bediscussed next.

= − +

= − −

33 Equation 1

43

2 Equation 24

y x

y x


Module 3

y

x

3

3x + 4y – 12 = 0

6x + 8y + 16 = 0

Type I: Independent System

System: 2x – y = –2x + y = – 4

In slope intercept form:

y = 2x + 2y = – x – 4

Graph:

Characteristics:• different slopes• can have the same or different y-intercepts• intersecting lines• one unique solution which is at the point of intersection

Type II: Dependent System

System: x – y = 13x – 3y = 3

In slope intercept form:

y = x – 1y = x – 1

l

l

l

x

y

y = x 4

y =

2x

+ 2

( 2, 2)


Module 3

Graph:

Characteristics:

• same slope• same y-intercept• same line• infinitely many solutions

Type III: Inconsistent System

System: 3x + 2y = 63x + 2y = –6

In slope-intercept form:

Graph:

l

l

x

y

l

l3x + 2y

= 6

3 x + 2y = 6

y x

y x

=−

+

=−

−

32

3

32

3

l

lx

y

y =

x

1


Module 3

Characteristics:

• same slope• different y-intercept• parallel lines• no solution, since the lines do not intersect

Systems of equations can be categorized according to whetherthey have solutions or not.

1. A consistent system of equations has at least one solution.

a) An independent system has a unique solution.

b) A dependent system has an infinite number of solutions.

2. An inconsistent system of equations has no solution.

Check the characteristics of each type of system as they areuseful in deciding which type of system you have.

Example 4

Check that (–2, 3) is a solution of the system.

x + 2y = 4 Equation 13x – 2y = –12 Equation 2

To check (–2, 3) as a solution, substitute –2 for x and 3 for y intoeach equation.

Equation 1, check:

Equation 2, check:

Because the coordinates satisfy both equations, the point (–2, 3)is the solution.

After finding the solution of a system of linear equations isfound graphically, check the solutions algebraically.

A linear system can be written to model a specific situation.

3 2 12

3 2 2 3 12

6 6 1212 12

x y− = −

− − = −

− − = −− = −

x y+ =− + =

=

2 4

2 2 3 4

4 4


Module 3

( )

( )( )

Example 5During a basketball game, your friend scored two 3-point shotsbut could not remember how many free throws (worth one pointeach) and field goals (worth 2 points each) she scored. Thescorekeeper said she scored 20 times for 34 points. How manyfield goals did she make? How many free throws did she make?

Solution

Number of free throws + Number of field goals + Number of 3point shots = 20

∴ Number of free throws + Number of field goals = 18, wherethe number of 3-point shots is subtracted from 20 (that is, 20 – 2 = 18)

Let x = number of free throws

y = number of field goals

∴ x + y = 18

1 (number of free throws) + 2 (number of field goals) + 2 (number of 3-point shots) = 34 points

∴ 1x + 2y = 28, because you can subtract 2 (3) or 6 points from34 for the 3-point shots

System of equations: x + y = 18x + 2y = 28

Solve the system using your graphing calculator (if you haveone) and check your answer algebraically.

Enter the following equations into the function register.

Press (CALC) and choose 5: Intersect.

From the graph: x = 8 free throws

y = 10 field goals

Check: x + y = 18 x + 2y = 288 + 10 = 18 8 + 2(10) = 28

2nd

= − +

− +=

1

2

18

282

y x

xy


Module 3

If you don't have a graphing calculator, you'll have to sketch thegraphs manually. The first line y = –1x + 18 will pass throughpoints (0,18) and (1,17), using the y-intercept and slope to locatetwo points. In a similar manner you can determine that thesecond line

If you sketch these lines carefully you will be able to locate thepoint of intersection at (8,10).


1. Match the linear system with its graph. (If you don’t have agraphing calculator, sketch graphs manually to answer thequestions.) How many solutions does the system have? Namethe type of system.

2. How many solutions does the linear system have? Nameeach type of system. Hint: change to y = mx + b form andcompare slopes and y-intercepts.

c) x yx y

+ =− + =

4 82 10

b) x yx y

− = −− + = −

3 32 6 30

a) 24 9 38 3 1

x yx y

− =− =

22

2

4

6

246 x

y

1

2

1

2

123 x

y

12

2

2

4

424

4

x

y

c) 2 54 2 10

x yx y

− =− + = −

b) − + =− =

2 3 122 3 6

x yx y

a) 2 52 0

x yx y

− = −+ =

= +1

– 14 will pass through the points (0,14) and (2,13).2

y x


Module 3

i) ii) iii)

3. Decide whether the ordered pair is a solution of the linearsystem. Check algebraically.

a) b) c)

4. Decide whether the ordered pair is a solution of the linearsystem. Use a graphic check.

a) b) c)

5. Solve each system by graphing and classify the system asbeing independent, inconsistent, or dependent. If the systemis independent, check the solution on both equations toverify that it is a solution.

Note: In some questions, you may have to simplify theequation before attempting to graph.

a) b)

c) d)

For questions 6, 7, and 8, set up the two equations that youwould use to solve the problem. If you have a graphingcalculator, solve the system using the Intersect function. If youdo not have a graphing calculator, wait until Lesson 4 when youwill learn algebraic methods for solving systems of twoequations.

6. Jim is thinking of joining a health club. Club A charges aone-time membership fee of $400 and a monthly fee of $5.Club B does not have a membership fee but charges amonthly fee of $30. After how many months will the chargesfor the two clubs be equal.

x y

y x2 3

1

=

− =

3 1

3 2 3

x y

y x

+ =

= −

y x

x y

= −

= −

2 323

13

2 3

3

x y

y x

+ =− = −

Point (2, –1)Point (4, 4)Point (0, 3)

− + = −= −

x y

y

2 4

1

− + =− + =

x y

x y

2 4

3 4 4x y

x y

+ =+ = −2 8

2 3

Point (5, 2)where (k, m) isthe ordered pair

Point (4, 0)Point (2, –2)

2 3 43 17k m

k m

− =+ =

− + =+ =

x y

x y

6 4

3 124 10

2 3 2

x y

x y

− =+ = −


Module 3

( )

7. A total of $4500 is invested in two funds paying 4% and 5%in annual interest. The combined annual interest is $210.How much of the $4500 is invested in each fund?

8. Your family is planning to open a restaurant. You need aninitial investment of $85 000. Each week your costs will beabout $7400. If your weekly income is $8000, how manyweeks will it take to break even?

9. Describe a technique other than graphing that could be usedto solve the following linear systems.

a) b)


y xx

= −=

2 42

y x

y

= +

=

4

2


Module 3

Module 3

Lesson 4Solving Systems of Equations Algebraically

OutcomesWhen you complete this lesson, you will be able to

• find the solution to a system of equations algebraically

• solve a linear system to answer questions about a real-lifesituation

OverviewIt is difficult to find a precise intersection point for hand-drawngraphs. The solution may be also approximate when you use agraphing calculator.

In this lesson, you will investigate two algebraic methods to findan exact solution to a system of linear equations.

I. Solving by Substitution

Follow these steps to solve a system:

1. Solve one of the equations for one of its variables.

2. Substitute this expression into the other equation and solvefor the other variable.

3. Substitute this value into either equation and solve.

4. Check the solution in each of the original equations.

Example 1Solve by substitution:

4x + y = 1 Equation 1

2x – 3y = 4 Equation 2

Solution

Solve Equation 1 for y in terms of x:y = –4x + 1 Revised Equation 1

Substitute this value for y in Equation 2 and solve for x:

Substitute y = –4x + 1Simplify

2 3 4

2 3 4 1 4

2 12 3 414 7

12

x y

x x

x x

x

x

− =− − + =

+ − ==

=


( )

Module 3

To solve for y, you can substitute x = into any of the above

equations. It is probably easier to use revised Equation 1.

The ordered pair is the solution.

Check in each equation:

When using the substitution method, you will obtain the samesolution (x, y) whether you solve for y first or x first. Thus, beginby solving for the variable that has a coefficient of 1 if possible.

Using the second equation: Using the first equation:

Solve for x first here: Solve for y first here:

If neither variable has a coefficient of 1, you can still use thesubstitution method by writing an equivalent equation thatdoes have a coefficient of 1.

− + =− = −

2 0

3 10

x y

x y

4 3 1

2 3

x y

x y

− =+ =

∴ −12

1, is the solution

2 3 4

212

3 1 4

1 3 4

4 4

x y− =

− − =

+ ==

4 1

412

1 1

2 1 1

1 1

x y+ =

+ − =

+ − =

=

12

1, −

12

1, −

becomesUsing x y x

y

y

y

= = − +

= − +

= − += −

12

4 1

412

1

2 11

,


12

( )( )

( )

Example 2Solve by substitution:

5x + 4y = 6 Equation 1–2x – 3y = –1 Equation 2

Solution

Since none of the x- and y-coefficients are 1, choose one of theequations and solve for one of its variables. Using Equation 1,solve for x in terms of y.

Substitute that value for x in Equation 2.

Substitute y = –1 into any of the original equations:

Solution is (2, –1). Check the original equations:

(2, –1) is the solution

− − = −

− − − = −

− + = −− = −

2 3 1

2 2 3 1 1

4 3 11 1

x y5 4 6

5 2 4 1 6

6 6

x y+ =+ − =

=

5 4 6

5 4 1 6

5 4 65 10

2

x y

x

x

x

x

+ =+ − =

− ===

Multiply by 5 to removedenominator

( ) ( )

( )

2 3 16 42 3 1

56 42 5 5 3 1 5

52 6 4 15 5

12 8 15 57 7

1

x yy y

y y

y yy y

yy

− − = −− − − = −

− − ⋅ − = − − − − = −

− + − = −− =

= −

5 6 46 4

5

x y

x y

= −= −


Module 3

( )

( )( ) ( )( )

Another method is the following:

II. Solving by Addition-Subtraction

Use the following steps to solve a system:

1. Arrange the equations with like terms in columns.

2. Make the coefficients of x or y the same by multiplying eachterm of one or both equations by an appropriate number.

3. Add or subtract the equations and solve for the remainingvariable.

4. Substitute the value obtained in step 3 into either of theoriginal equations and solve for the other variable.

5. Check the solution in each of the original equations.

Example 3Solve:

Solution

Rearrange so that the same variables are in the same columns:

5x + 4y = 6 Equation 1

–2x – 3y = –1 Equation 2

Multiply by 3: 5x + 4y = 6 15x + 12y = 18Multiply by 4: –2x – 3y = –1 –8x – 12y = – 4

7x = 14x = 2

Substitute x = 2 into either Equation 1 or 2.

Solution is (2, –1).

Because this is the same question as in Example 2, you havealready checked its solution.

5 4 6

5 2 4 6

10 4 6

4 4

1

x y

y

y

y

y

+ =+ =

+ == −= −

Add the equations

5 4 6

3 2 1

x y

y x

+ =− − = −


Module 3

( )

Note: We could have multiplied Equation 1 by 2 and Equation 2by 5 obtaining:

Both systems that you solved had exactly one solution. Thismeans they were independent systems. You will now examine alinear system that has no solution.

Example 4

Solve:

x – 2y = 3 Equation 1

–2x + 4y = 1 Equation 2

Multiply Equation 1 by –2.

Notice this results in a false equation. Conclusion: there is nosolution to the system if there is no “true” solution. Rememberthat means you have parallel lines that never intersect. Thesystem is inconsistent.

Subtract− + = −− + =

= −

2 4 62 4 1

0 7

x yx y

x y

x y

− =

− + =

2 3

2 4 1

AddSimplify

10 8 1210 15 5

7 71

2

x yx y

yyx

+ =− − = −

− == −=and


Module 3

Some linear systems have infinitely many solutions as indicatedin the following example.

Example 5Solve:

Clear the fractions by multiplying Equation 2 by 4.

0 = 0 is an equation that is always true. This shows that any (x, y) pair that solves the first equation also makes the secondequation true. Because the equations represent the same line,there are an infinite number of solutions. The system ofequations is dependent.

When solving a system by addition-subtraction, you may need tomultiply each of the equations by a different number before youcan eliminate one of the variables.

Example 6

Solve:

5x + 6y = 24 Equation 13x + 5y = 18 Equation 2

Solution

Subtract:

15 18 7215 25 90

7 18187

x yx y

y

y

+ =+ =

− = −

=

5 6 24

3 5 18

x y

x y

+ =

+ =Multiply by 3:Multiply by 5:

subtract

9 6 489 6 48

0 0

x yx y

+ =+ =

=

9 6 48

3 2 16

x y

x y

+ =

+ =Multiply by 3:

Equation 1

Equation 2

9 6 4834

12

4

x y

x y

+ =

+ =


Module 3

Substitute

Check in both equations.

Example 7Recall question 7 from the self-marking activity for Lesson 3:

A total of $4500 is invested in two funds paying 4% and 5% inannual interest. The combined annual interest is $210. Howmuch of the $4500 is invested in each fund?Let x = part of $4500 invested at 4%

Let y = part of $4500 invested at 5%

x + y = 4500 Equation 1

0.04x + 0.05y = 210 Equation 2

Multiply Equation 1 by 5: 5x + 5y = 22500Multiply Equation 2 by 100: 4x + 5y = 21000

Subtract x = 1500

3 5 1812 183 5 187 7

36 90 187 7

126 18718 18

x y+ = + =

+ =

=

=

5 6 24

5127

6187

24

607

1087

24

1687

24

24 24

x y+ =

+ =

+ =

=

=

The solution is 127

187

, .

183 5 187903 187

21 90 12621 36

36

x

x

xx

x

+ =

+ =

+ ==

=21

127

y = 187

into Equation 2


Module 3

Substitute 1500 in Equation 1.

1500 + y = 4500

y = 3000Check in both equations.

x + y = 4500 0.04x +0.05y= 210

1500 + 3000 = 4500 0.04 (1500) + 0.05 (3000) = 210

4500 = 4500 60 + 150 = 210

210 = 210$1500 at 4% interest.

$3000 at 5% interest

You can try similar methods to solve the systems you wrote forquestions 6 and 8 in Lesson 3.

Example 8Determine A and B so the graph of the given equation willcontain the points whose coordinates are given.

Solution


Module 3

y = Ax2

+ Bx P(1, –1) and Q(–2, –10)

− = +

− = − + −+ = −

− = −+ = −− = −

==

+ = −+ = −

= −

1 A 1 B 1

10 A 2 B 2

A B 14A 2B 104A 4B 44A 2B 10

6B 6B 1

A B 1A 1 1

A 2

2

2

Equation 1 4Equation 2

The equation is y = –2x2 + x.

Eqn (1)

Eqn (1)

Eqn (1)

Eqn (2)

Eqn (2)

Eqn (2)

( )

( ) ( )

( )


1. Use the substitution method to solve the following systems.Check your answers using graphing calculator or bysubstituting in both equations.

a) b) c)

d) e)

2. Use the addition subtraction method to solve the followingsystems.

a) b) c)

d) e) f)

g)

3. Solve each of the following systems choosing the method youconsider most appropriate for the system. Name the systemthat is represented.

a) b) c)

d) e) x yx y

+ =+ =

2 13 6 3

2 3 5

4 6 3

x y

x y

− =

− =

4 3 1

3 6 3

x y

x y

+ =

− − =

x y

x y2 2

32

2 3 1

+ =−

= −

2 3

3 2 6

x y

x y

+ =

+ =

=+ −

= +

3 42 1 3

8 3 10x y

x y

3 23

2

x yy

x yx

+=

−= −

8 3 3

3 2 5

x y

x y

− =

− = −2 3 17 0

9 5 16 0

x y

x y

+ + =

− + + =

3 5 1

10 6 0

y x

x y

+ =

+ =

3 2 4

3

x y

x y

+ =

− =

− + = −

+ =

x y

x y

1

1

4 2 6

2 3

x y

x y

− =

− =

53

23

1

x yy

x y

+=

+=

x yy

y x

+= −

− =

22

14

7

2 5 1

3

y x

x y

− =

+ = −2 3

2 3 6

x y

y x

+ =

− =


Module 3

4. Describe in words how you would use the substitutionmethod to solve the system

5. Describe in words how you would use the addition-subtraction method to solve the system

6. Suppose you used the substitution method to solve a linearsystem and obtained the equation –6 = –6. What would youconclude?

7. Suppose you used the substitution method to solve a linearsystem and obtained the result 10 = 0. What could youconclude?

8. State whether each statement is true or false. If thestatement is false, rewrite it so that it is true.

a) If you solve a system by addition-subtraction orsubstitution, you will find an approximate solution to thesystem.

b) If, while solving a system of linear equations, you are leftwith an equation that can never be true, you know thatthe equations are inconsistent.

c) If, while solving a system of linear equations, you are leftwith an equation that can never be false, you know thatthe system has no solution.

d) An inconsistent system of linear equations has infinitelymany solutions.

e) If the two linear equations in a system represent thesame line, the system is dependent.

f) Graphing a system of equations provides a method offinding the approximate solution of a system of linearequations.

9. In 1986-87, Michael Jordan became the first player to scoreover 3000 points in 24 years. He scored 3005 points on two-point baskets and one-point free throws. If he made a total of1919 two point baskets and free throws, how many freethrows did he make?

6 5 30

4 2 7

x y

x y

− =

− + =

12 4 1

3 9

x y

x y

+ =

− =


Module 3

10. Solve and simplify the following system of equations2(x – y) – 3(x + y) = –13 and 5 – 2(2x – y) = 3(x – 2y).

11. The paths of two ships are given by Ship A: x + y = 8 andShip B: x – y = 4. The paths of the two ships intersect at anisland. What are the coordinates of the location of theisland.

12. Determine A and B so that the graph of the given equationwill contain the points whose coordinates are given:

Ax + By = 10; P(2, –1) and Q(6, 2)



Module 3


Module 3

Notes

Module 3

Lesson 5

Systems of Equations Containing Three Variables

Outcomes


• solve a system of three linear equations in three variables

• model situations using systems of linear equations containingthree variables

Although the systems of equations that you have examined sofar have had two equations and two variables (known as a 2 x 2system), many important problems feature multiple variablesand equations.

Any equation of the form Ax + By + Cz = D where A, B, C, and Dare real numbers with A, B, C not all zero is called a linearequation in three variables, x, y, and z.

In a 2 x 2 system, the graph of Ax + By = C on a coordinateplane is a line. In a 3 x 3 system (3 equations with 3 variables),the graph of Ax + By + Cz = D is a plane in coordinate space.Coordinate space has three axes all perpendicular to each other.The solution of the system is the intersection of three planes.

There are four possibilities:

a) They might intersect at a common point called an orderedtriple (x, y, z).

b) They might intersect in a common line.

c) They might have no points in common.

d) They might coincide.


(a) A uniquesolution P(x, y, z)

(b) A solution ofpoints on a line

(c) No solution (d) A solution of allpoints on a plane

Module 3

To solve a linear system of three equations with three variablesuse elimination by addition/subtraction or substitution. In eachcase, your first goal is to reduce the system to two equationswith two variables. Then you can solve this reduced system withthe familiar technique used with 2 x 2 systems.

Example 1Solve this system for x, y, and z.

x + y – z = 2 .................................. (1)

x – 2y + z = –1 .............................. (2)

3x + y – 2z = 4 ............................. (3)

Solution

Working with two equations at a time, eliminate one of thevariables. Let us choose to eliminate z from Equations 1 and 2by adding them.

Eqn. (1) + Eqn. (2): 2x – y = 1 ................................... (4)

Now eliminate z from Equations 2 and 3.

Eqn. (2) x 2: 2x – 4y + 2z = –2 ....................... (5)Eqn. (3) + Eqn. (5): 5x – 3y = 2 ................................. (6)

Equations 4 and 6 contain only x and y. Now we can subtractthese to eliminate y and solve for x.

Eqn. (4) x 3: 6x – 3y = 3 ................................. (7)

Eqn. (6) – Eqn. (7): –x = –1

Therefore, x = 1.

Substitute x = 1 into Eqn. (4): 2(1) – y = 1

∴ y = 1

Substitute x = 1 and y = 1 into Eqn. (1): 1 + 1 – z = 2

∴ z = 0

The solution is (1, 1, 0).

You should check this solution in all three original equations.


Example 2Is (–1, 4, 2) a solution of the equation x + 2y – 3z = 1?

Solution

Substitute x = –1, y = 4, z = 2 into x + 2y – 3z = 1.

This is a true statement, so it is a solution.

Example 3Solve:

Solution

Eliminate y using Equations 1 and 2.

Eliminate y using Equations 2 and 3.

Eliminate z using Equations 4 and 5.

− − = −+ =

==

3 3 3 Eqn. (4)

5 3 9 Eqn. (5)Add2 6

3

x z

x zx

x

+ + =− + =

+ =

2 8 Eqn. (2)3 2 1 Eqn. (3) Add

5 3 9 Eqn. (5)

x y z

x y z

x z

+ − =+ + = ×

− − = −

2 13 Eqn. (1)4 2 2 16 Eqn. (2) 2

3 3 3 Subtract to get Eqn. (4)

x y z

x y z

x z

+ − =+ + =

− + =

2 13 Eqn. (1)

2 8 Eqn. (2)3 2 1 Eqn. (3)

x y z

x y zx y z

− + − =− + − =

=

1 2 4 3 2 1

1 8 6 1

1 1


Module 3

( ) ( )

Substitute x = 3 into Equation 4.

Substitute x = 3 and z = –2 into Equation 1 (or 2 or 3).

Example 4Solve:

Solution

Eliminate x using Equation 1 and 2.

This statement is never true so the system has no solution andis inconsistent.

Example 5

The sum of three numbers is 12. The second number is equal tothe sum of the first and third numbers. Twice the first numberexceeds the second number by 4. Find the numbers.

Solution

Let x = 1st numbery = 2nd numberz = 3rd number

Equation 1Equation 2Equation 3

x y zx y z

x y

+ + =− + − =

− =

120

2 4

x y z

y x zx y

+ + =

= +− =

12

2 4

+ − = ×+ − =

=

4 2 6 8 Eqn. (1) 24 2 6 7 Eqn. (2)

0 1 Subtract

x y zx y z

+ − =+ − =

+ − =

2 3 4 Eqn. (1)4 2 6 7 Eqn. (2)

2 3 5 Eqn. (3)

x y zx y z

x y z

+ + == = −

3 2 2 133 8 4 (3,4, 2)

yy y

− − = −− =

= −

3(3) 3 33 6

2

zzz


Module 3

Equation 1 + Equation 2

The numbers are (5, 6, 1).

Example 6

Find the equation of a parabola that has a vertical axis ofsymmetry and passes through the points (0, –4), (1, 1), and (2, 10).

Solution

Equation 4

Substitute y = 6 into Equation 3

Substitute x = 5, y = 6 into Equation 1

2 126

2 6 42 10

55 6 12

11 121

yy

xxxzzz

==

− ===

+ + =+ =

=


Module 3

The equation of a parabola with vertical axis of symmetry iswritten in the form

y = ax2

+ bx + c. Set up the three equations.

For (0, –4): –4 = a • (0)2 + b(0) + c

–4 = c ---------------------------------------- (1)

For (1, 1): 1 = a(1)2 + b(1) + c

1 = a + b + c ------------------------------- (2)

For (2, 10): 10 = a(2)2 + b(2) + c

10 = 4a + 2b + c ------------------------- (3)

(4)

(5)(6)Eqn (4) x 2:

Eqn (5) – Eqn (6):

Substitute c = –4 into Eqns (2) and (3): 1 45

10 4 2 414 4 210 2 2

4 22

= + −= += + −= += +==

a ba b

a ba ba ba

a


1. Which of the given ordered triples is a solution of the givenequation.

a) x + y – z = 2; (1, –2, –3), (0, –2, –4)

b) x – y + 2z = 3; (1, –1, –1), (2, –3, –1)c) x + 4y + 2z = 0; (0, 0, 4), (0, –2, 8)

d) 5x + y – z = 0; (–1, –1, 6), (2, –6, 4)

2. Evaluate the following expression:

2x + 3y + 3z when x = 2, y = –2, z = 5

3. Which values should be given to a, b, and c so that the linearsystem has (–1, 2, –3) as its only solution.

4. Solve the following systems:

a) b)

c)

5. The equation of a circle may be written as x2 + y2 + Dx + Ey + F = 0. Recalling that the coordinates ofany point on the circle must satisfy its equation, find thevalues of D, E, and F and thereby the circle which passesthrough (1, 1), (–2, 3), and (3, 4).

x y z

x y zx y z

+ + =

+ + =− + =

3 12

2 3 134 11

3 4 5 2

4 5 3 55 3 2 11

x y z

x y zx y z

− + =

+ − = −− + = −

3 2 2 11

2 12 2 4

x y z

x y zx y z

+ − =

+ + = −− + = −

x y z ax y z b

x y z c

+ − =− − + =+ − =

2 3

2 3 2


Module 3

∴ The equation is y = 2x2

+ 3x – 4.

Substitute a = 2, c = –4 into Eqn (2): 1 2 41 23

= + −= −=

bbb

6. Find A, B, and C so that the solution set of the equation

Ax + By + Cz = 1 will contain the given ordered triples:(4, 1, 2), (3, 2, 1), (–6, –1, –2).

7. The total revenue, R, is a quadratic function of the price, p,of books sold (R = ap2 + bp + c). Find the values of a, b, and cif the revenue is $6000 at a price of $30, $6000 at a price of$40, and $5000 at a price of $50.


Review Section 3 before attempting the review questions beginningon the next page. These questions should help you consolidate yourknowledge as you prepare Section Assignment 3.3.


Module 3


Module 3

Review

1. Find the centre and radius of the following circle:

2. Write the equation of the following circles:

a) Diameter with endpoints (–6, 5) and (4, –2)

b) Circle with radius 5 crosses the x-axis at (4, 0) and (10, 0)

c) Centre (2, –1) and circumference 10π

3. The two small circles have equations (x – 4)2 + y2 = 9 and(x – 10)2 + y2 = 9. Find the equation of the large circle.

4. A ship travels on a route represented by the line 2x – 2y + 7 = 0. A lighthouse is situated at point (5, –4). Ifthe lighthouse can be seen anywhere within a radius of10 km, will the ship see the light?

5. Given ∆ ABC with vertices at A(5, 4), B(7, –2), and C(–3, 4),find:

a) distance between the midpoints of AC and BC

b) length of median from C

c) length of altitude from B to ACd) area of triangle ABC

6. a) Find the distance between 3x – 4y = 12 and 3x – 4y = 24.

b) Find the horizontal distance between the two lines.

c) Find the vertical distance between the two lines.

+ + − − =2 2 12 4 24 0x y x y

Principles of Mathematics 11 Section 3, Review 189

Module 3

7. Solve by substitution and name the type of system.

8. Solve graphically and name the type of system.

9. Give an example of a system that is dependent.

10. Determine A and B so that the graph of y = Ax2

+ Bx willcontain the points P(–1, 5) and Q(2, 2).

11. Solve the following system.


Now do the Section Assignment which follow this section.

4 3 73 2 3 2

2

x y zx y z

x y z

+ − = −− + = −

+ − = −

2 3 62 3 12

x yx y

− = −− =

2 3 42 4

x yy x

+ = −− =

190 Section 3, Review Principles of Mathematics 11

Module 3

PRINCIPLES OF MATHEMATICS 11

Section Assignment 3.3

Module 3

Principles of Mathematics 11 Section Assignment 3.3 191

192 Section Assignment 3.3 Principles of Mathematics 11

Module 3

Module 3

Section Assignment 3.3

Analytical Geometry

1. Find the centre and radius of the circle

x2 + y2 – 6x + 4y + 7 = 0.

2. Write the equation of the circle having a diameter withendpoints (2, 1) and (4, –3).


Total Value: 40 marks(Mark values in brackets)

Use a scientific calculator and/or agraphing calculator if you have one

(2)

(3)

Module 3

3. Write the equation of a circle with centre (–2, 3) and acircumference of 20π units.

4. For the lines 2x + 3y = 6 and 2x + 3y = 12, find

a) horizontal distance between the lines

b) vertical distance between the lines


(2)

(1)

(1)

c) the shortest distance between the two lines


Module 3

(3)

5. In a triangle whose vertices are A (–2, 1), B (5, 2), andC (3, 7),

a) find the length of the altitude from A to BC

b) find the area of the triangle.


Module 3

(3)

(3)

6. a) Solve the system of equations graphically. Name the typeof system.

x – y = 1

2x + y = –4

b) Solve the system algebraically. Show steps.

x

y


Module 3

(3)

(3)

7. Solve the following systems of equations algebraically. Namethe type of systems.

a)

b) − + = −− =

4 6 86 9 12

x yx y

x yy

y x

+= −

− =

22

14

7


Module 3

(4)

(3)

8. John invests a total of $4500, one part of it earning simpleinterest at the rate of 5% and the remainder at 6%. His totalinterest for one year is $250. How much money did he investat 5% and how much at 6%?


Module 3

(4)

9. Solve for x, y, and z.

− + =+ − =− + = −

3 2 2 12 5

4 3 3

x y zx y zx y z


Module 3

(5)

(Total: 40)

Analytic Geometry I - Open School BCmedia.openschool.bc.ca/ocrmedia/pmath11/pdf/pma11... ·...

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