Pearson
Foundations and Pre-calculus Mathematics 10
Chapter 6
Linear Functions
Practice Test
(page 391)
1. For each line, count squares on the grid to determine the rise and the run.
rise
Use slope =
run
2
4
Slope of AB =
, or 
3
6
6
2
Slope of CD = , or
9
3
3
6
Slope of EF =
, or 
2
4
6
3
Slope of GH = , or
4
2
3
The line with slope  is EF, which is choice C.
2
2. Write the equation 2x – 3y + 2 = 0 in slope-intercept form.
2x – 3y + 2 = 0
Subtract 2x and 2 from each side.
–3y = –2x – 2
Divide each side by –3.
2
2
y=
x–
3
3
2
2
y= x+
3
3
2
The graph of this equation has slope .
3
2
From question 1, the line with slope
is CD, which is choice B.
3
3
a)
i) y = 
3
x+5
2
Since the equation is in slope-intercept form, I use the slope 
3
and the y-intercept
2
5 to graph the line.
3
, then move 3 squares down
2
and 2 squares right and mark a point. I draw a line through the points.
I mark a point at 5 on the y-axis. I write the slope as
1
(x + 2)
3
This equation is in slope-point form.
ii) y – 3 =
Practice Test
Copyright  2011 Pearson Canada Inc.
1
Pearson
Foundations and Pre-calculus Mathematics 10
Chapter 6
Linear Functions
To identify the coordinates of the point, I write the equation as:
1
y – 3 = (x – (–2))
3
A point on the line has coordinates (–2, 3), and the slope of the line is
1
.
3
1
to move 1 square up and 3 squares
3
right and mark another point. I draw a line through the points.
I mark a point at (–2, 3), then use the slope
iii) 3x – 4y – 12 = 0
Since the coefficients of x and y in this equation are factors of 12, I can use intercepts
to graph the equation.
For the x-intercept, substitute y = 0.
3x – 4(0) – 12 = 0
3x – 12 = 0
Solve for x. Add 12 to each side.
3x = 12
Divide each side by 3.
x=4
For the y-intercept, substitute x = 0.
3(0) – 4y – 12 = 0
–4y – 12 = 0
Solve for y. Add 12 to each side.
–4y = 12
Divide each side by –4.
y = –3
I mark a point at 4 on the x-axis and a point at –3 on the y-axis. I draw a line through
the points.
b) The graph of the equation y = 
will also have slope 
3
3
x + 5 has slope  , so any line parallel to this line
2
2
3
.
2
3
and passes through the point with coordinates (6, 2).
2
Use the slope-point form of the equation:
y – y1 = m(x – x1)
3
Substitute: y1 = 2, m =  , and x1 = 6
2
3
y – 2 =  (x – 6)
2
I know my equation is correct because the slope of the graph of this equation is equal to
The required line has slope 
Practice Test
Copyright  2011 Pearson Canada Inc.
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Pearson
Foundations and Pre-calculus Mathematics 10
Chapter 6
Linear Functions
the slope of the given parallel line, and the coordinates of the point on the graph given by
this equation are equal to the given coordinates.
1
1
(x + 2) has slope . The perpendicular line has a
3
3
1
slope that is the negative reciprocal of ; that is, its slope is –3.
3
The required line has slope –3 and passes through the point with coordinates (–1, 2).
Use the slope-point form of the equation:
y – y1 = m(x – x1)
Substitute: y1 = 2, m = –3, and x1 = –1
y – 2 = –3(x – (–1))
y – 2 = –3(x + 1)
Write this equation in general form. Remove brackets.
y – 2 = –3x – 3
Add 3x and 3 to each side.
3x + y + 1 = 0
This equation is in general form.
c) The graph of the equation y – 3 =
d) To determine the coordinates of a point P on the graph of 3x – 4y – 12 = 0, substitute a
value for x, then solve the equation for y.
Since the coefficient of y is 4 and the constant term, 12, is a multiple of 4, choose a value
of x that is a multiple of 4 so the value of y is an integer.
Substitute x = 8 in the equation:
3x – 4y – 12 = 0
3(8) – 4y – 12 = 0
Simplify.
24 – 4y – 12 = 0
Collect like terms.
12 – 4y = 0
Subtract 12 from each side of the equation.
–4y = –12
Divide each side by –4.
y=3
The coordinates of P are (8, 3).
The line passes through P and Q(1, 5).
Determine the slope of PQ.
change in y -coordinates
Use slope =
change in x-coordinates
53
Slope =
1 8
2
2
Slope =
, or 
7
7
Substitute the coordinates of Q and the slope in the slope-point form of the equation of a
line:
y – y1 = m(x – x1)
2
Substitute: y1 = 5, m =  , and x1 = 1
7
2
y – 5 =  (x – 1)
Solve for y. Add 5 to each side.
7
2
y =  (x – 1) + 5
Remove brackets.
7
2
2
y=  x+
+5
7
7
Practice Test
Copyright  2011 Pearson Canada Inc.
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Pearson
Foundations and Pre-calculus Mathematics 10
2
x+
7
2
y=  x+
7
y= 
2
35
+
7
7
37
7
Chapter 6
Linear Functions
This equation is in slope-intercept form.
4. a) Since I can identify the slope of the line and its y-intercept from the graph, I will write the
equation in slope-intercept form: y = mx + b
I count squares on the grid to determine the rise and the run.
rise
Slope =
run
4
Slope =
, or –2
2
From the graph, the y-intercept is –2.
So, an equation is: y = –2x – 2
b) The line is horizontal, so its equation is: y = –1
In general form, this is: y + 1 = 0
c) Since I can calculate the slope and identify the coordinates of a point on the line, I shall
use the slope-point form: y – y1 = m(x – x1)
The coordinates of two points on the line are (–1, –2) and (3, 1). I count squares on the
grid between these points to determine the rise and the run.
rise
Slope =
run
3
Slope =
4
3
Substitute the coordinates of the point (3, 1) and the slope
in the equation
4
y – y1 = m(x – x1)
3
So, an equation is: y – 1 = (x – 3)
4
5. The cost, C dollars, is a function of the number of people, n, who attend.
Use the given information as the coordinates of two points on a sketch of a graph of the
function.
Plot these points: (600, 11 250) and (400, 7650)
Practice Test
Copyright  2011 Pearson Canada Inc.
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Pearson
Foundations and Pre-calculus Mathematics 10
Chapter 6
Linear Functions
The cost per person is the slope of the line through the points on the graph.
change in C -coordinates
Slope =
change in n-coordinates
11 250  7650
Slope =
600  400
3600
Slope =
200
Slope = 18
The cost per person is $18.
Use the slope-point form of the equation:
C – C1 = m(n – n1)
Substitute: C1 = 7650, m = 18, and n1 = 400
C – 7650 = 18(n – 400)
This is an equation of the function.
a), c) To determine the cost for 340 people to attend, substitute n = 340 in the equation:
C – 7650 = 18(n – 400)
C – 7650 = 18(340 – 400)
Solve for C.
C – 7650 = 18(–60)
C – 7650 = –1080
C = 7650 – 1080
C = 6570
The cost for 340 people to attend is $6570.
b), c) To determine the number of people who can attend for a cost of $9810, substitute
C = 9810 in the equation:
C – 7650 = 18(n – 400)
9810 – 7650 = 18(n – 400)
Solve for n. Remove brackets and simplify.
2160 = 18n – 7200
9360 = 18n
n = 520
For a cost of $9810, 520 people can attend.
Practice Test
Copyright  2011 Pearson Canada Inc.
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