Unit 16
THE CARTESIAN COORDINATE
SYSTEM AND GRAPHS OF
LINEAR EQUATIONS
GRAPHING A LINEAR EQUATION


The graph of a straight line (linear) equation has an
infinite number of points that satisfy the equation
Procedure for graphing an equation:
1.
2.
3.
4.
5.
Write the equation in terms of either variable (x or
y)
a. Make a table of values. Head one column x
and the other column y
b. Select a least three convenient values for one
variable and find their corresponding values for
the second variable by substituting the first
variable values in the equation
Draw a label the x axis and the y axis on the
graph
Plot the coordinates on the graph
Connect the points. The straight line is the graph
of the equation
2
GRAPHING EXAMPLE

Graph the equation: 2x – y
=5
–
Write the equation in terms of x
or y. We will use y
–
–
Make a table of values that
satisfy the equation by
choosing a value for x and
finding the corresponding y
value
Graphing is shown on the next
slide
y = 2x – 5
x
y
0
-5
1
-3
3
1
3
EXAMPLE GRAPH
• Graph of the equation 2x – y = 5
• Draw and label the x and y axes.
• Plot the points on the graph.
• Connect the points
4
SLOPE OF A LINEAR EQUATION


The slope of a linear equation (a straight line)
refers to its steepness or inclination
Slope is defined as follows:
rise change in y
slope 

run change in x
5
SLOPE OF A LINEAR EQUATION

If a line rises moving from left to right, it
has a positive slope

If a line falls moving from left to right, it
has a negative slope

A horizontal line has a zero slope

A vertical line has an undefined slope
Positive
slope
Negative
slope
6
DETERMINING SLOPE

Determine the slope of the line that passes through the
points (1,3) and (–1,2):
change in y 3  2 1
slope 

 Ans.
change in x 1   1 2

Determine the slope of the line that passes through the
points (–2,4) and (3,–3):
change in y 4  3 7
7
slope 



Ans.
change in x 2  3 5 5
7
SLOPE-INTERCEPT FORM




The equation of the line can be determined if
the slope and the y-intercept of a straight line
are known
The y-intercept is the y-coordinate where the
line intercepts (crosses) the y axis
The value of the x at this point is 0
The general equation for the slope-intercept
of a straight line is:
y = mx + b
where m is the slope and b is the y-intercept
8
SLOPE-INTERCEPT FORM

Determine the equation of a line when the
slope is 2 and the y intercept is –5



Given: m = 2 and b = –5
Substituting these values into the slope-intercept
form: y = mx + b
Becomes y = 2x – 5 Ans
9
DETERMINING THE EQUATION OF
A LINE GIVEN TWO POINTS

If two points on a straight line are known,
an equation can be determined


Step 1: Find the slope of the line
Step 2: Find the y-intercept
10
DETERMINING THE EQUATION OF
THE LINE

Find the equation of a line that passes through (1,9)
and (–2,3).
93
6
slope 
 2
1  2 3

Find the slope:

Find the y-intercept. Use either point since they both lie on
the line
Substitute the point (1,9) (can use either point) and m = 2,
into the slope-intercept form
Solve for b y = mx + b becomes 9 = 2(1) + b; b = 7



Substitute the slope and y-intercept into the slope-intercept
form:
y = 2x + 7 Ans
11
POINT-SLOPE FORM


The equation of the line can be determined if
the slope and one point on the line are known
The general equation for the point-slope
equation of a line:
y  y1  m( x  x1 )

Where m is the slope and (x1,y1) is a point on
the line.
12
POINT-SLOPE FORM


Determine the equation of a line when its
slope is 2 and it passes through the point (2,–
5)
m = 2, x1 = 2, and y1 = –5
x  2  2( y  5)
x  2  2 y  10
x  2 y  12 Ans
13
Application

You rent a car from Rentus Car Co. They tell
you that the fee is $30 and then $.05 per mile
you drive during the rental.

What is the equation for the car rental?



$30 is a flat fee paid once…like a y-intercept….it is
where your bill will start if you never drive the car
$.05 is how much more you will pay for every mile you
drive…..like a slope….climbing
Y = .05x + 30
14
Application

You rent a car from Rentus Car Co. They tell you
that the fee is $30 and then $.05 per mile you drive
during the rental.

How much will it cost to drive 300 miles?
 You know that x is the miles driven and y is the price from
the original equation.
y  .05(300 )  30
y  45

So $45 for the car.
15
Application

You rent a car from Rentus Car Co. They tell you
that the fee is $30 and then $.05 per mile you drive
during the rental.

How much can you drive for $65?
 You know that x is the miles driven and y is the price from
the original equation.
65  .05 x  30
35  .05 x
700  x

So 700 miles
16
Application

You have found another company, Rentalot, that
you can rent from and they charge $20 and then
$.10 per mile.


What is the equation?
y  0.1x  20
What will it cost to drive 300 miles?


$50
So you can see this is more than Rentus was. So where is
the point that you would want to change companies?
17
Application

When you are looking for the point where a deal
switches companies, you are looking for an
intersection point. So where will that cost be the
same?


Remember cost is y…so when will y be the same?
We don’t care what that number is right now but it must be
the same…..so we have two equations.
y  .05 x  30 and y  0.1x  20

So if we want y to be the same in both equation we can
substitute.
0.05 x  30  0.1x  20
10  .05 x
200  x
18
Application

So at 200 miles is where the companies are even in
price. From the examples before we know that less
than 200 we will rent from Rentalot and for more than
200 miles Rentus to always get the best deal.
19
PRACTICE PROBLEMS

Graph the linear equations for problems 1
and 2 by plotting points:
1.
2.

2x – y = 6
6x + 3y = 15
Determine the slope of the lines that pass
through the points given in problems 3 and
4:
3.
4.
(–2, –3) and (4,5)
(5, –7) and (–3, –2)
20
PRACTICE PROBLEMS (Cont)

Write the equation of the line for problems 5
and 6 given the slope (m) and the y
intercept (b):
5.
6.

m = –2; b = 1
m = 2/3; b = –4
Find the equation for each of the lines
passing through the points given in
problems 7 and 8:
7.
8.
(–1,4) and (3,6)
(0, –6) and (10, –2)
21
PRACTICE PROBLEMS (Cont)
Two cell companies have plans that you are
debating over. The first is $50 for the phone
and $.01 per minute you talk a month. The
second is $40 per month and $.05 per
minute.

9.
10.
11.
What is the equation for each company?
How long can I talk on each plan if I only can
afford $60 per month?
How many minutes can I talk on both plans and
pay the same amount?
22
PROBLEM ANSWER KEY
1.
2.
23
PROBLEM ANSWER KEY
3.
4.
5.
6.
7.
8.
9.
10.
11.
4/3
–5/8
y = –2x + 1
y = 2/3x – 4
y = –x + 9/2
y = 2/5x – 6
y = 0.05x + 40, y = 0.01x + 50
400 and 1000 minutes
250 minutes
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