Section 1.3
Slope of a Line
Introduction
C o m p a r i n g t h e S t e e p n e s s o f Tw o O b j e c t s
Two ladders leaning
against a building.
Which is steeper?
We compare the vertical
distance from the base of
the building to the
ladder’s top with the
horizontal distance from
the ladder’s foot to the
building.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 2
Introduction
C o m p a r i n g t h e S t e e p n e s s o f Tw o O b j e c t s
Ratio of vertical distance
to the horizontal distance:
Latter A:
Latter B:
8 feet
4 feet
8 feet
2 feet


2
1
4
1
So, Latter B is steeper.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 3
Property of Comparing the Steepness of Two Objects
C o m p a r i n g t h e S t e e p n e s s o f Tw o O b j e c t s
Property
To compare the steepness of two objects such as two
ramps, two roofs, or two ski slopes, compute the
ratio
vertical distance
horizontal distance
for each object. The object with the larger ratio is
the steeper object.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 4
Comparing the Steepness of Two Roads
C o m p a r i n g t h e S t e e p n e s s o f Tw o O b j e c t s
Example
Road A climbs steadily for 135 feet over a horizontal
distance of 3900 feet. Road B climbs steadily for
120 feet over a horizontal distance of 3175 feet.
Which road is steeper? Explain.
Solution
• These figures are of the two roads, however they
are not to scale
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 5
Comparing the Steepness of Two Roads
C o m p a r i n g t h e S t e e p n e s s o f Tw o O b j e c t s
Solution Continued
A: = vertical distance = 135 feet ≈ 0.035
horizontal distance
3900 feet
1
B: =
vertical distance = 120 feet ≈
horizontal distance
3175 feet
0.038
1
• Road B is a little steeper than road A
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 6
Comparing the Steepness of Two Roads
F i n d i n g a L i n e ’s S l o p e
Definition
The grade of a road is the ratio of the vertical to the
horizontal distance written as a percent.
Example
What is the grade of roads A?
Solution
Ratio of vertical distance to horizontal distance is for
road A is 0.038 = 0.038(100%) = 3.8%.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 7
Slope of a Non-vertical Line
F i n d i n g a L i n e ’s S l o p e
We will now calculate the
steepness of a non-vertical line
given two points on the line.
Pronounced
Pronouncedxxsub
sub1
1and
and yysub
sub11
Let’s use subscript 1 to label x1 and y1 as the
coordinates of the first point, (x1, y1). And x2 and y2
for the second point, (x2, y2).
Run: Horizontal Change = x2 – x1
Rise: Vertical Change = y2 – y1
The slope is the ratio of the rise to the run.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 8
Slope of a Non-vertical Line
F i n d i n g a L i n e ’s S l o p e
Definition
Let (x1, y1) and (x2, y2) be two
distinct point of a non-vertical
line. The slope of the line is
vertical change
rise
m=
=
=
horizontal change
run
y 2 – y1
x 2 – x1
In words: The slope of a non-vertical line is equal to
the ratio of the rise to the run in going from one point
on the line to another point on the line.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 9
Slope of a Non-vertical Line
F i n d i n g a L i n e ’s S l o p e
Definition
A formula is an equation that contains two or more
variables. We will refer to the equation
a
m 
y 2  y1
x 2  x1
as the slope formula.
Sign of rise or run Direction (verbal)
run is positive
goes to the right
run is negative
goes to the left
rise is positive
goes up
rise is negative
goes down
Section 1.3
Lehmann, Intermediate Algebra, 4ed
(graphical)
Slide 10
Finding the Slope of a Line
F i n d i n g a L i n e ’s S l o p e
Example
Find the slope of the line that contains the points
(1, 2) and (5, 4).
Solution
(x1, y1) = (1, 2)
(x2, y2) = (5, 4).
m 
Section 1.3
42
5 1

2
4

1
2
Lehmann, Intermediate Algebra, 4ed
Slide 11
Finding the Slope of a Line
F i n d i n g a L i n e ’s S l o p e
Warning
A common error is to substitute the slope formula
incorrectly:
Correct
m 
Section 1.3
y 2  y1
x 2  x1
Incorrect
m 
Incorrect
y 2  y1
x1  x 2
Lehmann, Intermediate Algebra, 4ed
m 
x 2  x1
y 2  y1
Slide 12
Finding the Slope of a Line
F i n d i n g a L i n e ’s S l o p e
Example
Find the slope of the line that contains the points
(2, 3) and (5, 1).
Solution
By plotting points, the run
is 3 and the rise is –2.
m 
rise
run
Section 1.3

2
3

2
3
Lehmann, Intermediate Algebra, 4ed
Slide 13
Definition
Increasing and Decreasing Lines
Increasing: Positive Slope Decreasing: Negative Slope
Positive rise
m=
Positive run
= Positive slope
Section 1.3
negative rise
m=
positive run
= negative slope
Lehmann, Intermediate Algebra, 4ed
Slide 14
Finding the Slope of a Line
Increasing and Decreasing Lines
Example
Find the slope of the line that contains the points
(– 9 , –4) and (12, –8).
Solution
m 
8   4 
12    9 
–

8  4
12  9

4
21

4
21
• The slope is negative
• The line is decreasing
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 15
Comparing the Slopes of Two Lines
Increasing and Decreasing Lines
Example
Find the slope of the two
lines sketched on the right.
Solution
For line l1 the run is 1 and the
rise is 2.
m 
rise
run
Section 1.3

1
2
2
Lehmann, Intermediate Algebra, 4ed
Slide 16
Comparing the Slopes of Two Lines
Increasing and Decreasing Lines
Solution Continued
For line l2 the run is 1 and the
rise is 4.
m 
rise
run

4
4
1
Note that the slope of l2 is
greater than the slope of l1,
which is what we expected
because line l2 looks steeper
than line l1.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 17
Investigating Slope of a Horizontal Line
H o r i z o n t a l a n d Ve r t i c a l L i n e s
Example
Find the slope of the line that
contains the points (2, 3) and
(6, 3).
Solution
Plotting the points (above) and calculating the slope
we get
33 0
m 
62

0
4
The slope of the horizontal line is zero, no steepness.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 18
Investigating the slope of a Vertical Line
H o r i z o n t a l a n d Ve r t i c a l L i n e s
Example
Find the slope of the line that
contains the points (4, 2) and
(4, 5).
Solution
Plotting the points (above) and calculating the slope
we get
m
52
44

3
, division by zero is undefined.
0
The slope of the vertical line is undefined.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 19
Property
H o r i z o n t a l a n d Ve r t i c a l L i n e s
Property
• A horizontal line has slope of zero (left figure).
• A vertical line has undefined slope (right figure).
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 20
Finding Slopes of Parallel Lines
Parallel and Perpendicular Lines
Definition
Two lines are called parallel
if they do not intersect.
Example
Find the slopes of the lines l1
and l2 sketch to the right.
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 21
Finding Slopes of Parallel Lines
Parallel and Perpendicular Lines
Solution
• Both lines the run is 3, the
rise is 1
• The slope is, m 
rise
run

1
3
• It makes sense that the nonvertical parallel lines
have equal slope
• Since they have the same steepness
Section 1.3
Lehmann, Intermediate Algebra, 4ed
Slide 22