Fifth Edition
James Stewart  Lothar Redlin  Saleem Watson

2

2.4

Lines
In this section, we find equations for straight lines lying in a coordinate plane.
• The equations will depend on how the line is inclined.
• So, we begin by discussing the concept of slope.

Slope of a Line
We first need a way to measure the “steepness” of a line.
• This is how quickly it rises (or falls) as we move from left to right.

Slope of a Line
We define ‘run’ to be the distance we move to the right and ‘rise’ to be the corresponding distance that the line rises (or falls).
The slope of a line is the ratio of rise to run: slope = rise run

Slope of a Line
The figure shows situations where slope is important.
• Carpenters use the term pitch for the slope of a roof or a staircase.
• The term grade is used for the slope of a road.

Run & Rise
If a line lies in a coordinate plane, then:
• Run is the change in the x -coordinate between any two points on the line.
• Rise is the corresponding change in the y -coordinate.

Slope of a Line —Definition
That gives us the following definition of slope.
The slope m of a nonvertical line that passes through the points A ( x
1
, y
1
) and B ( x
2
, y
2
) is: m
 rise run
 y
2
 y
1 x
2
 x
1
• The slope of a vertical line is not defined.

Slope of a Line

Slope of a Line
We can see that this is true from the similar triangles in the figure: y x
2
2

 y x
1
1
 y x
2

2
 
1

 y x

1

Slope of a Line
The figure shows several lines labeled with their slopes.

Positive & Negative Slopes
• Lines with positive slope slant upward to the right.
• Lines with negative slope slant downward to the right.

Steep & Horizontal Lines
The steepest lines are those for which the absolute value of the slope is the largest.
A horizontal line has slope zero.

E.g. 1 —Finding Slope of a Line Through Two Points
Find the slope of the line that passes through the points
P (2, 1) and Q (8, 5)
• Since any two different points determine a line, only one line passes through these two points.

E.g. 1 —Finding Slope of a Line Through Two Points
From the definition, the slope is: m
 y
2
 y
1 x
2
 x
1

5 1

4 2
8 2 6 3
• This says that, for every 3 units we move to the right, the line rises 2 units.

E.g. 1 —Finding Slope of a Line Through Two Points

Equations of Lines
Now let’s find the equation of the line that passes through a given point
P ( x
1
, y
1
) and has slope m .

Equations of Lines
A point P ( x , y ) with x ≠ x
1 lies on this line if and only if:
• The slope of the line through P
1 and is equal to m ,
P that is, y
 y
1 x
 x
1
 m

Equations of Lines
That equation can be rewritten in the form y – y
1
= m ( x – x
1
)
• Note that the equation is also satisfied when x = x
1 and y = y
1
.
• Therefore, it is an equation of the given line.

Point-Slope Form of the Equation of a Line
An equation of the line that passes through the point ( x
1
, y
1
) and has slope m is: y – y
1
= m ( x – x
1
)

E.g. 2 —Equation of a Line with Given Point and Slope
(a) Find an equation of the line through
(1, –3 ) with slope –½.
(b) Sketch the line.

E.g. 2 —Equation of a Line
Example (a)
Using the point-slope form with m = –½ , x
1
= 1, y
1
= –3, we obtain an equation of the line as: y 3 1
2
 x

1

2 y
   
1 x

2 y
 
0

E.g. 2 —Equation of a Line
Example (b)
The fact that the slope is –½ tells us that:
• When we move to the right 2 units, the line drops 1 unit.

E.g. 2 —Equation of a Line
Example (b)

E.g. 3 —Equation of a Line Through Two Points
Find an equation of the line through the points
( –1, 2) and (3, –4)
• The slope of the line is: m

3
 
6 3
4 2

E.g. 3 —Equation of a Line Through Two Points
Using the point-slope form with x
1
= –1 and y
1
= 2, we obtain: y 2 3
2
 x

1

2 y 4 3 x

3
3 x

2 y
 

Equations of Lines
Suppose a nonvertical line has slope m and y -intercept b.
• This means the line intersects the y -axis at the point (0, b ).

Slope-Intercept Form of the Equation of a Line
So, the point-slope form of the equation of the line, with x = 0 and y = b , becomes: y – b = m ( x – 0)
• This simplifies to y = mx + b.
• This is called the slope-intercept form of the equation of a line.

Slope-Intercept Form of the Equation of a Line
An equation of the line that has slope m and y -intercept b is: y = mx + b

E.g. 4 —Lines in Slope-Intercept Form
(a) Find the equation of the line with slope 3 and y -intercept –2.
(b) Find the slope and y -intercept of the line 3 y – 2 x = 1.

E.g. 4 —Slope-Intercept Form
Example (a)
Since m = 3 and b = – 2, from the slopeintercept form of the equation of a line, we get: y = 3 x – 2

E.g. 4 —Slope-Intercept Form
Example (b)
We first write the equation in the form y = mx + b :
3 y

2 x

1
3 y

2 x

1 y

2
3 x

1
3
• From the slope-intercept form of the equation of a line, we see that: the slope is m = ⅔ and the y -intercept is b = ⅓.

Equation of Horizontal Line
If a line is horizontal, its slope is m = 0.
So, its equation is: where b is y = b the y -intercept.

Equation of Vertical Line
A vertical line does not have a slope.
Still, we can write its equation as: x = a where a is the x -intercept.
• This is because the x -coordinate of every point on the line is a .

Vertical and Horizontal Lines
An equation of the vertical line through
( a , b ) is: x = a
An equation of the horizontal line through
( a , b ) is: y = b

E.g. 5 —Vertical and Horizontal Lines
(a) An equation for the vertical line through
(3, 5) is x = 3.
(b) The graph of the equation x = 3 is a vertical line with x -intercept 3.

E.g. 5 —Vertical and Horizontal Lines
(a) The equation for the horizontal line through (8, –2) is y = –2.
(b) The graph of the equation y = – 2 is a horizontal line with y -intercept – 2.

Linear Equation
A linear equation is an equation of the form
Ax + By + C = 0 where:
• A , B , and C are constants.
• A and B are not both 0.

Equation of a Line
The equation of a line is a linear equation:
• A nonvertical line has the equation y = mx + b or – mx + y – b = 0
This is a linear equation with
A = –m , B = 1, C = – b
• A vertical line has the equation x = a or x – a = 0
This is a linear equation with
A = 1, B = 0, C = – a

Graph of a Linear Equation
Conversely, the graph of a linear equation is a line:
• If B ≠ 0, the equation becomes: y
 
A x

C
B B
This is the slope-intercept form of the equation of a line (with m =
–A
/ B and b =
–C
/ B ).
• If B = 0, the equation becomes: Ax + C = 0 or x = –C / A
This represents a vertical line.

General Equation of a Line
We have proved the following.
• The graph of every linear equation
Ax + By + C = 0 ( A , B not both zero) is a line.
• Conversely, every line is the graph of a linear equation.

E.g. 6 —Graphing a Linear Equation
x –
y –

E.g. 6 —Graphing a Linear Equation
Solution 1
Since the equation is linear, its graph is a line.
To draw the graph, it is enough to find any two points on the line.
• The intercepts are the easiest points to find.

E.g. 6 —Graphing a Linear Equation
Solution 1 x -intercept:
• Substitute y = 0.
• You get: 2 x – 12 = 0.
• Thus, x = 6.
y -intercept:
• Substitute x = 0.
• You get: –3 y
–
12 = 0.
• Thus, y = –4.

E.g. 6 —Graphing a Linear Equation
Solution 1
With those points, we can sketch the graph as shown.

E.g. 6 —Graphing a Linear Equation
Solution 2
We write the equation in slope-intercept form:
2 x

3 y

12

0
2 x

3 y

12

3 y
 
2 x

12 y

2
3 x

4
• This equation is in the form y = mx + b.
• So, the slope is m = ⅔ and the y -intercept is b =
–
4.

E.g. 6 —Graphing a Linear Equation
Solution 2
To sketch the graph, we:
1. Plot the y -intercept.
2. Move 3 units to the right and 2 units up.

Parallel Lines
Since slope measures the steepness of a line, it seems reasonable that parallel lines should have the same slope.
• In fact, we can prove this.

Parallel Lines
Two nonvertical lines are parallel if and only if:
• They have the same slope.

Parallel Lines —Proof
Let the lines l
1 and l
2 have slopes m
1 and m
2
.
• If the lines are parallel, the right triangles ABC and DEF are similar.
• Thus, m
1



,
,





,
,


 m
2

Parallel Lines —Proof
Conversely, if the slopes are equal, then the triangles will be similar.
• So, 
BAC
 
EDF and the lines are parallel.

E.g. 7 —Equation of a Line Parallel to a Given Line
Find an equation of the line through the point (5, 2) that is parallel to the line
4 x + 6 y + 5 = 0.

E.g. 7 —Equation of a Line Parallel to a Given Line
First, we write the equation of the given line in slope-intercept form:
4 x

6 y
 
0
6 y
 
4 x

5 y
 
2

3 x 5
6
• So, the line has slope m = –⅔.
• Since the required line is parallel to the given line, it also has slope m = –⅔.

E.g. 7 —Equation of a Line Parallel to a Given Line
From the point-slope form of the equation of a line, we get: y 2 2
3
 x

5

3 y 6 2 x

10
2 x

3 y

16

0
• The equation of the required line is:
2 x + 3 y – 16 = 0

Perpendicular Lines

Perpendicular Lines
Two lines with slopes m
1 and m
2 are perpendicular if and only if m
1 m
2
= –1, that is, their slopes are negative reciprocals: m
2
 
1 m
1
• Also, a horizontal line (slope 0) is perpendicular to a vertical line (no slope).

Perpendicular Lines —Proof
Here, we show two lines intersecting at the origin.
• If the lines intersect at some other point, we consider lines parallel to these that intersect at the origin.
• These lines have the same slopes as the original lines.

Perpendicular Lines —Proof
If the lines l
1 and l
2 have slopes m
1 and m
2
, then their equations are: y = m
1 x and y = m
2 x
Notice that:
• A (1, m
1
) lies on l
1
.
• B (1, m
2
) lies on l
2
.

Perpendicular Lines —Proof
By the Pythagorean Theorem and its converse, OA
┴
OB if and only if:

   2      2     2

Perpendicular Lines —Proof
By the Distance Formula, this becomes:

1 2  m
1
2
 
1 2  m
2
2
 
1 1
  m
2
 m
1

2
2
 m
1
2  m 2
2
 m 2
2

2 m m
1 2
 m
1
2 m m
1 2
2
 
2 m m
1 2
 
1

E.g. 8 —Perpendicular Lines
Show that the points
P (3, 3), Q (8, 17), R (11, 5) are the vertices of a right triangle.

E.g. 8 —Perpendicular Lines
The slopes of the lines containing PR and
QR are, respectively, m
1

5 3


1
11 3 4 m
2
  
4
• As m
1 m
2
= –1, these lines are perpendicular.
• So, PQR is a right triangle.

E.g. 8 —Perpendicular Lines

E.g. 9 —Equation of Line Perpendicular to Given Line
Find an equation of the line that is perpendicular to the line 4 x + 6 y + 5 =0 and passes through the origin.
• In Example 7, we found that the slope of the line 4 x + 6 y + 5 = 0 is –2/3.
• Thus, the slope of a perpendicular line is the negative reciprocal —that is, 3/2.

E.g. 9 —Equation of Line Perpendicular to Given Line
Since the required line passes through
(0, 0), the point-slope form gives: y – 0 = 3/2( x – 0) y = 3/2 x

E.g. 10 —Graphing a Family of Lines
Use a graphing calculator to graph the family of lines y = 0.5
x + b for b = – 2, – 1, 0, 1, 2.
• What property do the lines share?

E.g. 10 —Graphing a Family of Lines
The lines are graphed here in the viewing rectangle [ –6, 6] by [–6, 6].
• They all have the same slope.
• So, they are parallel.

Modeling with Linear Equations:
Slope as Rate of Change

Rate of Change
When a line is used to model the relationship between two quantities, the slope of the line is the rate of change of one quantity with respect to the other.

Rate of Change
For example, this graph gives the amount of gas in a tank that is being filled.
• The slope between the indicated points is: m

6 gallons
3 minutes

2 gal/min
• The slope is the rate at which the tank is being filled —2 gallons per minute.

Rate of Change
Here, the tank is being drained at the rate of 0.03 gallon per minute.
The slope is –0.03.

Slope as Rate of Change
The next two examples give other situations where the slope of a line is a rate of change.

E.g. 11 —Slope as Rate of Change
A dam is built on a river to create a reservoir.
The water level w in the reservoir is given by w = 4.5
t + 28 where:
• t is the number of years since the dam was constructed.
• w is measured in feet.

E.g. 11 —Slope as Rate of Change
(a) Sketch a graph of this equation.
(b) What do the slope and w -intercept of this graph represent?

E.g. 11 —Slope as Rate of Change
This equation is linear.
So, its graph is a line.
Example (a)
• Since two points determine a line, we plot two points that lie on the graph and draw a line through them.

E.g. 11 —Slope as Rate of Change
Example (a)
When t = 0, w = 4.5(0) + 28 = 28
• So, (0, 28) is on the line.
When t = 2, w = 4.5(2) + 28 = 37
• So, (2, 37) is on the line.

E.g. 11 —Slope as Rate of Change
Example (a)
The line determined by those points is shown here.

E.g. 11 —Slope as Rate of Change
The slope is m = 4.5.
Example (b)
It represents the rate of change of water level with respect to time.
• This means that the water level increases
4.5 ft per year.

E.g. 11 —Slope as Rate of Change
Example (b)
w
t =
• This represents the water level when the dam was constructed.

E.g. 12 —Temperature-Elevation Linear Relationship
(a) As dry air moves upward, it expands and cools.
• If the ground temperature is 20°C and the temperature at a height of 1 km is 10 °C, express the temperature T (in °C) in terms of the height h (in kilometers).
• Assume that the relationship between T and h is linear.

E.g. 12 —Temperature-Elevation Linear Relationship
(b) Draw the graph of the linear equation.
What does its slope represent?
(c) What is the temperature at a height of 2.5 km?

E.g. 12 —Temperature-Elevation
Example (a)
Because we are assuming a linear relationship between T and h , the equation must be of the form
T = mh + b where m and b are constants.

E.g. 12 —Temperature-Elevation
Example (a)
When h = 0, we are given that T = 20.
• Thus,
20 = m (0) + b b = 20
• So, we have:
T = mh + 20

E.g. 12 —Temperature-Elevation
Example (a)
When h = 1, we have T = 10.
• So,
10 = m (1) + 20 m = 10 – 20 = –10
• The required expression is:
T = –10 h + 20

E.g. 12 —Temperature-Elevation
Example (b)
• The slope is: m = –10°C/km
• This represents the rate of change of temperature with respect to distance above the ground.
• So, the temperature decrease s 10 °C per kilometer of height.

E.g. 12 —Temperature-Elevation
At a height of h = 2.5 km, the temperature is:
Example (c)
T = –10(2.5) + 20
= –25 + 20
= –5°C