§ 2.5
The Point-Slope Form of the Equation of a
Line
Point-Slope Form
Point-Slope Form of the Equation of a Line
The point-slope equation of a nonvertical line with
slope m that passes through the point  x1 , y1  is
y  y 1  m  x  x1 
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 2.5
144
Point-Slope Form
145
EXAMPLE
Write the point-slope form and then the slope-intercept form of
the equation of the line with slope -3 that passes through the
point (2,-4).
SOLUTION
y  y 1  m  x  x1 
y   4    3  x  2 
Substitute the given values
Simplify
y  4   3 x  2 
This is the equation of the line in point-slope form.
y  4  3 x  6
y  3x  2
Distribute
Subtract 4 from both sides
This is the equation of the line in slope-intercept form.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 2.5
Point-Slope Form
145
EXAMPLE
Write the point-slope form and then the slope-intercept form of
the equation of the line that passes through the points (2,-4) and
(-3,6).
SOLUTION
First I must find the slope of the line. That is done as follows:
m 
6   4 
32

10
5
 2
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 2.5
Point-Slope Form
145
CONTINUED
Now I can find the two forms of the equation of the line. In find
the point-slope form of the line, I can use either point provided.
I’ll use (2,-4).
y  y 1  m  x  x1 
y   4    2  x  2 
y  4  2x  2 
Substitute the given values
Simplify
This is the equation of the line in point-slope form.
y  4  2 x  4
y  2 x
Distribute
Subtract 4 from both sides
This is the equation of the line in slope-intercept form.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 2.5
Equations of Lines
146
Equations of Lines
Standard Form
Slope-Intercept Form
Horizontal Line
Vertical Line
Point-slope Form
Ax + By = C
y = mx + b
y=b
x=a
y  y 1  m  x  x1 
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 2.5
Deciding which form to use:
y = mx + b
146
y – y1 = m(x – x1)
Begin with the slope-intercept
form if you know:
Begin with the point-slope form
if you know:
The slope of the line and the yintercept
The slope of the line and a point
on the line other than the yintercept
or
or
Two points on the line, one of
which is the y –intercept
Two points on the line, neither of
which is the y-intercept
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 2.5
Modeling Life Expectancy
147
EXAMPLE 3
Go over example 3 See Figures 2.35(a) and 2.35(b)
Use the data points to write the slope-intercept form of the equation of this line. Use
the linear function to predict the life expectancy of an American man born in 2020.
Find the slope of 0.215. The slope indicates that for each subsequent
birth year, a man’s life expectancy is increasing by 0.215 years.
Use the point-slope form to write the equation (model).
y – 70.0 = 0.215(x – 20)
Change to slope-intercept form
y = 0.215x + 65.7
Life expectancy of men born in 2020
f(60) = 0.215(60) + 65.7 = 78.6 years
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 2.5
Modeling Life Expectancy
148
Check Points
Do Check Point 3 on page 148 See Figure 2.36
Use the data points to write the slope-intercept form of the equation of this line.
(round to 2 decimal places). Use the linear function to predict the life expectancy of
an American woman born in 2020.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 2.5
Modeling Life Expectancy
148
Check Points, continued
Do Check Point 3 on page 148 See Figure 2.36
Use the data points to write the slope-intercept form of the equation of this line. (round to 2
decimal places). Use the linear function to predict the life expectancy of an American woman
born in 2020.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 2.5
Modeling Public Tuition
Public Tuition: In 2005, the average cost of tuition
and fees at public four-year colleges was $6130,
and in 2010 it was $7610. Note that the known
value for 2008 is $6530.
Solution: The line passes through (2005, 6.1) and
(2010, 7.6). Find the slope.
m 
change in y

change in x
7610  6130
2010  2005

1480
 296
5
Thus, the slope of the line is 296; tuition and fees
on average increased by $296/yr.
Substitute 5 for 2005, 10 for 2010, and 8 for 2008.
m 
change in y
change in x

7610  6130
10  05

1480
5
 296
Figure not
in book
Modeling Public Tuition
Modeling public tuition: Write the slope-intercept
form of the of the line shown in the graph. What is
the y-intercept and does it have meaning in this
situation.
1480

y  y 1  m  x  x1 
y  6130  296  x  2005
 296
5

This is the equation of the line in point-slope form.
y  6130  296 x  593480
y  296 x  587350
Modeling public tuition: Substitute 5 for 2005,
10 for 2010, and 8 for 2008.
y  y 1  m  x  x1 
y  6130  296  x  5 
This is the equation of the line in slope-intercept form.
This is the equation of the line in point-slope form.
y  6130  296 x  1480
y  296 x  4650
This is the equation of the line in slope-intercept form.
Modeling Public Tuition
Using the slope-intercept form of the equation of
the line shown in the graph. Use the equation to
predict the average cost of tuition and fees at public
four-year colleges in 2008.
y  296 x  587350
y  296 ( 2008 )  587350
 7018
Substitute 2008 or 8 for x and compute y.
y  296 ( 8 )  4650
 7018
The model predicts that the tuition in 2008
will be $7018
Use the equation to predict the average cost of tuition
and fees at public four-year colleges in 2015.
Substitute 2015 or15 for x and compute y.
y  296 ( 2015 )  587350
y  296 (15 )  4650
 9090
 9090
The model predicts that the tuition in 2015
will be $9090.
Modeling the Graying of America
Write the slope-intercept form of the equation
of the line shown in the graph. Use the
equation to predict the median age of the
U.S. population in 2020.
Solution: The line passes through (10, 30.0)
and (30, 35.3). Find the slope.
m 
change in y
change in x

35 . 3  30 . 0
30  10

5 .3
 0 . 265
20
The slope indicates that each year the
median age of the U.S. population is
increasing by 0.265 year.
(30, 35.3)
(10, 30.0)
Modeling the Graying of America
Write the slope-intercept form of the equation
of the line shown in the graph. Use the
equation to predict the median age of the U.S.
population in 2020.
5 .3
m 
 0 . 265
20
The slope indicates that each year the median age
of the U.S. population is increasing by 0.265 year.
y  y 1  m  x  x1 
y  30 . 0  0 . 265  x  10 
This is the equation of the line in point-slope form.
y  30 . 0  0 . 265 x  2 . 65
y  0 . 265 x  27 . 35
This is the equation of the line in slope-intercept form.
A linear equation that models the median age of the
U.S. population, y, x years after 1970.
(30, 35.3)
(10, 30.0)
Modeling the Graying of America
Write the slope-intercept form of the equation
of the line shown in the graph. Use the
equation to predict the median age of the U.S.
population in 2020.
5 .3
m 
 0 . 265
20
The slope indicates that each year the median age
of the U.S. population is increasing by 0.265 year.
y  0 . 265 x  27 . 35
A linear equation that models the median age of the
U.S. population, y, x years after 1970.
Use the equation to predict the median age in 2020.
Because 2020 is 50 years after 1970, substitute 50 for
x and compute y.
y  0 . 265 ( 50 )  27 . 35
 40 . 6
The model predicts that the median age of the U.S.
population in 2020 will be 40.6.
(30, 35.3)
(10, 30.0)
Parallel and Perpendicular Lines
Slope and Parallel Lines
1) If two nonvertical lines are parallel, then they have
the same slope.
2) If two distinct nonvertical lines have the same slope,
then they are parallel.
3) Two distinct vertical lines, both with undefined
slopes, are parallel.
y = 2x + 6 and y = 2x – 4 are parallel.
y = -4x +5 and y = -4x + 3 are parallel.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 2.5
Parallel and Perpendicular Lines
Slope and Perpendicular Lines
1) If two nonvertical lines are perpendicular, then the
product of their slopes is -1.
2) If the product of the slopes of two lines is -1, then
the lines are perpendicular.
3) A horizontal line having zero slope is perpendicular
to a vertical line having undefined slope.
y = 2x + 6 and y = -(1/2)x – 4 are perpendicular.
y = -4x +5 and y = (1/4)x + 3 are perpendicular.
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 2.5
Parallel and Perpendicular Lines
148
EXAMPLE
Write an equation of the line passing through (2,-4) and parallel
to the line whose equation is y = -3x + 5.
SOLUTION
Since the line I want to represent is parallel to the given line, they have the
same slope. Therefore the slope of the new line is also m = -3. Therefore,
the equation of the new line is:
y – (-4) = -3(x –2)
y + 4 = -3(x –2)
Substitute the given values
Simplify
y + 4 = -3x + 6
y = -3x + 2
Distribute
Subtract 4 from both sides
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 2.5
Parallel and Perpendicular Lines
149-150
EXAMPLE
Write an equation of the line passing through (2,-4) and
perpendicular to the line whose equation is y = -3x + 5.
SOLUTION
The slope of the given equation is m = -3. Therefore, the slope
of the new line is 1 3 , since  3     1 . Therefore, the using the
slope m = 1 3 and the point (2,-4), the equation of the line is as
follows:
1
3
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 2.5
Parallel and Perpendicular Lines
CONTINUED
m=
y  y 1  m  x  x1 
y   4  
y4
1
3
1
3
y4
y
1
3
1
1
1
x  2 
Simplify
x
x
1
3
2
Distribute
3
2
4
3
x
2

4 3

1 3

12
3
x
3
y
and the point (2,-4),
Substitute the given values
3
y
3
x  2 
3
y
1
2
3
x
14
3
3
Subtract 4 from both sides
Common Denominators
Common Denominators
Simplify
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 2.5
Parallel and Perpendicular Lines
149 -150
Do Check Point 4 on page 149
Find the equation that passes through (-2, 5) and is parallel to the line y = 3x+1
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 2.5
Parallel and Perpendicular Lines
149 -150
Do Check Point 5 on page 150
Find the slope and equation of a line that passes through (-2, -6) and is perpendicular
to x+3y=12
Standard format must be changed
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.5
DONE
Parallel and Perpendicular Lines
One line is perpendicular to another line if its slope is the
negative reciprocal of the slope of the other line.
The following lines are perpendicular:
y = 2x + 6 and y = -(1/2)x – 4 are perpendicular.
y = -4x +5 and y = (1/4)x + 3 are perpendicular.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 2.5
Parallel and Perpendicular Lines
Two lines are parallel if they have the same slope.
The following lines are parallel:
y = 2x + 6 and y = 2x – 4 are parallel.
y = -4x +5 and y = -4x + 3 are parallel.
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 2.5
Example Modeling female officers
In 1995, there were 690 female officers in the Marine Corps, and by 2010 this number had
increased to about 1110. Refer to graph in Figure 3.48 on page 214.
a) The slope of the line passing through (1995, 690) and (2010,1110) is
m 
1110  690
2010  1995
 28
b) The number of female officers increased, on average
by about 28 officers per year.
y  y 1  m  x  x1 
y  690  28  x  1995
(2010, 1110)

(1995, 690)
c) Estimate how many female officers there were in 2006.
y  690  28  2006  1995
y  28 11   690  998
OR
y  28 x  55170
y  28 ( 2006 )  55170
y  56168  55170  998

//
Write the slope-intercept
form of the of the line
shown in the graph.
y  690  28 x  55860
y  28 x  55170