13.4 – Slope and Rate of Change
Slope is a rate of change.
Y
rise
slope 
run
change in y
slope 
change in x
y1  y2
slope 
x1  x2


 x2 , y2 
 x1, y1 
X
13.4 – Slope and Rate of Change
change in y
slope 
change in x
y1  y2
slope 
x1  x2
slope  5  0
4 1
5
slope 
3
Y


1,0
 4,5
X
13.4 – Slope and Rate of Change
change in y
slope 
change in x
y1  y2
slope 
x1  x2
5 1
slope 
4  5
4
4

slope 
9
9
 4,5
Y

 5,1

X
13.4 – Slope and Rate of Change
Slope of any Vertical Line
x2
y1  y2
slope 
x1  x2
Y

 2,3
3   4 
slope 
22
7
slope 
0
X

undefined
 2, 4
13.4 – Slope and Rate of Change
Slope of any Horizontal Line
Y
y  3
y1  y2
slope 
x1  x2
3   3
slope 
4   3
0
slope 
0
7
 3, 3

X

 4, 3
13.4 – Slope and Rate of Change
Find the slope of the line defined by: 5 x  4 y  10
x6
5  6  4 y  10
30  4 y  10
4 y  20
y  5
 6, 5
x2
5  2  4 y  10
10  4 y  10
4y  0
y0
 2, 0 
slope  5  0
62
5
slope 
4
13.4 – Slope and Rate of Change
Alternative Method to find the slope of a line
If a linear equation is solved for y, the coefficient of the x
represents the slope of the line.
5 x  4 y  10
5 x  4 y  10
5
slope 
4
4 y  5 x  10
5
10
y  x
4
4
5
5
y  x
4
2
13.4 – Slope and Rate of Change
If a linear equation is solved for y, the coefficient of the x
represents the slope of the line.
2x  y  7
 y  2 x  7
y  2x  7
slope  2
5 x  7 y  2
7 y  5x  2
5
2
y  x
7
7
5
slope 
7
13.4 – Slope and Rate of Change
Parallel Lines are two or more lines with the same slope.
x y 5
y  x  5
slope  1
2x  2 y  3
2 y  2 x  3
3
y  x 
2
slope  1
These two lines are parallel.
13.4 – Slope and Rate of Change
Perpendicular Lines exist if the product of their slopes is –1.
x y 5
5 y  2x  3
2
3
y  x
5
5
2
slope 
5
5x  2 y  1
2 y  5 x  1
5
1
y   x
2
2
5
slope  
2
2  5
     1
5  2
These two lines are perpendicular.
13.4 – Slope and Rate of Change
Are the following lines parallel, perpendicular or neither?
3x  9 y  5  0
9 y  3 x  5
3
5
y  x
9
9
1
5
y  x
3
9
1
slope  
3
x  3y  2
3y  x  2
1
y  x 1
3
1
slope 
3
NEITHER
13.4 – Slope and Rate of Change
Are the following lines parallel, perpendicular or neither?
6 x  12 y  4
12 y  6 x  4
6
4
y
x
12
12
1
1
y  x
2
3
1
slope 
2
2x  y  3
y  2 x  3
slope  2
1
  2   1
2
These two lines are perpendicular.
13.4 – Slope and Rate of Change
For every twenty horizontal feet a road rises 3 feet. What is
the grade of the road?
rise
slope 
run
3 feet
slope 
20 feet
grade  %  slope  100%
3
grade  %  
 100%
20
grade  %  15%
13.4 – Slope and Rate of Change
The pitch of a roof is a slope. It is calculated by using the
vertical rise and the horizontal run. If a run rises 7 feet for
every 10 feet of horizontal distance, what is the pitch of the
roof?
rise
pitch  slope 
run
7 feet
pitch 
10 feet
7
pitch 
10
13.5 – Equations of Lines
Slope-Intercept Form– requires the
y-intercept and the slope of the line.
m = slope of line
y  mx  b
2
y  x4
3
b = y-intercept


13.5 – Equations of Lines
Slope-Intercept Form:
m = slope of line
b = y-intercept

y  mx  b
6
y   x2
5

13.5 – Equations of Lines
Slope-Intercept Form:
m = slope of line
b = y-intercept

y  mx  b

3
y  x 3
2
13.5 – Equations of Lines
Slope-Intercept Form:
m = slope of line
y  mx  b
3
y  x4
4

b = y-intercept

13.5 – Equations of Lines
Write an equation of a line given the slope and the y-intercept.
y  mx  b
9
m
y  intercept =  2
11
9
y
x 2
11
7
m
3
7
y   x 5
3
y  intercept  5
21 
3
m
 0,  
13 
4
21
3
y
x
4
13
13.5 – Equations of Lines
Point-Slope Form – requires the
coordinates of a point on the line and
the slope of the line.
y  y1  m  x  x1 
1
y 1   x  2
3
1
m
 2,1
3


13.5 – Equations of Lines
Point-Slope Form – requires the
coordinates of a point on the line and
the slope of the line.
y  y1  m  x  x1 
3
y  2    x  3
4
3
 3, 2 m  
4


13.5 – Equations of Lines
Point-Slope Form – requires the
coordinates of a point on the line and
the slope of the line.
y  y1  m  x  x1 
4
y  4   x  2
5
 2, 4
4
m
5


13.5 – Equations of Lines
Writing an Equation Given Two Points
1. Calculate the slope of the line.
2. Select the form of the equation.
a. Standard form
Ax  By  C
b. Slope-intercept form
c. Point-slope form
y  mx  b
y  y1  m  x  x1 
3. Substitute and/or solve for the selected form.
13.5 – Equations of Lines
Writing an Equation Given Two Points
Given the two ordered pairs, write the equation of the line
using all three forms.
1,3 5, 2
Calculate the slope.
3   2 
5

m
4
1 5
5

4
or
2  3
m
5 1
5

4
5

4
13.5 – Equations of Lines
Writing an Equation Given Two Points
1,3 5, 2
5
m
4
Point-slope form
y  y1  m  x  x1 
5
y  3    x  1
4
5
y  2    x  5
4
13.5 – Equations of Lines
Writing an Equation Given Two Points
1,3 5, 2
5
m
4
5
y  3    x  1
4
5
5
y 3   x
4
4
5
5
y 33   x  3
4
4
5
17
y  x
4
4
Slope-intercept form
5
y  2    x  5
4
5
25
y2  x
4
4
5
25
y22   x 2
4
4
5
17
y  x
4
4
13.5 – Equations of Lines
Writing an Equation Given Two Points
1,3 5, 2
5
m
4
5
17
y  x
4
4
LCD: 4
5
17
 4 y   4  x   4
4
4
Standard form
4 y  5 x  17
5 x  4 y  17
13.5 – Equations of Lines
Solving Problems
The pool Entertainment company learned that by pricing
a pool toy at $10, local sales will reach 200 a week.
Lowering the price to $9 will cause sales to rise to 250 a
week.
a. Assume that the relationship between sales price and
number of toys sold is linear. Write an equation that
describes the relationship in slope-intercept form. Use
ordered pairs of the form (sales price, number sold).
b. Predict the weekly sales of the toy if the price is
$7.50.
13.5 – Equations of Lines
Solving Problems
 sales price, number sold 
10, 200  9, 250
250  200
m
9  10
50
m
1
m  50
y  200  50  x  10
y  250  50  x  9
y  200  50 x  500
y  250  50 x  450
y  50 x  700
y  50 x  700
13.5 – Equations of Lines
Solving Problems
 sales price, number sold 
Predict the weekly sales of the toy if the price is $7.50.
x  7.50
y  50 x  700
y  50  7.50  700
y  375  700
y  325 items sold