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3.3
Linear Equations in Two Variables
Slope-Intercept Form
y b
m
x0
y b
m
x
mx  y  b
mx  b  y
y  mx  b
Slope-Intercept Form
The slope-intercept form of the equation of a line
with slope m and y-intercept (0, b) is
y = mx + b.
Slope y-intercept is (0, b).
Slide 3.3- 2
EXAMPLE 1
Find an equation of the line with slope 2 and
y-intercept (0, –3).
m=2
b = –3
Substitute these values into the slope-intercept form.
y = mx + b
y = 2x – 3
Slide 3.3- 3
EXAMPLE 2
Graph the line, using the slope and y-intercept.
x + 2y = –4
Write the equation in slope-intercept form by solving
for y.
x + 2y = –4
2y = –x – 4
Subtract x.
1
y   x2
2
Slope
Divide by 2
y-intercept (0, –2)
Slide 3.3- 4
continued
1
Graph: y   x  2
2
1. Plot the y-intercept.
(0, -2)
1
1
2. The slope is
or
.
2
2
3. Using (-1/2), begin at (0,-2)
and move 1 unit down and 2
units right.
4. The line through these two points is the required
graph.
Slide 3.3- 5
Point-Slope Form
The point-slope form of the equation of a line with
slope m passing through the point (x1, y1) is
Slope
y – y1 = m(x – x1).
Given point
If you do not like to deal with fractions,
you can use your slope formula as well.
Slide 3.3- 6
EXAMPLE 3
Find an equation of the line with slope 2/5 and passing
through the point (3, –4).
Use the point-slope form with (x1, y1) = (3, –4) and
m = 2/5.
y  y1  m( x  x1 )
2
y  (4)  ( x  3)
5
2
y  4  ( x  3)
5
5 y  20  2 x  6
2 x  5 y  26
2 x  5 y  26
Substitute
Multiply by 5.
Subtract 2x and 20.
Multiply by -1.
Slide 3.3- 7
EXAMPLE 4
Find an equation of the line passing through the points
(–2, 6) and (1, 4). Write the equation in standard form.
First find the slope by the slope formula.
46
2
2
m


1  (2) 3
3
Use either point as (x1, y1) in the point-slope form of
the equation of a line.
Using the point (1, 4): x1 = 1 and y1 = 4
Slide 3.3- 8
continued
m = -2/3; x1 = 1 and y1 = 4
y  y1  m( x  x1 )
2
y  4   ( x  1)
3
Substitute
3 y  12  2 x  2
Multiply by 3.
2 x  3 y  14
Add 2x and 12.
If the other point were used, the same equation
would result.
Slide 3.3- 9
Equations of Horizontal and Vertical
Lines
The horizontal line through the point (a, b) has
equation y = b.
The vertical line through the point (a, b) has equation
x = a.
Slide 3.3- 10
EXAMPLE 5
Find an equation of the line passing through the point
(–8, 3) and
a. parallel to the line 2x – 3y = 10;
b. perpendicular to the line 2x – 3y = 10.
Write each equation in slope-intercept form.
a. Find the slope of the line 2x – 3y = 10
by solving for y.
2 x  3 y  10
3 y  2 x  10
2
10
y  x
3
3
Slide 3.3- 11
Find an equation of the line
passing through the point (–8, 3).
continued
2
3
The slope is
Parallel lines have the same slope. Use point slope form
and the given point.
y  y1  m( x  x1 )
2
y  3  [ x  (8)]
3
2
y  3  ( x  8)
3
2
16
y 3  x 
3
3
2
16 9
y  x 
3
3 3
2
25
y  x
3
3
The fractions were not
cleared because we want the
equation in slope-intercept
form instead of standard
form.
Slide 3.3- 12
continued
Find an equation of the line
passing through the point (–8, 3).
b. Perpendicular lines. The slope is the negative
2
reciprocal of .
3
3
Use point slope form and the given point. m = 
2
y  y1  m( x  x1 )
3
y  3   [ x  (8)]
2
3
y  3   ( x  8)
2
3
y  3   x  12
2
3
y   x9
2
Slide 3.3- 13
Slide 3.3- 14
EXAMPLE 6
Suppose there is a flat rate of $0.20 plus a charge of
$0.10 per minute to make a telephone call. Write an
equation that gives the cost y for a call of x minutes.
y = $0.20 + $0.10x
or
y = $0.10x + 0.20
Slide 3.3- 15
EXAMPLE 7
The percentage of the U.S. population 25 years and
older with at least a high school diploma is shown in
the table for selected years. Find an equation that
models the data, using x = 0 to represent 1940, x = 10
to represent 1950, and so on.
Year
Percent
1940
24.5
1950
34.3
1960
41.4
1970
52.3
1980
66.5
1990
75.2
2000
80.4
Slide 3.3- 16
continued
Choose two data points and find the slope. Use 1940
and 2000.
80.4  24.5 55.9
m

 0.9316
2000  1940
60
The y-intercept is (0, 24.5).
The equation is:
y = 0.93x + 24.5
Selecting two different ordered
pairs will lead to a different equation.
Year
Percent
1940
24.5
1950
34.3
1960
41.4
1970
52.3
1980
66.5
1990
75.2
2000
80.4
Slide 3.3- 17
EXAMPLE 8
Use the ordered pairs (11, 164) and (13, 203) to find an
equation that models the data in the graph below.
203  164 39
m

 19.5
13  11
2
Use the point-slope form with (11, 164).
y  y1  m( x  x1 )
y  164  19.5( x  11)
y  164  19.5 x  214.5
y  19.5 x  50.5
Slide 3.3- 18
Homework
pg 186 # 4, 6-90 m6, 97, 98
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 3.3- 19