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Chapter 10
Graphing
Equations and
Inequalities
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
10.5
Equations of Lines
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Slope-Intercept Form
When a linear equation in two variables is written in
slope-intercept form,
y = mx + b
slope
(0, b), y-intercept
then m is the slope of the line and (0, b) is the
y-intercept of the line.
Note: This form is useful for graphing, since
you have a point and the slope readily visible.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Martin-Gay, Developmental Mathematics, 2e
3
Example
Use the slope-intercept form to graph the equation
3
y  x  2.
5
The slope is 3/5.
The y-intercept is –2.
Begin by graphing (0, –2),
move up 3 units and right 5 units.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Martin-Gay, Developmental Mathematics, 2e
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Example
Find the slope and y-intercept of the line –3x + y = –5.
First, we need to solve the linear equation for y.
By adding 3x to both sides, y = 3x – 5.
Once we have the equation in the form of y = mx + b,
we can read the slope and y-intercept.
slope is 3
y-intercept is (0, –5)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Martin-Gay, Developmental Mathematics, 2e
5
Example
Find an equation of the line with y-intercept (0, –2)
and slope of 1 . We are given the slope and the
3
y-intercept. We let m = 1 and b = –2 and write the
3
equation in slope-intercept form, y = mx + b.
y = mx + b
1
y  x  ( 2)
3
1
y  x2
3
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Martin-Gay, Developmental Mathematics, 2e
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Point-Slope Form
The point-slope form of the equation of a line is
y  y1  m( x  x1 ),
where m is the slope of the line and (x1, y1) is a
point on the line.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Martin-Gay, Developmental Mathematics, 2e
7
Example
Find an equation of the line with slope –2 that passes through
(–11, –12). Write the equation in slope-intercept form, y = mx + b,
and in standard form, Ax + By = C.
We substitute the slope and point into the point-slope form of an
equation.
y – (–12) = – 2(x – (– 11))
Let m = –2 and (x1, y1) = (–11, –12).
y + 12 = –2x – 22
y = –2x – 34
Use the distributive property.
2x + y = –34
Add 2x to both sides and we have
standard form.
Slope-intercept form.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Martin-Gay, Developmental Mathematics, 2e
8
Example
Find the equation of the line through (–4, 0) and
(6, –1). Write the equation in standard form.
First, find the slope.
y2  y1
1  0
1
m


x2  x1 6  (4) 10
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Martin-Gay, Developmental Mathematics, 2e
continued
9
continued
Now substitute the slope and one of the points into the
point-slope form of an equation.
1
y  0   ( x  (4))
10
10y = –1(x + 4)
Clear fractions by multiplying both
sides by 10.
10y = –x – 4
Use the distributive property.
x + 10y = –4
Add x to both sides.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Martin-Gay, Developmental Mathematics, 2e
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