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2.5
Other Equations of Lines

Point-Slope Form

Parallel and Perpendicular Lines
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Point-Slope Form
Suppose that a line of slope m passes through
the point (x1, y1). For any other point (x, y) to
lie on this line, we must have
(x, y)
y– y 1
(x1, y1)
y  y1
 m.
x  x1
x– x 1
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Point-Slope Form
Any equation y – y1 = m(x – x1) is said to be
written in point-slope form and has a graph
that is a straight line.
The slope of the line is m.
The line passes through (x1, y1).
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Example
Find an equation of the line passing
through (1, 4) with slope –7.
Solution
Use the point-slope form equation:
y – y1 = m(x – x1)
y – 4 = –7(x – 1)
Substituting
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Parallel and Perpendicular
Lines
Two lines are parallel if they lie in the same
plane and do not intersect no matter how
far they are extended.
Two lines are perpendicular if they intersect
at a right angle. If one line is vertical and
another is horizontal, they are perpendicular.
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Slope and Parallel Lines
Two lines are parallel if they have the
same slope or if both lines are vertical.
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Example
Determine whether the line passing through (1, 2)
and (7, 5) is parallel to the line given by
1
f ( x )  x  1.
2
Solution
The slope of the line passing through (1,2) and (7,5)
is given by
52 3 1
m
  .
7 1 6 2
Since the graph of f (x) = (½)x – 1 also has a slope
of ½, the lines are parallel.
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Slope and Perpendicular Lines
Two lines are perpendicular if the product
of their slopes is –1 or if one line is vertical
and the other is horizontal.
Thus, if one line has slope m (m  0), the
slope of a line perpendicular to it is –1/m.
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Slide 2- 9
Example
Consider the line given by the equation
5y = 3x + 10.
a) Find an equation for a parallel line
passing through (0, 6).
b) Find an equation for a perpendicular
line passing through (0, 6).
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Solution
Both parts (a) and (b) require find the
slope of the line given by 5y = 3x + 10.
Solve for y to find the slope-intercept
form:
5 y  3x  10
3
y x2
5
The slope is 3/5.
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a) The slope of any parallel line will be 3/5.
The slope-intercept form yields:
3
y  x  6.
5
b) The slope of any perpendicular line will
be –5/3 (the opposite of the reciprocal of
3/5). The slope-intercept form yields:
5
y   x  6.
3
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