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Title: Parallel Lines
Given two lines
If
is perpendicular to
,then:
.
Example 1:
y = 3x + 14 is parallel to y = 3x - 72
If two lines are not parallel, there exist a point of intersection. This point can be found by solving the two equations simultaneously.
Example2:
Determine whether the following pairs of lines are parallel.
l1: y = x + 6
l2: the line joining A (1, 4) and B (-4, -1)
Solution:
Gradient of
Gradient of
Since the two gradients are the same, the pair of lines is parallel.
Perpendicular Lines
Given two lines
If
is perpendicular to
, then:
.
Example 3:
y = 3x + 14 is perpendicular to y=
x - 72
Example 4:
Given the line 2x 3y = 9 and the point (4, 1), find lines through the point that are
1: parallel to the given line and
2: perpendicular to it.
Solution for parallel line:
Clearly, the first thing we need to do is solve "2x 3y = 9" for "y=", so that we can find the reference slope:
2x 3y = 9 3y = 2x + 9 y = (2/3)x 3
So the reference slope from the reference line is m = 2/3.
Since a parallel line has an identical slope, then the parallel line through (4, 1) will have slope m = 2/3. Hey, now I have a point and a slope! So I'll use the point-slope form to find the line:
y (1) = ( 2/3)(x 4)
y + 1 = ( 2/3)x 8/3
y = ( 2/3)x 8/3 3/3
y = ( 2/3)x 11/3
This is the parallel line that they asked for.
Solution for perpendicular line:
For the perpendicular line, we have to find the perpendicular slope. The reference slope is m = 2/3, and, for the perpendicular slope, we'll flip this slope and change the sign. Then the perpendicular slope is m = 3/2. Now we'll use the slope-intercept form.
y (1) = ( 3/2)(x 4)
y + 1 = ( 3/2)x + 6
y = ( 3/2)x + 5
Example 5:
Find the perpendicular bisector of the line segment joining A (-3, 4) and B (2, -1).
Solution:
Gradient of AB =
Gradient of perpendicular bisector:
Midpoint of AB =
Equation:
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