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Analytic Geometry: (lesson 3 of 4)

## Parallel and Perpendicular 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:

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: