There is one other consideration
for straight-line equations: finding parallel
and perpendicular lines.
Here is the usual format for the question:

Given the line 2x – 3y = 9 and
the point (4,
–1), find lines through
the point that are:

(a) parallel to
the given line and
(b) perpendicular to it.

In other words, they've
given me a reference line — 2x – 3y = 9 — that I'll
be comparing to, and some point somewhere else on the plane — namely, (4, –1).
Then they want me to find the line through (4,
–1) that is parallel
to (that has the same slope as) 2x – 3y = 9.
On top of that, they then want me to find the line through (4,
–1) that is perpendicular
to (that has a slope that is the negative reciprocal of the slope of) 2x – 3y = 9.

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:

ADVERTISEMENT

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.

For the perpendicular
line, I have to find the perpendicular slope. The reference slope is m = ^{2}/_{3},
and, for the perpendicular slope, I'll flip this slope and change the
sign. Then the perpendicular slope is m = – ^{3}/_{2}.
So now I can do the point-slope form. Note that the only change from
the calculations I just did is that the slope is different now.

y – (–1) = ( –^{ 3}/_{2 })(x – 4) y + 1 = ( –^{ 3}/_{2 }) x + 6

y =
( –^{ 3}/_{2 }) x + 5

Then the full solution
to this exercise is:

parallel: y = (^{ 2}/_{3 }) x – ^{11}/_{3
}perpendicular:
y = ( –^{ 3}/_{2 }) x + 5

Warning: If a question
asks you whether two given lines are "parallel, perpendicular, or
neither", you must answer that question by finding their slopes, not by drawing a picture! Pictures can only give you a rough idea
of what is going on, but you cannot tell "by looking" that lines
with slopes of, say, m_{1} = 1.00 and m_{2} = 0.99 are NOT parallel,
because they'll sure look parallel on their graphs. But since 1.00 does not equal 0.99,
the lines are not parallel. Find the slopes; don't just draw the pictures.

You can use the Mathway widget below to
practice finding a parallel line through a given point. Try the entered
exercise, or type in your own exercise. Then click the "paper-airplane" button to compare your answer to Mathway's. (The next
widget is for finding perpendicular lines.)

(Clicking on "Tap to view steps"
on the widget's answer screen will take you to the Mathway site for a paid upgrade.)

You can use the Mathway widget below to practice finding a perpendicular line through a given point. Try
the entered exercise, or type in your own exercise. Then click the "paper-airplane" button to compare your answer to Mathway's.

Stapel, Elizabeth.
"Straight-Line Equations: Parallel and Perpendicular Lines." Purplemath. Available from http://www.purplemath.com/modules/strtlneq3.htm.
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