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### Slopes of Perpendicular Lines

```
Date: 09/17/98 at 22:32:59
From: sarah mielke
Subject: Slopes of Perpendicular Lines

Our class has been working on graphing lines. I understand that the
product of the opposite reciprocals of the slopes of perpendicular
lines that are not vertical and horizontal is -1. I see how it works,
but I don't understand how you come to that. Could you show me the
proof that justifies that? Thanks a lot.

Sarah Mielke
```

```
Date: 09/18/98 at 12:56:07
From: Doctor Peterson
Subject: Re: Slopes of Perpendicular Lines

Hi, Sarah. You've combined two different ways of saying the same thing,
but I understand what you are saying. You can say either that the
slopes of perpendicular lines are "opposite reciprocals" of one another
(that is, m2 = -1/m1), or that the product of their slopes is -1 (that
is, m1*m2 = -1). These equations say the same thing.

Let's just draw a quick picture and see what this means:

B        C
+--------+
\   a   |
\      |
\     |                  B'
\    |b                 +
L1\   |               /  |
\  |       L2  /      |a
\ |       /          |
\|   /       b      |
+------------------+
A                  C'

Lines L1 and L2 are perpendicular. I've drawn a right triangle ABC with
its legs parallel to the axes and its hypotenuse along L1, then rotated
it 90 degrees so that the hypotenuse of AB'C' is along L2. Since each
leg has been rotated 90 degrees, AC' and B'C' are now parallel to the
axes - but they have switched axes. That means that the slope of L1 is
- b/a (negative because when we go right a distance of a, we go down a
distance of b), and the slope of L2 is a/b. So the product of the
slopes is

b     a
m1 * m2 = - --- * --- = -1
a     b

There's your proof. Whenever I think of the slopes of perpendicular
lines, I think of this picture.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Equations, Graphs, Translations
High School Linear Equations

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