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/