Slopes of Perpendicular LinesDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/