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Gradients of Perpendicular Lines

Date: 01/16/2003 at 09:12:05
From: Anonymous
Subject: Gradients of perpendicular lines

Why is the product of the two gradients of perpendicular lines equal 
to -1?


Date: 01/16/2003 at 09:23:44
From: Doctor Jerry
Subject: Re: Gradients of perpendicular lines

Hi,

There are several arguments that can be given. Here's a short 
argument:

If a1 and a2 are the angles of inclination of two lines with slopes m1
and m2, then

tan(a1-a2) = [tan(a1)-tan(a2)]/[1+tan(a1)*tan(a2)].
  
           = [ m1 - m2 ]/[1+m1*m2].

If the two lines are perpendicular, then a1-a2 or a2-a1 is pi/2. So,
the left side of the above equation is not defined. The only way this
can happen is if the denominator on the right side is 0, which happens
when m1*m2 = -1.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Calculus

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