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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.