proving right triangle with complex points

• Mar 31st 2009, 05:06 PM
morganfor
proving right triangle with complex points
i have to prove that a triangle is a right triangle given these points: 3+i, 6, and 4+4i. I know if you show that two lines have perpendicular slope that proves it but I don't understand how to do it with complex points. Thank you!
• Mar 31st 2009, 05:11 PM
TheEmptySet
Quote:

Originally Posted by morganfor
i have to prove that a triangle is a right triangle given these points: 3+i, 6, and 4+4i. I know if you show that two lines have perpendicular slope that proves it but I don't understand how to do it with complex points. Thank you!

These are just the same as point in \$\displaystyle \mathbb{R}^2\$

\$\displaystyle 3+i=(3,1),6=(6,0),4+4i=(4,4)\$

I hope this helps.
• Apr 1st 2009, 12:29 PM
Opalg
In the complex plane, multiplication by i has the effect of rotation through a right angle. If the complex numbers u,v,w represent the vertices of a triangle then the side from u to v is represented by the number v–u. So the condition for the angle at u to be a right angle is that w–u should be a real multiple of i(v–u).