# Rotating point about the line y=-x

• December 3rd 2011, 11:25 AM
benny92000
Rotating point about the line y=-x
If you have the point 4,3 and you have to rotate about the line y=-x, how do you do it? I thought it would be either -4,-3 or -3, -4.
• December 3rd 2011, 12:25 PM
Plato
Re: Rotating point about the line y=-x
Quote:

Originally Posted by benny92000
If you have the point 4,3 and you have to rotate about the line y=-x, how do you do it? I thought it would be either -4,-3 or -3, -4.

It would be $x=-3,~y=-4$.
• December 4th 2011, 08:41 AM
HallsofIvy
Re: Rotating point about the line y=-x
I assume you are working in two dimensions so you are rotating through 180 degrees. Drop a perpendicular from (4, 3) to the line. The line, y= -x has slope -1 so the line perpendicular to it has slope 1. The line perpendicular to it and through (4, 3) has equation y= (x- 4)+ 3= x- 1. That crosses the line y= -x where y= -x= x-1 or 2x= 1 so x= 1/2, y= -1/2. To go from (4, 3) to (1/2, -1/2) x decreases by 4- 1/2= 7/2 and y decreases by 3- (-1/2)= 7/2 also. To continue on the other side by the same distance, subtract 7/2 from each again: 1/2- 7/2= -3 and -1/2- 7/2= -4.