what is the image of (1,-6) for a 90 degree counterclockwise rotation about the origin?
i guess it is (-1.6) is it?
That point is in quadrant 4, down to the right.(1 unit right of -y axis and 6 down)
Spin it 90 degrees and it ends up above the x axis a distance 1 and a distance 6 right of the origin
so I get
(1,6)
Sorry, I made a typo
one up and 6 right is
(6,1)
which you can also get by he matrix operation
0 -1
1 +0
times your (1,-6) vector
no,
did you make a sketch?
I see it as (-6,-1)
Proof: slope of original line = (-6-0)/(1-0) = -6
slope of new line = (-1-0)/(-6-0) = 1/6
they are negative recipricals so they form a 90º angle
also you can verify that their lengths are the same.
btw, are you using a rotation matrix ?
what grade level is this in ?
R(theta)=
│costheta -sintheta │
│sintheta costheta] │
your best bet might be to make a sketch on graph paper, and use your intuition.
Notice that the x and y values contain the same numbers, except their signs might have switched as well as their positions, (-1,6) became (-6,-1)
in general (a,b) becomes (-b,a) after a 90º counter-clockwise rotation and (b,-a) for a clockwise rotation.
Damon, I apologize
You went the correct way, I messed up and went the wrong way.