Posted by jessica on Monday, February 23, 2009 at 5:15pm.
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)
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] │
i am in grade 11 but this stuff is from grade 10. i am a new student and have not studied it back in my country. the paper is due tomorrow and i don't have a clue how to do it.... the substitute handed us the paper today. so i cannot even ask my teacher till tomorrow....
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.
thank you so much....i think i got it. thanks again!!!
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
Damon, I think she wanted to rotate counter-clockwise, you went clockwise
Damon, I apologize
You went the correct way, I messed up and went the wrong way.
thank you both....so the correct answer is (6,1) Thanks!!! :)
I cheated and checked the matrix in my thick math book :)
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