Asked by Ashley
Point A(-5,-40) is rotated 90 degrees counterclockwise about the point (-2,-3). What are the coordinate of A' after the rotation? Answer: A'(-1,-6)
Please explain how they got the answer.
Please explain how they got the answer.
Answers
Answered by
Reiny
that answer cannot possibly be correct
let C(-2,-3) be the centre of rotation.
then CA = <-5+2, -40+3> , where <...> is a vector
= <-3,-37>
using the rotation matrix
cosØ -sinØ
sinØ cosØ
times <-3,-37>
=
0 -1
1 0
times
-3
-37
= <37, -3>
but we have to "move" this vector to its starting point of (-2,-3)
so A' = (35, -6) <-----------
check
slope CA = 37/3
slope CA' = -3/37 , which is the negative reciprocal of the other.
So they are perpendicular.
CA = √( (-37)^2 + 3^2) = √1378
CA' = √1378
so my answer is correct, since I have shown that CA and CA' have the same length and form a 90° angle.
I just noticed that you must have a typo and point A should have been (-5,-4)
If you follow my steps above you will get A' to be (-1,-6)
let C(-2,-3) be the centre of rotation.
then CA = <-5+2, -40+3> , where <...> is a vector
= <-3,-37>
using the rotation matrix
cosØ -sinØ
sinØ cosØ
times <-3,-37>
=
0 -1
1 0
times
-3
-37
= <37, -3>
but we have to "move" this vector to its starting point of (-2,-3)
so A' = (35, -6) <-----------
check
slope CA = 37/3
slope CA' = -3/37 , which is the negative reciprocal of the other.
So they are perpendicular.
CA = √( (-37)^2 + 3^2) = √1378
CA' = √1378
so my answer is correct, since I have shown that CA and CA' have the same length and form a 90° angle.
I just noticed that you must have a typo and point A should have been (-5,-4)
If you follow my steps above you will get A' to be (-1,-6)
Answered by
Ashley
Oh ok, it makes sense now. Thank you very much for your help!
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