Rotate a point in circle about an angle

How should I rotate a point $(x,y)$ to location $(a,b)$ on the co-ordinate by any angle?

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The rotation transformation of a point, treated as a vector, would be (you seem to be going clockwise so I'll swing with that):

$\bigl(\begin{smallmatrix} x'\\ y' \end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix} \cos(\theta)&\sin(\theta)\\ -\sin(\theta)&\cos(\theta) \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} x\\ y \end{smallmatrix}\bigr)$

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Bleh, I took time typing out the rotation matrix! :) –  Guess who it is. Aug 14 '10 at 1:14

Assuming that you ask where will x end up after being rotated clockwise for an angle $\theta$ around the origin, then the answer is as given by George S. and J.M. I'd like to give a brief explanation of why that is the answer.

The key is that rotation is a linear transformation. If we write Ax for the point x rotated, then then $A(ax+bx) = a(Ax) + b(Ax)$, where a and b are some real numbers. Here ax is a point obtained by stretching both coordinates, and lies on the line that contains x and the origin. It's a scaling operation. Also, $x+y$ is obtained by adding the coordinates, which you can imagine as the diagonal of a parallelogram whose edges are x and y. So the equation $A(ax+bx) = a(Ax) + b(Ax)$ says that if you scale a figure with two points, then `add' the points by taking the parallelogram diagonal, and then rotate you get the same result you get if you start by rotating, and then you scale and add.

Now, assuming that equation is OK when A represents a rotation, note that every vector $x=(a,b)$ can be written as $a(1,0)+b(0,1)$, so its rotation will be $a(A(1,0))+b(A(0,1))$. In other words, all you need to know is where do the points (1,0) and (0,1) end up after being rotated (namely, you need the points A(1,0) and A(0,1), each having two coordinates). So four numbers are enough to figure out where all points end up after being rotated.

Linear transformations are the reason matrices are important.

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That's funny (but not surprising due to equivalence), I on the other hand started from the Argand plane to get to my answer; essentially expanding a point z=x+iy into polar form: $r\exp(i\theta)$, and considered the real and imaginary parts of when I subtract an angle $\theta'$ from the point (corresponding to clockwise rotation). –  Guess who it is. Aug 14 '10 at 7:02

See:

http://en.wikipedia.org/wiki/Rotation_matrix#Two_dimensions

Here since the rotation is clockwise rather than counter clockwise, you would have to substitute $- \theta$ in place of $\theta$.

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I am completely baffled at the downvote this received (which I just canceled). Can someone explain? –  Jason DeVito Aug 14 '10 at 1:36
@Jason: I believe the downvote was because of the feeling that it's better to have answers self-contained (unless referring to something too complicated to type out here), and answers that are nothing more than a link are better left as a comment. –  ShreevatsaR Aug 14 '10 at 7:38
Thank you for the clarification. I'm still learning the etiquette of the site. –  Jason DeVito Aug 14 '10 at 18:11