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):

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


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.

  • $\begingroup$ 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). $\endgroup$ – J. M. is a poor mathematician Aug 14 '10 at 7:02



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


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