# Find a point position on a rotated rectangle

I am programming a $2D$ game, and I have noticed I am having trouble getting the position $(X, Y)$ of a rectangle's corner, when such rectangle is rotated. The position I am seeking is absolute in the $2D$ space.

As you can see, I need help finding the correct formula to find the bottom-right corner of my rectangle when it is rotated. Given the rectangle's dimensions are $100\times 50$, and it has been rotated by degrees.

Also, the $Y$ position in this language is increased the lower the point is. Basically, the top of the screen is $Y=0$ and the bottom is $Y=600$

Thank you.

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Rotating with respect to what? Where is the origin of your coordinate system? – J. M. May 7 '11 at 0:31
The origin? Do you mean, the top-left of the rectangle? Sorry I forgot to mention that. – Zol Tun Kul May 7 '11 at 0:35
The origin: you're rotating with respect to some point. What's that point? Note that if your (horizontal) rectangle were laying on the horizontal axis and the lower left corner is $(0,0)$, the coordinate of your lower right corner is $(100,0)$. – J. M. May 7 '11 at 0:37
(0,50) is the origin. Sorry, my mistake twice :( – Zol Tun Kul May 7 '11 at 0:41

The idea is to first translate your coordinates to a new coordinate system such that the origin is at $(0,0)$ (subtract $(0,50)$ from all your original coordinates), perform the rotation, and then translate back.
So, in this new coordinate system, the bottom of your rectangle has endpoints at $(0,0)$ and $(100,0)$. You can then use a(n anticlockwise) rotation matrix to transform the point $(100,0)$ to its rotated image:
\begin{align*}100\cos45^\circ-0\sin45^\circ&=50\sqrt{2}\\100\sin45^\circ+0\cos45^\circ&=50\sqrt{2}\end{align*}
and then undo the translation you made before rotating, which yields the point $(50\sqrt{2},50+50\sqrt{2})$ as the coordinates for the upper left corner in its new position.