analytic geometrygeometry
Initially the position of the shape was in (100, 100). I am rotating (say 30 degrees) the shape as shown in the image below. I have found the starting point of the rotated object. Is there a formula to find the point of the shape when the rotation was 0 degree? Any help?
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.
Let point $C$ be the center of rotation.
It should belong to the perpendicular bisector of the segment $[P, P']$.
So if we imagine it as a triangle, $CPP'$ should be a equilateral triangle where the angle $C = 15^0$ (Or whatever your rotation angle below 180).
Let's assemble it in a drawing:
Your task can be minimized to computing $M$!:
Knowns:
Solution:
Notice that: $\frac{d}{M} = tan(\frac{\theta}{2})$
So $M = \frac{d}{tan(\frac{\theta}{2})}$
Now you have all the necessary information to find your center of rotation $C$, I will left to you to transform everything to $(x,y)$ coordinates as an exercise ;)
Best Answer
Rotating any point $(x,y)$ on a plane about the origin by $\theta$ degrees (counter-clockwise) is given by $$ \left[ \begin{array}{c} x' \\ y' \\ \end{array} \right] = \left[ \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \end{array} \right], $$ where $(x',y')$ denotes your new position.