TITLE: rotating a rectangle via a rotation matrix
QUESTION [5 upvotes]: I want to rotate a 2D rectangle using a rotation matrix. After the rotation, I want the points (x, y) of the rectangle to be:
(625.49939, 632.40015) top left
(636.49939, 632.40015) top right
(636.49939, 672.40015) bottom right
(625.49939, 672.40015) bottom left

Before the rectangle is rotated, it appears 40 units wide and 11 units tall. After the rotation, it will appear 11 units wide and 40 units tall.
I know that I have to rotate it 90 degrees, but how can I find exactly where to place the rectangle before the rotation so that it ends up in my desired position?

The rectangle starts in position ??? so that it appears 40 units wide and 11 units tall
The rectangle will be rotated 90 degrees clockwise (at its center point).
The rectangle ends with the coordinates listed above.

How do I find the starting coordinates?

REPLY [2 votes]: The calculation follow the steps of @mixedmath:
Let's denote the points of the rectangle with
\begin{align*}
A&=(625.49939,632.40015)=(x_A,y_A)\\
B&=(636.49939,632.40015)=(x_B,y_A)\\
C&=(636.49939,672.40015)=(x_B,y_C)\\
D&=(625.49939,672.40015)=(x_A,y_C)\\
\end{align*}

1. Step: Center $M$


We calculate the center $M$ of the rectangle $\Box(A,B,C,D)$:
\begin{align*}
M=\frac{1}{2}(A+C)=\frac{1}{2}(B+D)=\frac{1}{2}(x_A+x_B,y_A+y_C)
\end{align*}


2. Step: Shift $M$ to $(0,0)$


We shift the rectangle so that $M$ becomes the origin and obtain a rectangle $\Box(A^\prime, B^\prime, C^\prime, D^\prime)$
\begin{align*}
A^\prime&=A-M=\frac{1}{2}(x_A-x_B,y_A-y_C)\\
B^\prime&=B-M=\frac{1}{2}(-x_A+x_B,y_A-y_C)\\
C^\prime&=C-M=\frac{1}{2}(-x_A+x_B,-y_A+y_C)\\
D^\prime&=D-M=\frac{1}{2}(x_A-x_B,-y_A+y_C)\\
\end{align*}

Intermezzo: Rotation matrix
We consider the rotation matrix $\Phi$ clockwise through $90^\circ$ which corresponds to $-\frac{\pi}{2}$
\begin{align*}
\Phi=\begin{pmatrix}
\cos \phi&-\sin \phi\\
\sin \phi& \cos \phi
\end{pmatrix}
=
\begin{pmatrix}
\cos\left(-\frac{\pi}{2}\right)&-\sin\left( -\frac{\pi}{2}\right)\\
\sin\left( -\frac{\pi}{2}\right)& \cos\left( -\frac{\pi}{2}\right)
\end{pmatrix}
=
\begin{pmatrix}
0&1\\
-1& 0
\end{pmatrix}
\end{align*}
A point $P=(x_P,y_P)$ will be transformed via $\Phi$ to
\begin{align*}
\Phi P=\begin{pmatrix}
0&1\\
-1& 0
\end{pmatrix}
\begin{pmatrix}
x_P\\
y_P
\end{pmatrix}
=
\begin{pmatrix}
y_P\\
-x_P
\end{pmatrix}
\end{align*}

3. Step: Rotate rectangle clockwise through $-\frac{\pi}{2}$


We now rotate the rectangle $\Box(A^\prime, B^\prime, C^\prime, D^\prime)$ and obtain a rectangle $\Box(\Phi A^\prime, \Phi B^\prime, \Phi C^\prime ,\Phi D^\prime)$
\begin{align*}
\Phi A^\prime&=\frac{1}{2}(y_A-y_C,-x_A+x_B)\\
\Phi B^\prime&=\frac{1}{2}(y_A-y_C,x_A-x_B)\\
\Phi C^\prime&=\frac{1}{2}(-y_A+y_C,x_A-x_B)\\
\Phi D^\prime&=\frac{1}{2}(-y_A+y_C,-x_A-x_B)\\
\end{align*}


4. Step: Shift rectangle back from origin to $M$


We reverse step 2 by adding $M$ to the points of $\Box(\Phi A^\prime, \Phi B^\prime, \Phi C^\prime ,\Phi D^\prime)$
\begin{align*}
\Phi A^\prime+M&=\frac{1}{2}(x_A+x_B+y_A-y_C,-x_A+x_B+y_A+y_C)=(610.99939,657.90015)\\
\Phi B^\prime+M&=\frac{1}{2}(x_A+x_B+y_A-y_C,x_A-x_B+y_A+y_C)=(610.99939,646.90015)\\
\Phi C^\prime+M&=\frac{1}{2}(x_A+x_B-y_A+y_C,x_A-x_B+y_A+y_C)=(650.99939,646.90015)\\
\Phi D^\prime+M&=\frac{1}{2}(x_A+x_B-y_A+y_C,-x_A-x_B+y_A+y_C)=(650.99939,657.90015)\\
\end{align*}

Note: A code can be easily derived from the calculations

[2015-07-28] Add-on: According to @BROYs comment here's the rotational displacement of the rectangle using homogeneous matrices.


A translational displacement by a vector $U=\begin{pmatrix}x_U\\y_U\end{pmatrix}$ can be represented by the homogeneous matrix $T_U$ with
\begin{align*}
  T_U=\begin{pmatrix}
    1&0&x_U\\
    0&1&y_U\\
    0&0&1\\
    \end{pmatrix}
  \end{align*}

The point $A=(x_A,y_A)$ in homogeneous coordinates is $\begin{pmatrix}x_A\\y_A\\1\end{pmatrix}$. The translation of $A$ in direction $U$ is
\begin{align*}
  T_UA=\begin{pmatrix}
    1&0&x_U\\
    0&1&y_U\\
    0&0&1\\
    \end{pmatrix}
  \begin{pmatrix}
    x_A\\
    y_A\\
    1
  \end{pmatrix}
 =
 \begin{pmatrix}
   x_A+x_U\\
   y_A+y_U\\
   1
   \end{pmatrix}
    \end{align*}
Observe $T_UA$ corresponds to $A+U$.

A rotational displacement by a vector $\Phi=-\frac{\pi}{2}$ can be represented by the homogeneous matrix $R_\Phi=R_{-\frac{\pi}{2}}$ with
\begin{align*}
  T_\Phi&=\begin{pmatrix}
    \cos(\Phi)&-\sin(\Phi)&0\\
    \sin(\Phi)&\cos(\Phi)&0\\
    0&0&1\\
    \end{pmatrix}\\
T_{-\frac{\pi}{2}}&=
  \begin{pmatrix}
    0&1&0\\
    -1&0&0\\
    0&0&1\\
    \end{pmatrix}  
  \end{align*}

The rotation of $A$ through an angle $\Phi=-\frac{\pi}{2}$
\begin{align*}
 T_{-\frac{\pi}{2}}A= \begin{pmatrix}
    0&1&0\\
    -1&0&0\\
    0&0&1\\
    \end{pmatrix}
  \begin{pmatrix}
    x_A\\
    y_A\\
    1
  \end{pmatrix}
 =
 \begin{pmatrix}
   y_A\\
   -x_A\\
   1
   \end{pmatrix}
    \end{align*}
Note: In the following it's more convenient to use vector notation for the points. I will omit the transpose symbol.
Homogeneous matrices in action: The players: The four points of the rectangle $A,B,C,D$ and the center $M$ of the rectangle \begin{align*}
A&=\begin{pmatrix}x_A\\
y_A\\
1\end{pmatrix}
\qquad
B=\begin{pmatrix}
x_B\\ y_B\\1
\end{pmatrix}\qquad
C=\begin{pmatrix}
x_C\\ y_C\\ 1\end{pmatrix}\qquad
D=\begin{pmatrix} x_D\\y_D\\ 1\end{pmatrix}\\
M&=\frac{1}{2}(A+C)=\frac{1}{2}(B+D)=\frac{1}{2}\begin{pmatrix} x_A+x_B\\y_A+y_C\\2
\end{pmatrix}
\end{align*}
Note: You could also use 1/4 (A+B+C+D) which will always work irrespective of the order of the coordinates.

First we want to move $M$ to the origin. This is a translational displacement by $-M$ represented by the homogeneous matrix $T_{-M}$. Next we rotate through $-\frac{\pi}{2}$ represented by $R_{-\frac{\pi}{2}}$ and at the end we shift back by $T_M$. The complete displacement is given by


\begin{align*}
   T_M R_{-\frac{\pi}{2}} T_{-M}&=
   \begin{pmatrix}
   1&0&\frac{1}{2}(x_A+x_B)\\
   0&1&\frac{1}{2}(y_A+y_C)\\
   0&0&1
   \end{pmatrix}
  \begin{pmatrix}
    0&1&0\\
    -1&0&0\\
    0&0&1\\
    \end{pmatrix}
   \begin{pmatrix}
   1&0&-\frac{1}{2}(x_A+x_B)\\
   0&1&-\frac{1}{2}(y_A+y_C)\\
   0&0&1
   \end{pmatrix}\\
   &=
   \begin{pmatrix}
   1&0&\frac{1}{2}(x_A+x_B)\\
   0&1&\frac{1}{2}(y_A+y_C)\\
   0&0&1
   \end{pmatrix}
   \begin{pmatrix}
     0&1&-\frac{1}{2}(y_A+y_C)\\
     -1&0&-\frac{1}{2}(x_A+x_B)\\
   0&0&1
   \end{pmatrix}\\
   &=
   \begin{pmatrix}
     0&1&-\frac{1}{2}(x_A+x_B-y_A-y_C)\\
     -1&0&-\frac{1}{2}(x_A+x_B+y_A+y_C)\\
   0&0&1
\end{pmatrix}
   \end{align*}

The complete displacement of the point $A$ is obtained by
\begin{align*}
     T_MR_{-\frac{\pi}{2}}T_{-M}A&=T_MR_{-\frac{\pi}{2}}T_{-M}\begin{pmatrix}
   x_A\\
   y_A\\
   1
   \end{pmatrix}=
     \frac{1}{2}\begin{pmatrix}
       x_A+x_B+y_A-y_C\\
       -x_A+x_B+y_A+y_C\\
       2
       \end{pmatrix}=
     \begin{pmatrix}
       610.99939\\
       657.90015\\
       1
       \end{pmatrix}
  \end{align*}
and similarly for $B,C,D$.