Rotation in 2D

A rotation of a point $(x, y)$ about a point $(h, k)$ by an angle $\theta$ to a new point $(x', y')$ is given by:

$x' = (x - h) \cos \theta + (y - k) \sin \theta + h$

$y' = -(x - h) \sin \theta + (y - k) \cos \theta + k$

If $A(1, 0)$ rotates to $A'(-2, 3)$ , then:

$-2 = (1 - h) \cos \theta + (0 - k) \sin \theta + h$

$3 = -(1 - h) \sin \theta + (0 - k) \cos \theta + k$

And if $B(-2, -3)$ rotates to $B'(1, 6)$ , then:

$1 = (-2 - h) \cos \theta + (-3 - k) \sin \theta + h$

$6 = -(-2 - h) \sin \theta + (-3 - k) \cos \theta + k$

These four equations solve to $h = -\frac{1}{2}$ , $k = \frac{3}{2}$ , $\cos \theta = -1$ , and $\sin \theta = 0$ , so that this particular rotation is given by:

$x' = (x + \frac{1}{2})(-1) + (y - \frac{3}{2})(0) - \frac{1}{2} = -x - 1$

$y' = -(x + \frac{1}{2})(0) + (y - \frac{3}{2})(-1) + \frac{3}{2} = -y + 3$

Therefore, $C(4, 5)$ will rotate to:

$x' = -x - 1 = -4 - 1 = -5$

$y' = -y + 3 = -5 + 3 = -2$

in other words, to $C'(-5, -2)$ . Therefore, $x = -5$ , $y = -2$ , and $x + y = \boxed{-7}$ .

The overall transformation matrix is

$M=T^{-1}RT$ , where $T=\begin {pmatrix} 1 & 0 & -x_1\\0 & 1 & -y_1\\0 & 0 & 1\end {pmatrix}$

is the translation matrix, $T^{-1}$ is it's inverse, and

$R=\begin {pmatrix} \cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\0 & 0 & 1\end {pmatrix}$

is the matrix of rotation.

Using this and the transformation equation

$X'=MX$ where $X'=\begin {pmatrix} x\\y\\1\end {pmatrix},X=\begin {pmatrix}4\\5\\1\end {pmatrix}$

we get

$M=\begin {pmatrix} -1 & 0 & -1\\0 & -1 & 3\\0 & 0 & 1\end {pmatrix}$ ,

and $X'=\begin {pmatrix} -5 \\ -2 \\1\end {pmatrix}$

So, $x=-5,y=-2,x+y=\boxed {-7}$ .

Let $D=A-\overrightarrow{AB}=(4,3)$ then $D'=A'-\overrightarrow{A'B'}=(-5,0)$

Let $F=A-\frac{5}{3}\overrightarrow{AB}=(6,5)$ then $F'=A'-\frac{5}{3}\overrightarrow{A'B'}=(-7,-2)$

As $C'D'=CD$ so $(x+5)^2+y^2=(4-4)^2+(5-3)^2=4$

As $C'F'=CF$ so $(x+7)^2+(y+2)^2=(4-6)^2+(5-5)^2=4$

After subtracting this equation we get $2(2x+12)+2(2y+2)=0 \Rightarrow 4(x+y+7)=0 \Rightarrow x+y=-7$