How to find the rotation of $A$, $B$ and $C$ about a fixed point like $(x_1,y_1)$? The problem is as follows:

Let the points:
$A=(-6,-1)$
$B=(-5,1)$
$C=(-2,-2)$
$D=(-3,2)$
These represent a quadrilateral on the $x-y$ coordinate system. Find the sum of the coordinates from all the vertex of such quadrilateral after the figure has been moved $6$ units to the right and rotated $90^{\circ}$ degrees about point $E=(5,-4)$ in the counterclockwise direction.

The choices given by my workbook are as follows:
$\begin{array}{ll}
1.&-25\\
2.&-22\\
3.&-23\\
4.&-24\\
\end{array}$
I'm not very sure how to solve this problem. But I can tell that when you first move six units to the right the new coordinates become as follows:
$A_2=(0,-1)$
$B_2=(1,1)$
$C_2=(4,-2)$
$D_2=(3,2)$
Then when it indicates that it has been rotated $\frac{\pi}{2}$ to the counterclockwise direction about point $(5,-4)$.
$A_3=(2,-9)$
$B_3=(0,-8)$
$C_3=(3,-5)$
$D_3=(-1,-6)$
In short I ommited the steps on how I did find these, but to my intuition I believe that if you make a right triangle you can then use this as an aid to rotate such right triangle to get the new coordinate.
For example what I did for point $A_3$.
With respect of point $E=(5,-4)$, point $A_2=(0,-1)$ means that $A_2$ is five units to the left of point $E$ on the $\textrm{x-direction}$, therefore when it is rotated $\frac{\pi}{2}$ to the counterclockwise direction it will became the new $\textrm{y-direction}$ thus this will be added to the previous $y$ coordinate on point $E$ moving it further downwards.
$y=-4-5=-9$
Conversely this point $A_2$ is located three units above point $E$ thus when rotated $\frac{\pi}{2}$ to the counterclockwise direction it will become the new $\textrm{x-direction}$. But because it is to the left of point $E$, it will be moved three units to the left of that point.
$x=5-3=2$
Therefore $A_3=(2,-9)$.
Thus adding all of these results into:
$2-9+0-8+3-5-1-6=-24$
Thus I believe the answer is$ -24$ or choice 4.
But I cannot be sure if this is the correct answer. Since I'm not good with spatial abilities. I really beg someone could instruct me on how to see these in the cartesian plane?. Can someone put some drawing for these rotations?. By the way does it exist a formula for these?
I can't see it clearly so I need help on this one.
 A: I'm going to continue from here.
$$A_2=(0,−1)$$
$$B_2=(1,1)$$
$$C_2=(4,−2)$$
$$D_2=(3,2)$$
To rotate these points 90º degrees about point $E=(5,−4)$ in the counterclockwise direction you could use a rotation matrix, or even better, treat vectors as complex numbers. That's what I'm going to do. That means $\vec{EA_3}=i\vec{EA_2}$ and so forth.
We have
\begin{align*}
\vec{EA_2}&=(0,−1)-(5,-4)=-5+3i &\Rightarrow& \vec{EA_3}=-3-5i\\
\vec{EB_2}&=-4+5i &\Rightarrow& \vec{EB_3}=-5-4i\\
\vec{EC_2}&=-1+2i &\Rightarrow& \vec{EC_3}=-2-i\\
\vec{ED_2}&=-2+6i &\Rightarrow& \vec{ED_3}=-6-2i
\end{align*}
Therefore, the sums of the coordinates of $A_3,B_3,C_3$ and $D_3$ is $-3-5-5-4-2-1-6-2+\underbrace{4(5-4)}_{\text{4 times the sum of the coordinates of $E$.}} = -24$, as you've found.
A: 
You have calculated correctly.
As for visualization, the rotation through $90º$ is one of the easiest angles to work with.  For example, to move $B_2$ to $B_3$, I noted that $B_2$ was $4$ units left and $5$ units up from the rotation center, so I started from the rotation center and moved $4$ units down and $5$ units left.  Up $\rightarrow$ Left $\rightarrow$ Down $\rightarrow$ Right $\rightarrow$ Up...and so on.
