How to find the rotation of $A$, $B$ and $C$ about a fixed point like $(x_1,y_1)$?

Question

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.

Answer 1

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.

Answer 2

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.