Rotation on the coordinate plane This question was taken from Khan Academy's Prealculus section

My main issue here is figuring out the right plan of attack for this type of question. This is how I went about it.
1) I found the midpoints of both lines, $AB$ and $MN$ which happened to be $(1.5, -2.5)$ and $(-1.5, 3.5)$, respectively. 
2) Then using the two given points of each line I found the equations of the lines. 
$AB$ was $y=-3 x+2$ and $MN$ was $y=-\frac { 1 }{ 5 } x+\frac { 16 }{ 5 } $
3) Next I found the perpendicular bisectors of both lines by using their midpoints. These came out to be $y=\frac { 1 }{ 3 } x-3$ and $y=5x+11$
4) Then I calculated the point where these two lines intersect each other by setting both equations equal to each other and this point equaled to $(-3,-4)$
Now I am not sure whether or not I am going about this the right way and how to get the coordinates of the point he gave under the $90$ degree rotation
 A: Your methodology for finding the center of rotation is correct. 
$\textbf{Hint}$ for the second part: Look up rotation matrices. Your image point $ \boldsymbol{r} = (x', y')$ satisfies the linear transformation equation:
$$
\boldsymbol{r} = \boldsymbol{x}_0 + R \boldsymbol{\Delta \boldsymbol{x}}
$$
This is basically a multiplication of a position vector $\Delta\boldsymbol{x}$ by a rotation matrix $R$ followed by a shift by a constant vector $\boldsymbol{x}_0$. You have all the information you need to determine $\boldsymbol{x}_0$, $R$ and $\Delta \boldsymbol{x}$. 
A: Let us call the point of intersection of perpendicular bisectors C$(-3,-4)$, and P$(-6,-2)$.
If we want to rotate P$(-6,-2)$ counterclockwise by 90 with center at C, we can shift P by a distance of PC on the perpendicular to PC at C.
Equation of Perpendicular to PC at C is , $-3x+2y-1=0$, Now tricky part is how to use the distance form , $\dfrac{x-x_1}{\cos\theta}=\dfrac{y-y_1}{\sin\theta}=\pm r$. If the slope of line is + then if we go toward +y i.e. +r we will actually be shifting the point clockwise so will use the distance  -r.
$\dfrac{x+3}{\cos\theta}=\dfrac{y+4}{\sin\theta}=- r$
A: Here's how to do this without matrices, using only the elementary tools you used so far:
Let's say that O is the center of rotation, X is the point you want to rotate, and Y is the point it rotates to. Since this is a 90 degrees rotation, the angle between lines OX and OY should be 90 degrees. You know the coordinates of O and X, so you can find the equation of the line OX, and then the equation of the line that is perpendicular to OX at O. Knowing the equation of this line, you can now find the point Y on it that has the particular distance from O equal to OX (which you can also calculate). You will find two possible points Y, of which one will correspond to clockwise and the other to counterclockwise rotation.
