Finding coordinates for 3rd point in isosceles right triangle, knowing 2 other points and lengths of sides I have the following triangle

The length of AB, AC, AC', BC and BC' are known and AB = AC = AC', BC = BC'.
Coordinates of A and B are known.
What is a fast way of finding coordinates for C and C'?
I've tried using the distance formula for AB and AC and tried to get the coordinates from the resulting equation system, but I think I messed up somewhere because the C coordinates were off. Tested with A(1, 0) and B(5, 1) and I got C' y coordinate at 3.05 and that didn't add up.
I'm looking for a general formula if there is one, not one just for the set of points I mentioned above. Haven't tried any trigonometry formulas for this because I didn't know which one to use.
 A: From $A$ and $B$ coordinates you can compute vector $B-A=(x,y)$. Then:
$$
C-A=(-y,x),\quad C'-A=(y,-x),
$$
because they are rotated 90° with respect to $B-A$.
Add these to $A$ and you're done.
A: You have the points $a, b$ and thus the vectors to these points that are just their coordinates
$$\overrightarrow{a}=(a_x ~a_y)^T, \overrightarrow{b}=(b_x ~b_y)^T$$
You can establish a vector going from $a$ to $b$, by subtracting them:
$$\overrightarrow{ab}= \overrightarrow{b} -  \overrightarrow{a}$$
To rotate a vector $(x~y)^T$ by some angle $\alpha$, multiply it with a matrix like so:
$$
\left(
\begin{matrix}
x_{new}\\
y_{new}
\end{matrix}
\right)
=
\left(
\begin{matrix}
\cos \alpha & -\sin \alpha\\
\sin \alpha &\cos \alpha
\end{matrix}
\right)
\left(
\begin{matrix}
x\\
y
\end{matrix}
\right)
$$
Let $\overrightarrow{ab}$ start from $a$ and go to $b$, now rotate the vector counterclockwise 90° and it will go from $a$ to $c$. Counterclockwise rotation is usually defined as positive. The matrix is very easy, because of the 90°:
$$
\overrightarrow{ac}
=
\left(
\begin{matrix}
0 & -1\\
1 & 0
\end{matrix}
\right)
\overrightarrow{ab}
$$
Then you can find $c$ via $a$
$$\overrightarrow{c} = \overrightarrow{a} + \overrightarrow{ac}$$
You could rotate the other way around to find $c'$, but because it is exactly the opposite direction, you can as easily negate the direction vector
$$\overrightarrow{c'} = \overrightarrow{a} - \overrightarrow{ac}$$
