Finding the third vertex of a right isosceles triangle

Question

Okay so, here is what I was stuck on:

Two vertices of isosceles right triangle ABC are located at A(2,2) and B(5,2), where segment AB is a leg of the triangle.
Find the coordinates of C.
Find the coordinates of a different C.
How many different locations are there for C.

I've tried everything, but I just don't know how to go about this. . .

Could someone please point me in the right direction???

Answer 1

HINT: We are told that $AB$ is a leg of a right isosceles triangle. So the other leg, of equal length (3 units) and perpendicular to $AB$, is either $BC$ or $AC$. $AB$ is horizontal, so the other leg must be vertical. Can you see how to continue from here?

If you don't, refer to the diagram below:

P   Q
|\ /|
| X |
|/ \|
A---B
|\ /|
| X |
|/ \|
R   S

Here $C$ can be any one of $P,Q,R,S$.

Answer 2

Let us use complex numbers. $A$ is $2 + 2i$ and $B$ is $5+2i$. The other vertex can be obtained by rotating $AB$ about $A$ or $B$ by $90^\circ$ either clockwise or anticlockwise. Thus the four possibilities for $C$ are \begin{align*} 2+2i + i(5+2i - 2 - 2i) &= 2+5i\\ 2+2i - i(5+2i - 2 - 2i) &= 2-i\\ 5+2i+ i(2+2i-5-2i) &= 5-i \\ 5+2i - i(2+2i-5-5i) &= 5+5i \end{align*} Thus the required vertex positions for $C$ are $(2,5), (2,-1), (5,-1), (5,5)$