Triangle $ABC$ has 2 given vertices, $A(1,1)$ and $B(5,3)$. Also, AC=BC and $\angle ACB = \,^{\circ}\mathrm{90}$.
The triangle is in the first quadrant entirely. What are the coordinates of vertex C?
I could only figure out that AB = $2\sqrt{5}$ and that Line $ = 0,5x + 0,5 $
But I don't know what to do after that? Can anyone help? What also confuses me is $\angle ACB = \,^{\circ}\mathrm{90}$. What exactly is $\angle ACB$