Finding coordinate of vertice of triangle given 2 other vertices and angle (Coordinate Geometry)

Question

Two points $A$ and $B$ and have coordinates (-1, 0) and (7, 2) respectively. The point $C(x,y)$ where $x>0$ and $y>0$, lies on the perpendicular bisector of $AB$ and $ACB=90^\circ$ .

Find the coordinates of $C$.


I know that $CA$ and $CB$ will have the same length, so I tried using the distance formula to come up with 2 equations whereby $CA$ = $CB$, but I ended up with both $x^2$ and $y^2$, at this point which I' stuck.

Any hint will be appreciated.

Answer 1

for the Point $$C(x;y)$$ you have two conditions: $$(x+1)+y^2=(x-7)^2+(y-2)^2$$ and $$y=-4x+13$$ can you solve it now?

Answer 2

The mid point $D$ of $AB$ is $(3,1)$ and the point $C$ can be got by rotating $DA$ by $90^\circ$. Hence it is $3+i \pm i((3+i) - (-1)) = 3+i \pm (4i -1)$. Hence the possible coordinates of $C$ are $(2,5)$ or $(4,-3)$