[Math] Find third point to make isosceles triangle with a specific area

Question

Using points A(1,2) and B(-2,-2), find a third point, with a positive y-value, that makes ABC an isosceles triangle with area 10 units${^2}$.

I have found AB to be 5 and used this as $r^2$ below..

So using A as the centre of the circle, I have equation $(x-1)^2+(y-2)^2 = 25$

So I'd love to just pick a positive, arbitrary y-value and find x, but I need the triangle's area to be 10. But until I find the third piont, C(x,y), I'm not sure how to find base/height in order to fix the area.

Answer 1

Hint: Consider the line $\ell$ orthogonal to the segment $AB$ and passing through its midpoint. What can you tell about the triangle $\Delta ABC$ when $C$ is a point of $\ell$? Can you use this to find what you're looking for?