Find the third vertex if the vertices of the hypotenuse of an isosceles right triangle are given 
If the vertices of the hypotenuse of an isosceles right triangle are $(9,-9)$ and $(1,-3)$, then find the third vertex

The fomula for this is given by $(x,y)=\bigg(\dfrac{x_1+x_2\pm(y_1-y_2)}{2},\dfrac{y_1+y_2\mp(x_1-x_2)}{2}\bigg)$
I really don't seem to find any reason to simply just bi-hart the equation. What is the easier way to solve such problems ?. Can I do transformations of the triangle and solve the problem ?
Attempt

$A(9,-9)$, $B(1,-3)$, $C(a,b)=\;?$
First the triangle is transformed such that the third vertex becomes zero.ie, $A(9-a,-9-b)$, $B(1-a,-3-b)$, $C(0,0)$. Now I think I can rotate AC and BC so that they coincide with the $x$ and $y$ axes, then from there I can find the third vertex ?
How do I do it mathematically ?
 A: One easy way to see this is to add in the midpoint $M$ of $AB$. Now since the angle at $C$ is $90^\circ$, $MA=MB=MC$, and since it is isosceles $MC$ is perpendicular to $AB$. The vector $\vec{AM}=\frac12\vec{AB}=\frac12\binom{x_2-x_1}{y_2-y_1}$, and $\vec{MC}$ is equal in magnitude and perpendicular, so must be $\pm\frac12\binom{y_2-y_1}{x_1-x_2}$. Thus $$\vec{OC}=\binom{x_1}{y_1}+\frac12\binom{x_2-x_1}{y_2-y_1}\pm\frac12\binom{y_2-y_1}{x_1-x_2}.$$
A: If $(x_1,y_1);(x_2,y_2)$ are the two vertices of hypotenuse of  an isosceles right triangle 
and $(h,k)$ be third vertex
$$\dfrac{y_1-k}{x_1-h}\cdot\dfrac{y_2-k}{x_2-h}=-1\ \ \ \ (1)$$
and $$(x_1-h)^2+(y_1-k)^2=(x_2-h)^2+(y_2-k)^2$$
$$\iff x_1^2-2x_1h+y_1^2-2y_1k=x_2^2-2x_2h+y_2^2-2y_2k\ \ \ \  (2)$$
Solve the two simultaneous equations for $h,k$ 
A: Hint:
If $(x_1,y_1);(x_2,y_2)$ are the two vertices of hypotenuse of  an isosceles right triangle 
and $(h,k)$ be third vertex
$$\dfrac{k-y_1}{\cos t}=\dfrac{h-x_1}{\sin t}=r$$
where $2r^2=(x_1-x_2)^2+(y_1-y_2)^2$  where $t=\pm45^\circ$
