# 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 ?

• Do you know the angles of the triangle? – Varun Vejalla Jul 24 '19 at 11:36

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 $$\vec{OC}=\binom{x_1}{y_1}+\frac12\binom{x_2-x_1}{y_2-y_1}\pm\frac12\binom{y_2-y_1}{x_1-x_2}.$$

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 $$h,k$$

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 $$t=\pm45^\circ$$