Triangle with Special Coordinates

Starting with cross-multiplication, we get $\begin{aligned}(u-x)(u+x) & = (y-v)(y+v) \\ u^2 - x^2 & = y^2 - v^2 \\ u^2 + v^2 & = x^2 + y^2 \\ |OA|^2 & = |OB|^2 \\ |OA| & = |OB|.\end{aligned}$ Thus, the triangle is isosceles, with $O$ as the vertex.

Alternatively, consider the parallelogram $OAPB$ , where $P\:(u+x,v+y)$ . Then the fractions $s_1 = \frac{y-v}{x-u},\ s_2 = \frac{y+v}{x+u}$ describe the slopes of the diagonals $AB$ and $OP$ , respectively. The given equation says that $-1/s_1 = s_2$ ; thus, the diagonals are perpendicular. But a parallelogram with perpendicular diagonals is a rhombus, so that $|OA| = |OB|$ .