Question

Find the locus of the third vertex of a right angled triangle, the ends of whose hypotenuse are $(4,0),(0,4)$.

Answer 1

$mAB\times mBC=-1$

$\Rightarrow \frac{-k}{4-h}\times \frac{4-k}{-h}=-1$

$\Rightarrow -4k+k^2=4h-h^2$

$\Rightarrow h^2+k^2-4h-4k=0$

$\therefore x^2+y^2-4x-4y=0$