Question

If $A(7,9),B(3,-7)$ and $C(-3,3)$ are the vertices of a triangle and $S(\alpha ,\beta ),O(l,m)$ are its circumcentre and orthocentre respectively, then the square of the distance between $P(m,\alpha )$ and $Q(\beta ,l)$ is

Answer 1

Given vertices: $A(7,9),B(3,-7)$ and $C(-3,3)$
$AB=\sqrt{16+256}=\sqrt{272}BC=\sqrt{36+100}=\sqrt{136}AC=\sqrt{100+36}=\sqrt{136}$
$\therefore AC=BC$ and $A{B}^2=B{C}^2+A{C}^2$
Hence, $△ABC$ is right angled isosceles triangle with $\angle C={90}^{\circ }$
$\therefore S(\alpha ,\beta )=$ Mid point of $AB$ i.e., $(5,1)$
$O(l,m)=$ Vertex $C=(-3,3)$

$P(m,\alpha )=(3,5)Q(\beta ,l)=(1,-3)\therefore P{Q}^2=4+64=68$