Question

The coordinates of the foot of the perpendiculars from the vertices of a triangle on the opposite sides are $(20,25),(8,16)$ and $(8,9).$ If the orthocentre of the triangle is $(h,k),$ then $k^3+h$ is equal to

Answer 1

Let $A(20,25),B(8,16)$ and $C(8,9)$
Then
$a=BC=\sqrt{{(8-8)}^2+{(9-16)}^2}=7b=AC=\sqrt{{(8-20)}^2+{(9-25)}^2}=20c=AB=\sqrt{{(8-20)}^2+{(16-25)}^2}=15$
As orthocentre of triangle is the incentre of the pedal triangle
$h=\frac{7.20+20.8+15.8}{7+20+15},k=\frac{7.25+20.16+15.9}{7+20+25}\Rightarrow k^3+h=3385$