Question

If the coordinates of the orthocentre of the triangle formed by the lines $y=0,37x-36y+37\times 36=0$ and $64x-63y+64\times 63=0$ is $(a,b),$ then $a-b$ is equal to

Answer 1

Let the equation of $BC$ be $y=0$, then the coordinates of $B$ are $(-36,0)$ and of $C$ are $(-63,0)$.

Equation of $AB$ is $37x-36y+37\times 36=0$ and of $AC$ is $64x-63y+64\times 63=0$.
Equation of the altitudes through $B$ and $C$ are
$y=\frac{-63}{64}(x+36),y=\frac{-36}{37}(x+63)$
Solving these equations we get the coordinates of the orthocentre of the $△ABC$
So, $a=36\times 63,b=-36\times 63$
$\Rightarrow a-b=2\times 36\times 63=4536$