Question

The locus of the orthocentre of the triangle formed by the lines $(1+p)x-py+p(1+p)=0$, $(1+q)x-qy+q(1+q)=0$ and $y=0$, where $p\ne q$, is

Answer 1

REF. Image.

$q(1+n)x-qny+pq(1+p)$

$\therefore A(pq,(p+1\left)\right(q+1\left)\right)$

Equation of ${}^{\prime }A{D}^{\prime }$

$y-(p+1)(q+1)=\frac{-1}{0}(x-pq)$

$\therefore x=pq$......(i)

Equation of ${}^{\prime }B{F}^{\prime }$

$y-0=\frac{-q}{(1+q)}(x+p)$

$\Rightarrow y(1+q)=-q.p(q+1)$

$\therefore y=-pq$........(ii)

From (i) & (ii)

$x=-y$

$\Rightarrow x+y=0$

which is a straight line.