JEE Advanced 2009 Paper 2Question 1
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 $p\neq q$, is
Points and Straight Lines
Solution
Vertices $(-p,0)$, $(-q,0)$ and $C=(pq,\,pq+p+q+1)$. The altitude from $C$ is $x=pq$; the altitude from $(-p,0)$ is perpendicular to the second line: $q(x+p)+(1+q)y=0$. Solving gives the orthocentre $(pq,-pq)$, which lies on the line $x+y=0$.