JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Straight Lines

The first line intersects $y=0$ at $x=-p$, giving $B=(-p,0)$.

The second line intersects $y=0$ at $x=-q$, giving $A=(-q,0)$.

The two lines intersect at $C=(pq,(p+1)(q+1))$.

The altitude from $C$ to $AB$ (which is the $x$-axis) is the vertical line $x=pq$.

The altitude from $B$ to $AC$ has slope perpendicular to $AC$. The slope of $AC$ is $\frac{(p+1)(q+1)}{pq+q}=\frac{p+1}{q}$.

The altitude from $B$ has slope $-\frac{q}{p+1}$ and passes through $(-p,0)$:


At $x=pq$:


Thus the orthocentre is at $(pq,-pq)$, which satisfies $x+y=0$, a straight line.