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

• A. $y=-2x$
• B. $y=2x$
• C. $y=x$
• D. $y=-x$

Answer 1

The correct option is D $y=-x$

Let $L_1:(1+p)x-py+p(1+p)=0,$
$L_2:(1+q)x-qy+q(1+q)=0$ and
$L_3:y=0$
The coordinates of vertices are
$A(-p,0),B(-q,0),C(pq,(1+p\left)\right(1+q\left)\right)$


The equation of the altitude through $C$ is
$x=pq\cdots \left(1\right)$
The equation of the altitude through $B$ is
$px+(1+p)y+\lambda =0$
Putting $B(-q,0)$, we get
$-pq+\lambda =0\Rightarrow \lambda =pq$
Now, equation of altitude through $B$
$px+(1+p)y+pq=0\cdots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get
$p^{2}q+(1+p)y+pq=0\Rightarrow y=-pq$

Let $(h,k)$ be the coordinates of the orthocentre
$h=pq,k=-pq\Rightarrow k=-h$

Hence, the locus of orthocentre is $y=-x$.