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 $4y=0,$ where $p\ne q$ is

• A. $x-y=0$
• B. $x+y=0$
• C. $x+2y=0$
• D. $x-2y=0$

Answer 1

The correct option is B $x+y=0$

Equations of sides of triangle are
$(1+p)x-py+p(1+p)=0\cdots \left(1\right)$
$(1+q)x-qy+q(1+q)=0\cdots \left(2\right)$
$y=0\cdots \left(3\right)$
Solving $\left(1\right)$ and $\left(3\right),$ we get the intersection point of the two lines.
$B\equiv (-p,0)$

Solving $\left(2\right)$ and $\left(3\right),$ we get $C\equiv (-q,0)$

Solving $\left(1\right)$ and $\left(2\right),$ we get $A\equiv (pq,(p+1)(q+1))$

Now, slope of $BC$ is $0$
So, equation of altitude from $A$ to $BC$ is
$x=pq\cdots \left(4\right)$

Slope of $AC$ is $\frac{(p+1)(q+1)}{pq+q}=\frac{1+q}{q}$
So, equation of altitude from $B(-p,0)$ to $AC$ is
$y=\frac{-q}{1+q}(x+p)\cdots \left(5\right)$

As orthocentre is the point where all the three altitudes of a triangle intersect each other,
solving $\left(4\right)$ and $\left(5\right),$ we get orthocentre as $(pq,-pq)$
Thus, the locus of the orthocentre is $x+y=0$