NIMCET 2018 Question Paper with solutions

Calculation: Let's say that the lines are:
$\mathit{L}1\to \left(1+\mathit{p}\right)\mathit{x}-\mathit{p}\mathit{y}+\mathit{p}\left(1+\mathit{p}\right)=0$
$\mathit{L}2\to \left(1+\mathit{q}\right)\mathit{x}-\mathit{q}\mathit{y}+\mathit{q}\left(1+\mathit{q}\right)=0$
$\mathit{L}3\to \mathit{y}=0$
Multiplying ${\mathit{L}}_1$ by $\mathit{q}$ and ${\mathit{L}}_2$ by $\mathit{p}$ and subtracting will give us:
$\mathit{x}\left(\mathit{q}+\mathit{p}\mathit{q}-\mathit{p}-\mathit{p}\mathit{q}\right)+\left(\mathit{p}\mathit{q}+{\mathit{p}}^2\mathit{q}-\mathit{p}\mathit{q}-\mathit{p}{\mathit{q}}^2\right)=0$
$\Rightarrow \mathit{x}\left(\mathit{q}-\mathit{p}\right)+\mathit{p}\mathit{q}\left(\mathit{p}-\mathit{q}\right)=0$
$\Rightarrow \mathit{x}=\mathit{p}\mathit{q}.$
Substituting this in either ${\mathit{L}}_1$ or ${\mathit{L}}_2$, we will get:
$\left(1+\mathit{p}\right)\left(\mathit{p}\mathit{q}\right)-\mathit{p}\mathit{y}+\mathit{p}\left(1+\mathit{p}\right)=0$
$\Rightarrow \mathit{y}=\left(1+\mathit{p}\right)\left(1+\mathit{q}\right)$
Therefore:
${\mathit{L}}_1$ and ${\mathit{L}}_2$ intersect at $\mathit{A}\left[\mathit{p}\mathit{q},\left(1+\mathit{p}\right)\left(1+\mathit{q}\right)\right]$ .
$\mathit{L}1$ and $\mathit{L}3$ intersect at $\mathit{B}\left(-\mathit{p},0\right)$ .
$\mathit{L}2$ and L3 intersect at C $\left(-\mathit{q},0\right)$ .
Using the formula for line perpendicular to two points:
Equation of altitude from A:
$\left(-\mathit{q}+\mathit{p}\right)\mathit{x}+\left(0\right)\mathit{y}=\left(-\mathit{q}+\mathit{p}\right)\left(\mathit{p}\mathit{q}\right)+\left(0\right)\left(1+\mathit{p}\right)\left(1+\mathit{q}\right)$
$\Rightarrow \mathit{x}=\mathit{p}\mathit{q}$
Equation of altitude from B:
$\left(\mathit{p}\mathit{q}+\mathit{q}\right)\mathit{x}+\left[\left(1+\mathit{p}\right)\left(1+\mathit{q}\right)-0\right]\mathit{y}=\left(\mathit{p}\mathit{q}+\mathit{q}\right)\left(-\mathit{p}\right)+\left[\left(1+\mathit{p}\right)\left(1+\mathit{q}\right)-0\right]\left(0\right)$
$\Rightarrow \mathit{q}\left(1+\mathit{p}\right)\mathit{x}+\left(1+\mathit{p}\right)\left(1+\mathit{q}\right)\mathit{y}=\left(-\mathit{p}\mathit{q}\right)\left(1+\mathit{p}\right)$
$\Rightarrow \mathit{q}\mathit{x}+\left(1+\mathit{q}\right)\mathit{y}+\mathit{p}\mathit{q}=0...\left(2\right)$
To find the ortho-center, we solve the equations (1) and (2) of the two altitudes:
$\mathit{q}\left(\mathit{p}\mathit{q}\right)+\left(1+\mathit{q}\right)\mathit{y}+\mathit{p}\mathit{q}=0$
$\Rightarrow \left(1+\mathit{q}\right)\mathit{y}+\mathit{p}\mathit{q}\left(1+\mathit{q}\right)=0$
$\Rightarrow \mathit{y}=-\mathit{p}\mathit{q}$
$\therefore$ The co-ordinates of the ortho-center are $\mathit{x}=\mathit{p}\mathit{q}$ and $\mathit{y}=-\mathit{p}\mathit{q}$.
The locus of the ortho-center will be $\mathit{x}+\mathit{y}=0$,
which is a straight line.