Explanation

(d)
Given, lines are $(1 + p)x - py + p(1 + p) = 0$ …(i)
$and\ (1 + q)x - qy + q(1 + q) = 0$….(ii)
$and\ y = 0$
on solving Eqs. (i) and (ii), we get
$C\left\{ pq,(1 + p)(1 + q) \right\}$
$\therefore$ Equation of altitude $CM$ passing through $C$ and perpendicular to $AB$ is
$x = pq$ …(iii)
$\because Slope\ of\ line\ (ii)is\ \left( \frac{1 + q}{q} \right)$
$\because Slope\ of\ altitude\ BN\ (as\ shown\ in\ figure)is\frac{- q}{1 + q}$
[image:10_image17.emf]
$\therefore Equation\ of\ BN\ is\ y - 0 = \frac{- q}{1 + q}(x + p)$
$\Rightarrow \ y = \frac{- q}{(1 + q)}(x + p)\ \ \ \ldots(iv)$
Let orthocentre of triangle be $H(h,k)$, which is the point of intersection of Eqs. (iii) and (iv).
$\therefore$ On solving Eqs. (iii) and (iv), we get
$x = pq\ and\ y = - pq$
$\Rightarrow h = pq\ and\ k = - pq$
$\therefore\ h + k = 0$
$\therefore\$Locus of $H(h,k)$ is $x + y = 0$.