Q. The locus of the orthocentre of the triangle formed by the lines $(1 + p )x -p y + p (1 + p) = 0 (1+ q)x-qy + q (1+ q) = 0$ and $y = 0$, where $p\ne q$ is
IIT JEEIIT JEE 2009Straight Lines
Solution:
Given, lines are $(1+p) x-p y+p(1+p)=0 ....$ (i)
and $(1+q) x-q y+q(1+q)=0....$(ii)
On solving Eqs. (i) and (ii), we get
$C\{p q,(1+p)(1+q)\}$
$\therefore$ Equation of altitude $C M$ passing through $C$ and perpendicular to $A B$ is
$x=p q .....$(iii)
$\because$ Slope of line (ii) is $\left(\frac{1+ q}{q}\right)$
$\therefore$ Slope of altitude $B N$ (as shown in figure) is $\frac{-q}{1+q}$.
$\therefore$ Equation of $B N$ is $y-0=\frac{-q}{1+q}(x+p)$
$\Rightarrow y=\frac{-q}{(1+q)}(x+p) .....$(iv)
Let orthocentre of triangle be $H(h, k)$, which is the point of intersection of Eqs. (iii) and (iv).
On solving Eqs. (iii) and (iv), we get
$x=p q$
$y=-p q$
$\Rightarrow h=p q$
and $ k=-p q$
$\therefore h+k=0$
$\therefore$ Locus of $H(h, k)$ is $x+y=0$.
