Straight Line and Pair of Straight Lines 1 Question 24
24. The locus of the orthocentre of the triangle formed by the lines $(1+p) x-p y+p(1+p)=0$,
$(1+q) x-q y+q(1+q)=0$ and $y=0$, where $p \neq q$, is
(2009)
(a) a hyperbola
(b) a parabola
(c) an ellipse
(d) a straight line
Show Answer
Solution:
- Given, lines are $(1+p) x-p y+p(1+p)=0$
and
$(1+q) x-q y+q(1+q)=0$
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
$$ x=p q $$
$\because$ Slope of line (ii) is $\frac{1+q}{q}$.
$\therefore$ Slope of altitude $B N$ (as shown in figure) is $\frac{-q}{1+q}$.
$\therefore \quad$ Equation of $B N$ is $y-0=\frac{-q}{1+q}(x+p)$
$\Rightarrow \quad y=\frac{-q}{(1+q)}(x+p)$
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
$$ \begin{array}{ll} & x=p q \text { and } y=-p q \\ \Rightarrow & h=p q \text { and } k=-p q \\ \therefore & h+k=0 \end{array} $$
$\therefore$ Locus of $H(h, k)$ is $x+y=0$.
