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:

  1. 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 $x+y=0$.



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