Question

A line $L_1$ intersects $x$ and $y$ axes at $P$ and$Q$ respectively. Another line $L_2,$ perpendicular to $L_1$ cuts $x$ and $y$ axes at $R$ and $S$ respectively. The locus of the point of intersection of the lines $PS$ and $QR$ is a circle passing through the

• A. origin
• B. point P
• C. point Q
• D. point R

Answer 1

Answer: (A), (B), (C)

origin
point P
point Q

Let the equation of $L_1$ be $\frac{x}{a}+\frac{y}{b}=1$. so that the coordinates of $P$ and $Q$ are $(a,0)$ and $(0,b).$

respectively, $(Fig.)$. Now, the slope of $L_1$ is$-\frac{b}{a}$ and therefore, that of $L_2$ is $\frac{a}{b}$

Let the equation of $L_2$ be

$\frac{x}{bk}-\frac{y}{ak}=1$

Then the coordinates of $R$ and $S$ are $(bk,0)$ and $(0,-ak)$

respectively. Also, the equation of $PS$ is $\frac{x}{a}-\frac{y}{ak}=1,$ and that of $RQ$ is $\frac{x}{bk}+\frac{y}{b}=1.$ Therefore, eliminating

$k,$ we get the required locus as

$\frac{x-a}{y}=\frac{b-y}{x}$ $\Rightarrow x(x-a)+y(y-b)=0$

$\Rightarrow x^2+y^2-ax-by=0$

which is a circle passing through the origin, points $P$ and $Q.$