Question

A line cuts the x-axis at $A(5,0)$ and the y-axis at $B(0,–3)$. A variable line $PQ$ is drawn perpendicular to $AB$ cutting the x-axis at $P$ and the y-axis at $Q$. If $AQ$ and $BP$ meet at $R$, then the locus of $R$ is

• A. $x^2+y^2–5x+3y=0$
• B. $x^2+y^2+5x+3y=0$
• C. $x^2+y^2+5x-3y=0$
• D. $x^2+y^2–5x–3y=0$

Answer 1

The correct option is A $x^2+y^2–5x+3y=0$

Equation of line $AB$ is, $\frac{x}{5}+\frac{y}{-3}=1\Rightarrow 3x-5y=15$

Any line perpendicular to $AB$ is,
$5x+3y=c$
So, coordinates of $P$ and $Q$ are $(\frac{c}{5},0),(0,\frac{c}{3})$ respectively.

Equation of $AQ$ is,
$\frac{x}{5}+\frac{y}{c/3}=1\Rightarrow \frac{3y}{c}=1-\frac{x}{5}\Rightarrow \frac{1}{c}=\frac{1}{3y}(1-\frac{x}{5})...\left(1\right)$

Equation of $BP$ is,
$\frac{x}{c/5}-\frac{y}{3}=1\Rightarrow \frac{1}{c}=\frac{1}{5x}(1+\frac{y}{3})...\left(2\right)$

From eqn(1) and (2), we have
$\frac{1}{3y}(1-\frac{x}{5})=\frac{1}{5x}(1+\frac{y}{3})$
$\Rightarrow 5x-x^2=3y+y^2\Rightarrow x^2+y^2-5x+3y=0$