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 $x^2+y^2-5x+3y=0$

Line $AB$ is $\frac{x}{5}+\frac{y}{-3}=1\Rightarrow 3x-5y=15$
any perpendicular line to $AB$
$5x+3y=\lambda$ So $P(\frac{\lambda }{5},0),Q(0,\frac{\lambda }{3})$
$AQ$ is $\frac{x}{5}+\frac{y}{\lambda /3}=1\Rightarrow \frac{3y}{\lambda }=1-\frac{x}{5}\Rightarrow \frac{1}{\lambda }=\frac{1}{3y}(1-\frac{x}{5})...\left(1\right)$
and $BP$ is $\frac{x}{\lambda /5}-\frac{y}{3}=1\Rightarrow \frac{5x}{\lambda }=1+\frac{1}{\lambda }=\frac{1}{5}(1+\frac{y}{3})....\left(2\right)$
$\frac{1}{3y}(1-\frac{x}{5})=\frac{1}{5x}(1+\frac{y}{3})\Rightarrow 5(1-\frac{x}{5})=3y(1+\frac{y}{3})\Rightarrow 5x-x^2=3y+y^2\Rightarrow x^2+y^2-5x+3y=0$