Question

A line cuts $x$-axis at $A(7,0)$ and the $y$-axis at $B(0,-5)$. A variable line $PQ$ is drawn perpendicular to $AB$ cutting the $x$-axis at $P(a,0)$ and the $y$-axis at $Q(0,b)$. If $AQ$ and $BP$ intersect at $R$, the locus of $R$ is

• A. $x^2+y^2+7x+5y=0$
• B. $x^2+y^2+7x-5y=0$
• C. $x^2+y^2-7x+5y=0$
• D. $x^2+y^2-7x-5y=0$

Answer 1

The correct option is C $x^2+y^2-7x+5y=0$

Slope of the line $AB$ is $\frac{5}{7}$
Slope of the line $PQ$
$\frac{-b}{a}=\frac{-7}{5}\frac{b}{a}=\frac{7}{5}$
Equation of the line
$BP:\frac{x}{a}-\frac{y}{5}=1\frac{1}{a}=\frac{5+y}{5x}\cdots \left(1\right)$
Equation of the line
$AQ:\frac{x}{7}+\frac{y}{b}=1\frac{1}{b}=\frac{7-x}{7y}\cdots \left(2\right)$
solving above equations we get
$x^2+y^2-7x+5y=0$