Question

A line cuts the x-axis at $A\equiv (7,0)$ and the y-axis at $B\equiv (0,-5)$. A variable line $PQ$ is drawn perpendicular to $AB$ cutting the x-axis in $P$ and the y-axis in $Q$. 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 B $x^2+y^2-7x+5y=0$

Equation of line $AB$ is
$\frac{x}{7}+\frac{y}{-5}=1\Rightarrow 5x-7y=35$ ...(1)
Equation of line perpendicular to $AB$ is
$5x+7y=\lambda$ ...(2)
It meets x-axis at $P(\frac{\lambda }{7},0)$ and y-axis at $Q(0,\frac{\lambda }{5})$
The equation of lines $AQ$ and $BP$ are $\frac{x}{7}+\frac{5y}{\lambda }=1$ and $\frac{7x}{\lambda }-\frac{y}{5}=1$ respectively
Let $R(h,k)$ be their point of intersection of lines $AQ$ and $BP$
Then $\frac{h}{7}+\frac{5k}{\lambda }=1$ and $\frac{7h}{\lambda }-\frac{k}{5}=1$
$\Rightarrow \frac{1}{5k}(1-\frac{h}{7})=\frac{1}{7h}(1+\frac{k}{5})\Rightarrow h(7-h)=k(5+k)\Rightarrow h^2+k^2-7h+5k=0$
Hence, the locus of a point is $x^2+y^2-7x+5y=0$