Question

Let coordinates of the points $A$ and $B$ are $(5,0)$ and $(0,7)$ respectively. $P$ and $Q$ are the variable points lying on the $x$ -and $y$ -axis respectively so that $P Q$ is always perpendicular to the line $A B$. Then locus of the point of intersection of $B P$ and $A Q$ is

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

Answer 1

Answer: (D)

Let the equation of $P Q$ be $\frac{x}{a}+\frac{y}{b}=1 \dots$(i)
Then equation of $B P$ is $\frac{x}{a}+\frac{y}{7}=1 \dots$(ii)
and equation of $A Q$ is $\frac{x}{5}+\frac{y}{b}=1 \dots$(iii)
Since $A B \perp P Q$,
therefore, $-\frac{b}{a} \times-\frac{7}{5}=-1$
$\Rightarrow 5 a+7 b=0 \dots$(iv)

Now, slope of $B P$ is $-\frac{7}{a}$ and
slope of $A Q$ is $-\frac{b}{5}$
So, product of these slopes is
$\frac{7 b}{5 a}=-1[$ from (iv) $]$
$\therefore B P$ and $A Q$ intersect each other at right angle. So, the locus of the point of intersection of $B P$ and $A Q$ is a circle with $A B$ as diameter.
Thus equation of locus is
$x(x-5)+y(y-7)=0 $
$\Rightarrow x^{2}+y^{2}-5 x-7 y=0$