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 $PQ$ is always perpendicular to the line $AB$. Then locus of the point of intersection of $BP$ and $AQ$ is

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

Answer 1

Answer: (D)

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

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