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A line cuts the $X$ -axis at $A(5,0)$ and the $Y$ -axis at $B(0,-3)$. A variable line $P Q$ is drawn perpendicular to $A B$ cutting the $X$ -axis at $P$ and the $Y$ -axis at $Q$. If $A Q$ and $B P$ meet at $R$, then the locus of $R$ is
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Verified Answer
The correct answer is:
$x^{2}+y^{2}-5 x+3 y=0$
Equation of line $A B$ is
$$
\begin{array}{l}
\frac{x}{5}+\frac{y}{-3}=1 \\
3 x-5 y=15
\end{array}
$$
Perpendicular line to $A B$ is
$$
5 x+3 y=\lambda
$$
Coordinate of $P$ is $\left(\frac{\lambda}{5}, 0\right)$ and coordinate of $Q$ is $(0, \lambda / 3)$
Now, equation of line $A Q$ is
$x / 5+\frac{y}{\lambda / 3}=1$
$\Rightarrow \quad \frac{x}{5}+\frac{3 y}{\lambda}=1$
$\Rightarrow \quad \frac{3 y}{\lambda}=1-\frac{x}{5}$
$\Rightarrow \quad \frac{1}{\lambda}=\frac{1}{3 y}\left(1-\frac{x}{5}\right)$
and equation of line $B P$ is
$\frac{x}{\lambda / 5}+\frac{y}{-3}=1$
$\Rightarrow \quad \frac{5 x}{\lambda}-\frac{y}{3}=1$
$\Rightarrow \quad \frac{1}{\lambda}=\frac{1}{5 x}\left(\frac{y}{3}+1\right)$
From Eqs. (i) and (ii),
$\frac{1}{3 y}\left(1-\frac{x}{5}\right)=\frac{1}{5 x}\left(\frac{y}{3}+1\right)$
$\Rightarrow \quad 5 x\left(1-\frac{x}{5}\right)=3 y\left(\frac{y}{3}+1\right)$
$\Rightarrow \quad 5 x-x^{2}=y^{2}+3 y$
$\Rightarrow x^{2}+y^{2}-5 x+3 y=0$
which is a circle.
$$
\begin{array}{l}
\frac{x}{5}+\frac{y}{-3}=1 \\
3 x-5 y=15
\end{array}
$$
Perpendicular line to $A B$ is
$$
5 x+3 y=\lambda
$$
Coordinate of $P$ is $\left(\frac{\lambda}{5}, 0\right)$ and coordinate of $Q$ is $(0, \lambda / 3)$
Now, equation of line $A Q$ is
$x / 5+\frac{y}{\lambda / 3}=1$
$\Rightarrow \quad \frac{x}{5}+\frac{3 y}{\lambda}=1$
$\Rightarrow \quad \frac{3 y}{\lambda}=1-\frac{x}{5}$
$\Rightarrow \quad \frac{1}{\lambda}=\frac{1}{3 y}\left(1-\frac{x}{5}\right)$
and equation of line $B P$ is
$\frac{x}{\lambda / 5}+\frac{y}{-3}=1$
$\Rightarrow \quad \frac{5 x}{\lambda}-\frac{y}{3}=1$
$\Rightarrow \quad \frac{1}{\lambda}=\frac{1}{5 x}\left(\frac{y}{3}+1\right)$
From Eqs. (i) and (ii),
$\frac{1}{3 y}\left(1-\frac{x}{5}\right)=\frac{1}{5 x}\left(\frac{y}{3}+1\right)$
$\Rightarrow \quad 5 x\left(1-\frac{x}{5}\right)=3 y\left(\frac{y}{3}+1\right)$
$\Rightarrow \quad 5 x-x^{2}=y^{2}+3 y$
$\Rightarrow x^{2}+y^{2}-5 x+3 y=0$
which is a circle.
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