Question

The locus of mid points of the perpendiculars drawn from points on the line $x=2y$ to the line $x=y$ is

• A. $2x-3y=0$
• B. $3x-2y=0$
• C. $5x-7y=0$
• D. $7x-5y=0$

Answer 1

The correct option is C

$5x-7y=0$

Explanation for the correct answer

Locus of midpoint :

Let $PQ$ be the perpendicular drawn from the line $x=2y$ to the line $x=y$

$R$ is the midnight of $PQ$.

The equation of the line $PQ$ can be written as $y=-x+c$

$y=-x+c$ intersects $y=x$ at $Q:\left(\frac{c}{2},\frac{c}{2}\right)$ [$R$ is the midnight of $PQ$]

$y=-x+c$ intersects $x=2y$ at $P:\left(\frac{2c}{3},\frac{2c}{3}\right)$

Midpoint formula for points $\left(x_1,y_1\right),\left(x,y\right),\left(x_2,y_2\right)$ is $\mathbf{x}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}}{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\frac{{\mathbf{y}}_{\mathbf{1}}\mathbf{+}{\mathbf{y}}_{\mathbf{2}}}{\mathbf{2}}$

Using midpoint formula for $P:\left(\frac{2c}{3},\frac{c}{3}\right)$ , $Q:\left(\frac{c}{2},\frac{c}{2}\right)$ and $R$ is

$$\begin{gathered} x=\frac{{\displaystyle \frac{2c}{3}}+{\displaystyle \frac{c}{2}}}{2},y=\frac{{\displaystyle \frac{c}{3}}+{\displaystyle \frac{c}{2}}}{2} \\ x=\frac{{\displaystyle \frac{4c+3c}{6}}}{2},y=\frac{{\displaystyle \frac{2c+3c}{6}}}{2} \\ x=\frac{7c}{12},y=\frac{5c}{12} \end{gathered}$$

By making an equation with points

So, the locus of $R$ is $5x-7y=0$

Hence, option C is correct.