Question

The locus of the 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 $5x-7y=0$


Let $R$ be the mid-point of $PQ$ whose locus is $(h,k)$
$PQ$ is perpendicular to line $y=x$
$\therefore$ Equation of the line $PQ$ can be written as $y=-x+c$
$y=-x+c$ intersects $y=x$ at $Q(\frac{c}{2},\frac{c}{2})$
$y=-x+c$ intersects $x=2y$ at $P(\frac{2c}{3},\frac{c}{3})$
$\therefore$ Coordinates of midpoint is$R(\frac{7c}{12},\frac{5c}{12})$
Locus of $R:h=\frac{7c}{12},k=\frac{5c}{12}$
$\Rightarrow 5h-7k=0$
$\therefore$ locus of required equation is $5x-7y=0$