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

Suppose that the coordinatres of the point on the line $x=2y$ is $\left(2\alpha ,\alpha \right)$ and suppose that  the foot of the perpendciular from $P\left(2\alpha ,\alpha \right)$ on the line $x=y$ is  $Q\left(\beta ,\beta \right)$. Let $R\left(h,k\right) be the midpoint of PQ.$

 Slope of PQ $=\frac{k-\alpha }{h-2\alpha }=-1$, since $PQ$ is perpendicualr to the line $y=x$

$\Rightarrow k-\alpha =-h+2\alpha$

$\Rightarrow \alpha =\frac{h+k}{3}$…………. (1)

Also $2h=2\alpha +\beta$ and $2k=\alpha +\beta$

$\Rightarrow 2h=\alpha +2k$

$\Rightarrow \alpha =2h-2k$…………. (2)

From (1) and (2)

$\frac{h+k}{3}=2\left(h-k\right)$

$\therefore$ locus is $6x-6y=x+y\Rightarrow 5x=7y$