The locus of midpoints of the perpendicular drawn from the points on the line  $\mathrm{x}=3\mathrm{y}$ to the line x=y is

The general point on the line $x=3y$  is   $\left(3\mathrm{h},\mathrm{h}\right)$
The foot of the perpendicular of $\left(3\mathrm{h},\mathrm{h}\right)$ on the line  $\mathrm{x}=\mathrm{y}$ is  $\left(\mathrm{l},\mathrm{m}\right)$
 $\frac{\mathrm{l}-3\mathrm{h}}{1}=\frac{\mathrm{m}-\mathrm{h}}{-1}=\frac{-\left(3\mathrm{h}-\mathrm{h}\right)}{1^2+1^2}$ 
Simplify 
$\frac{\mathrm{l}-3\mathrm{h}}{1}=-\mathrm{h}\Rightarrow \mathrm{l}=2\mathrm{h}$  
And 
 $\frac{\mathrm{m}-\mathrm{h}}{-1}=-\mathrm{h}\Rightarrow \mathrm{m}=2\mathrm{h}$ 
Hence the foot of the perpendicular of $\left(3\mathrm{h},\mathrm{h}\right)$ on the line $\mathrm{x}-\mathrm{y}=0$ is  $\left(2\mathrm{h},2\mathrm{h}\right)$
Suppose that $\mathrm{P}\left({\mathrm{x}}_{1,}{\mathrm{y}}_1\right)$ be any point on the locus.

The point  $\mathrm{P}\left({\mathrm{x}}_{1,}{\mathrm{y}}_1\right)$ is midpoint of $\left(3\mathrm{h},\mathrm{h}\right)$  and  $\left(2\mathrm{h},2\mathrm{h}\right)$

It implies that  $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(\frac{5\mathrm{h}}{2},\frac{3\mathrm{h}}{2}\right)$
Eliminate h  to get the locus of the point  $\mathrm{P}\left({\mathrm{x}}_{1,}{\mathrm{y}}_1\right)$ 
Therefore, the locus is $\frac{2\mathrm{x}}{5}=\frac{2\mathrm{y}}{3}\Rightarrow \boxed{3\mathrm{x}-5\mathrm{y}=0}$