The locus of the mid-points of the perpendiculars drawn from points on the line $x+y+5=0$  to the line $2x-y+3=0$  is

 $\frac{y-(2\alpha +3)}{x-\alpha }\times 2=-1$
$$\begin{gathered} 2(y-2\alpha -3)=\alpha -x \\ 2y-4\alpha -6=\alpha -x \\ 2y-6+x=5\alpha \mathrm{.........}\left(1\right) \end{gathered}$$ 

$$\begin{gathered} \frac{\beta +\alpha }{2}=x,\frac{-\beta -5+2\alpha +3}{2}=y \\ \beta +\alpha =2x,2y-2\alpha +2+\alpha =2x \\ 2y-2x+2=\alpha \\ -\beta +2\alpha -2=2y,\beta =2y-2\alpha +2 \end{gathered}$$ 
So, equation  $2y-6+x=5(2y-2x+2);2y-6+x=10y-10x+10;11x-8y-16=0$