If $\left(-2,6\right)$  is the image of the point $\left(4,2\right)$  with respect to the line $L=0$  then $L=0$  is

Midpoint of $\overline{AB}=\left(\frac{-2+4}{2}\frac{6+2}{2}\right)=\left(1,4\right)$

Slope of $AB=\frac{2-6}{4+2}=\frac{-4}{6}=\frac{-2}{3}$

Slope of $L=0$  is $\frac{3}{2}$

Equation of $L=0$  is

$\left(y-4\right)=\frac{3}{2}\left(x-1\right)$

$2y-8=3x-3$

$3x-2y+5=0$  

$\therefore$  required equation is $3x-2y+5=0$