If the family of lines $ax+3y-6=0 $ (a is a variable) intersect the lines $x-2y+3=0$ and & $x-y+1=0$ at P and Q respectively, then the locus of the mid point of P and Q is?r
My attempt : Solving $x-2y+3=0$ and $ ax+3y-6=0$, I got $$ P= (\frac {3}{3+2a},\frac{6+3a}{3+2a}) $$. Solving $x-y+1=0$ and $ax+3y-6=0 $, I got $$ Q= (\frac {3}{3+a},\frac{6+a}{3+a}) $$. How do I eliminate a from these equations?
Hint: Find Midpoint of P and Q = $(M_x,M_y) = (x,y)$,
Now $M_x = x$, solve a in terms of x
and put a in terms of x in $M_y = y$