Let $L$ be the line $x+3y=2$. Find the equation of the line $M$ such that $F_M F_L = T_{(1,3)}$.
I'm not very sure how to find this line. Here's the approach I've tried:
(1) Find $F_{x+3y-2=0} (u,v) = (u,v)-\frac15 (u+3v-2)(1,3)=T_{(1,3)}(u,v)=(1+u, 3+v)$.
(2) Express $F_{ax+by+c=0}(F_L(u,v))=\text{ (quite a messy expression) }=(1+u, 3+v)$.
The idea is to solve for $a,b,c$ in $M$, but the expression in $(2)$ is so tedious that I don't know how to solve this problem. Would appreciate your advice.