2. Given points $A(1,3), B(5,4), C(3,7), D(7,1), E(10,2), F(8,6)$, if line $l \parallel DF$, and the shortest distance from line $l$ to any point within (including the boundary) $\triangle ABC$ equals the shortest distance from line $l$ to any point within (including the boundary) $\triangle DEF$, then the equation of line $l$ is $\qquad$ .
Given, $k_{D F}=5 \Rightarrow D F: y=5 x-34$, thus $B G: y=-\frac{1}{5} x+5 \Rightarrow G\left(\frac{15}{2}, \frac{7}{2}\right)$. Then $M\left(\frac{25}{4}, \frac{15}{4}\right) \Rightarrow l: y=5 x-\frac{55}{2}$, so $l: 10 x-2 y-55=0$.