smc

The trisection points of $(-4,5)$ and $(5,-1)$ can be found by trisecting the x-coordinates and the y-coordinates separately. The difference of the x-coordinates is $9$, so the trisection points happen at $-4+\frac{9}{3}$ and $-4+\frac{9}{3}+\frac{9}{3}$, which are $-1$ and $2$. Similarly, the y-coordinates have a difference of $6$, so the trisections happen at $3$ and $1$. So, the two points are $(-1,3)$ and $(2,1)$. We now check which line has both $(3,4)$ and one of the two trisection points on it. Plugging in $(x,y)=(3,4)$ into all five of the equations works. The point $(2,1)$ doesn't work in any of the five lines. However, $(-1,3)$ works in line $E$. Thus, the answer is $\boxed{x-4y+13=0}$