Problem 71

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 $\fbox{E}$