Find the points of trisection of the line segment joining $(-2,-1)$ and $(7,8)$.

(N/A) Let $A(-2,-1)$ and $B(7,8)$ be the given points and let $P$ and $Q$ be the points of trisection of $\overline{AB}$.
Here,$x_1 = -2, y_1 = -1, x_2 = 7$ and $y_2 = 8$.
For point $P$ on $\overline{AB}$,the ratio $m:n = 1:2$.
Using the section formula,the coordinates of $P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.
$P = \left(\frac{1(7) + 2(-2)}{1+2}, \frac{1(8) + 2(-1)}{1+2}\right)$.
$P = \left(\frac{7-4}{3}, \frac{8-2}{3}\right) = \left(\frac{3}{3}, \frac{6}{3}\right) = (1, 2)$.
Now,$Q$ is the midpoint of $\overline{PB}$,or it divides $\overline{AB}$ in the ratio $2:1$.
Using the section formula for $Q$ with ratio $2:1$:
$Q = \left(\frac{2(7) + 1(-2)}{2+1}, \frac{2(8) + 1(-1)}{2+1}\right)$.
$Q = \left(\frac{14-2}{3}, \frac{16-1}{3}\right) = \left(\frac{12}{3}, \frac{15}{3}\right) = (4, 5)$.
Thus,the points of trisection of the line segment joining $(-2,-1)$ and $(7,8)$ are $(1, 2)$ and $(4, 5)$.