Find the coordinates of the points of trisection (i.e.,points dividing in three equal parts) of the line segment joining the points $A (2, -2)$ and $B (-7, 4)$.

(N/A) Let $P$ and $Q$ be the points of trisection of $AB$ such that $AP = PQ = QB$.
Point $P$ divides $AB$ internally in the ratio $1:2$. Using the section formula,the coordinates of $P$ are:
$P = \left(\frac{1(-7) + 2(2)}{1 + 2}, \frac{1(4) + 2(-2)}{1 + 2}\right) = \left(\frac{-7 + 4}{3}, \frac{4 - 4}{3}\right) = \left(\frac{-3}{3}, \frac{0}{3}\right) = (-1, 0)$.
Point $Q$ divides $AB$ internally in the ratio $2:1$. Using the section formula,the coordinates of $Q$ are:
$Q = \left(\frac{2(-7) + 1(2)}{2 + 1}, \frac{2(4) + 1(-2)}{2 + 1}\right) = \left(\frac{-14 + 2}{3}, \frac{8 - 2}{3}\right) = \left(\frac{-12}{3}, \frac{6}{3}\right) = (-4, 2)$.
Therefore,the coordinates of the points of trisection are $(-1, 0)$ and $(-4, 2)$.