To find the coordinates of the points of trisection, we extend the logic used for finding the midpoint of a segment. The midpoint formula finds the average of the x-coordinates and y-coordinates of the two endpoints. For trisection, we divide the segment into three equal parts, meaning we need to find the average of three coordinates.

Since point P is closer to $A$, it divides $AB$ in the ratio $1{:}2$. We can find $P$ by taking the average of the coordinates of $A$, $A$, and $B$ (counting $A$ twice because it is closer). Similarly, point Q is closer to $B$, dividing $AB$ in the ratio $2{:}1$, so we average the coordinates of $A$, $B$, and $B$ (counting $B$ twice).

Step 1: Calculate coordinates of P

$$P_x = \frac{4 + 4 + 16}{3} = \frac{24}{3} = 8$$ $$P_y = \frac{7 + 7 + (-2)}{3} = \frac{12}{3} = 4$$

So, $P = (8,\,4)$.

Step 2: Calculate coordinates of Q

$$Q_x = \frac{4 + 16 + 16}{3} = \frac{36}{3} = 12$$ $$Q_y = \frac{7 + (-2) + (-2)}{3} = \frac{3}{3} = 1$$

So, $Q = (12,\,1)$.

Thus, the points of trisection of the segment $AB$ are $P\,(8,\,4)$ and $Q\,(12,\,1)$. This method effectively generalizes the midpoint averaging concept to divide the segment into three equal parts.