Let P and Q be the points of trisection of AB, so AP = PQ = QB.
Point P divides AB in the ratio 1 : 2 internally.
$$P = \left(\frac{1(-3) + 2(-1)}{1+2},\ \frac{1(-2) + 2(4)}{1+2}\right) = \left(\frac{-3-2}{3},\ \frac{-2+8}{3}\right) = \left(\frac{-5}{3},\ 2\right)$$
Point Q divides AB in the ratio 2 : 1 internally.
$$Q = \left(\frac{2(-3) + 1(-1)}{2+1},\ \frac{2(-2) + 1(4)}{2+1}\right) = \left(\frac{-6-1}{3},\ \frac{-4+4}{3}\right) = \left(\frac{-7}{3},\ 0\right)$$
The coordinates of the points of trisection are $\left(\dfrac{-5}{3},\ 2\right)$ and $\left(\dfrac{-7}{3},\ 0\right)$.
Source: Chapter 7, Section 7.3 (Section Formula)
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