Find the co-ordinates of the points of trisection of the segment joining A(2, –2) and B(–7, 4).

Answer: P(–1, 0) and Q(–4, 2)

Let P and Q be the points of trisection, so AP = PQ = QB. Then P divides AB in the ratio 1 : 2 and Q divides AB in the ratio 2 : 1.

For P (ratio 1 : 2): \(x = \dfrac{1(-7) + 2(2)}{3} = \dfrac{-3}{3} = -1\), \(y = \dfrac{1(4) + 2(-2)}{3} = \dfrac{0}{3} = 0\). So P(–1, 0).

For Q (ratio 2 : 1): \(x = \dfrac{2(-7) + 1(2)}{3} = \dfrac{-12}{3} = -4\), \(y = \dfrac{2(4) + 1(-2)}{3} = \dfrac{6}{3} = 2\). So Q(–4, 2).

(Check: Q is the midpoint of PB: \(\left(\dfrac{-1 - 7}{2}, \dfrac{0 + 4}{2}\right) = (-4, 2)\).)