Find the co-ordinates of the points of trisection of the line segment joining the points A(5, -3) and B(2, -9).

Let point P is the first point of trisection, meaning it divides the segment AB internally in the ratio m₁ : m₂ = 1 : 2

By section-formula,

(x, y) = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get :

$\Rightarrow P(x, y) = \Big(\dfrac{1 \times 2 + 2 \times 5}{1 + 2}, \dfrac{1 \times -9 + 2 \times -3}{1 + 2}\Big) \\[1em] \Rightarrow P(x, y) = \Big(\dfrac{2 + 10}{3}, \dfrac{-9 - 6}{3}\Big) \\[1em] \Rightarrow P(x, y) = \Big(\dfrac{12}{3}, \dfrac{-15}{3}\Big) \\[1em] \Rightarrow P(x, y) = (4, -5).$

Let point Q is the second point of trisection, meaning it divides the segment AB internally in the ratio m₁ : m₂ = 2 : 1

Substituting values we get :

$\Rightarrow Q(a, b) = \Big(\dfrac{2 \times 2 + 1 \times 5}{2 + 1}, \dfrac{2 \times -9 + 1 \times -3}{2 + 1}\Big) \\[1em] \Rightarrow Q(a, b) = \Big(\dfrac{4 + 5}{3}, \dfrac{-18 - 3}{3}\Big) \\[1em] \Rightarrow Q(a, b) = \Big(\dfrac{9}{3}, \dfrac{-21}{3}\Big) \\[1em] \Rightarrow Q(a, b) = (3, -7).$

Hence, the coordinates of trisection are P(4, -5) and Q(3, -7).