Find the coordinates of the points of trisection of the line segment joining the points (3, -3) and (6, 9).

Let P(x₁, y₁) and Q(x₂, y₂) be the points of trisection of the points A(3, -3) and B(6, 9).

AP = PQ = QB ⇒ 2AP = PB

⇒ $\dfrac{AP}{PB} = \dfrac{1}{2}$ ⇒ P divides AB in the ratio 1 : 2, so coordinates of P are

$\Rightarrow \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] = \Big(\dfrac{1 \times 6 + 2 \times 3}{1 + 2}, \dfrac{1 \times 9 + 2 \times (-3)}{1 + 2}\Big) \\[1em] = \Big(\dfrac{6 + 6}{3}, \dfrac{9 - 6}{3}\Big) \\[1em] = \Big(\dfrac{12}{3}, \dfrac{3}{3}\Big) \\[1em] = (4, 1).

Q divides AB in the ratio 2 : 1, so coordinates of Q are

$\Rightarrow \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] = \Big(\dfrac{2 \times 6 + 1 \times 3}{2 + 1}, \dfrac{2 \times 9 + 1 \times (-3)}{2 + 1}\Big) \\[1em] = \Big(\dfrac{12 + 3}{3}, \dfrac{18 - 3}{3}\Big) \\[1em] = \Big(\dfrac{15}{3}, \dfrac{15}{3}\Big) \\[1em] = (5, 5).

Hence, (4, 1) and (5, 5) are the coordinates of the points of trisection of the line segment joining the points (3, -3) and (6, 9).