Show that A(3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8). Also, find the co-ordinates of the other points of trisection.

Let A divide line-segment joining the points (2, 1) and (5, -8) in m₁ : m₂.

By formula,

$x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] \Rightarrow 3 = \dfrac{m1 \times 5 + m2 \times 2}{m1 + m2} \\[1em] \Rightarrow 3m1 + 3m2 = 5m1 + 2m2 \\[1em] \Rightarrow 5m1 - 3m1 = 3m2 - 2m2 \\[1em] \Rightarrow 2m1 = m2 \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{1}{2} \\[1em] \Rightarrow m1 : m2 = 1 : 2.

Since, A divides line-segment joining the points (2, 1) and (5, -8) in 1 : 2.

Hence, proved A is a point of tri-section.

Let another point of tri-section be B(a, b). So, it will divide the line segment in 2 : 1.

$x = \dfrac{m$1x2 + m2x1}{m1 + m2}

Substituting values we get,

$\Rightarrow a = \dfrac{2 \times 5 + 1 \times 2}{2 + 1} \\[1em] \Rightarrow a = \dfrac{10 + 2}{3} \\[1em] \Rightarrow a = 4.$

$y = \dfrac{m$1y2 + m2y1}{m1 + m2}

Substituting values we get,

$\Rightarrow b = \dfrac{2 \times -8 + 1 \times 1}{2 + 1} \\[1em] \Rightarrow b = \dfrac{-16 + 1}{3} \\[1em] \Rightarrow b = -\dfrac{15}{3} = -5.$

B = (a, b) = (4, -5).

Hence, co-ordinate of other point of trisection = (4, -5).