Show that the line segment joining the points A(-5, 8) and B(10, -4) is trisected by the coordinate axes. Also, find the points of trisection of AB.

Let x-axis divide AB in the ratio k : 1 at the point P(x, 0).

By section-formula,

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

Substituting values we get,

$\Rightarrow 0 = \Big(\dfrac{k(-4) + 1(8)}{k + 1}\Big) \\[1em] \Rightarrow 0 = \Big(\dfrac{-4k + 8}{k + 1}\Big) \\[1em] \Rightarrow 0 = -4k + 8 \\[1em] \Rightarrow 4k = 8 \\[1em] \Rightarrow k = \dfrac{8}{4} = 2.$

The x-axis divides AB in the ratio k : 1 = 2 : 1.

Thus, m₁ : m₂ = 2 : 1.

By section-formula,

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

Substituting values we get,

$\Rightarrow x = \Big(\dfrac{2(10) + 1(-5)}{2 + 1}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{20 - 5}{3}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{15}{3}\Big) \\[1em] \Rightarrow x = 5.$

The coordinates of P = (x, 0) = (5, 0).

Let the y-axis divide AB in the ratio p : 1 at the point Q(0, y).

By section-formula,

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

Substituting values we get,

$\Rightarrow 0 = \Big(\dfrac{p(10) + 1(-5)}{p + 1}\Big) \\[1em] \Rightarrow 0 = \Big(\dfrac{10p - 5}{p + 1}\Big) \\[1em] \Rightarrow 10p - 5 = 0 \\[1em] \Rightarrow 10p = 5 \\[1em] \Rightarrow p = \dfrac{5}{10} \\[1em] \Rightarrow p = \dfrac{1}{2} \\[1em] \Rightarrow p : 1 = \dfrac{1}{2} : 1 = 1 : 2.$

By section-formula,

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

Substitute values we get,

$\Rightarrow y = \Big(\dfrac{1 \times -4 + 2 \times 8}{1 + 2}\Big) \\[1em] = \Big(\dfrac{-4 + 16}{3}\Big) \\[1em] = \dfrac{12}{3} \\[1em] = 4.$

The coordinates of Q = (0, y) = (0, 4).

Hence, points of trisection of AB are Q(5, 0) and P(0, 4).