Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.

Let A = (-5, 8) and B = (10, -4)

Let P and Q be points which trisects AB.

Let P (a, b) divide AB in 1 : 2 and Q (c, d) in 2 : 1.

By section formula (for P),

$x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] \Rightarrow a = \dfrac{1 \times 10 + 2 \times -5}{1 + 2} \\[1em] \Rightarrow a = \dfrac{10 - 10}{3} \\[1em] \Rightarrow a = \dfrac{0}{3} = 0. \\[1em] y = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] \Rightarrow b = \dfrac{1 \times -4 + 2 \times 8}{1 + 2} \\[1em] \Rightarrow b = \dfrac{-4 + 16}{3} \\[1em] \Rightarrow b = \dfrac{12}{3} = 4.

P = (a, b) = (0, 4).

By section formula (for Q),

$x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] \Rightarrow c = \dfrac{2 \times 10 + 1 \times -5}{2 + 1} \\[1em] \Rightarrow c = \dfrac{20 - 5}{3} \\[1em] \Rightarrow c = \dfrac{15}{3} = 5. \\[1em] y = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] \Rightarrow d = \dfrac{2 \times -4 + 1 \times 8}{2 + 1} \\[1em] \Rightarrow d = \dfrac{-8 + 8}{3} \\[1em] \Rightarrow d = \dfrac{0}{3} = 0.

Q = (c, d) = (5, 0).

Since, x co-ordinate of P = 0, it means P lies on y-axis and y co-ordinate of Q = 0, it means Q lies on x-axis.

Hence, proved that line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.