In what ratio does the line x – 5y + 15 = 0 divide the join of A(2, 1) and B(–3, 6)? Also, find the co-ordinates of their point of intersection.

Let P divides line AB in the ratio m : n.

By section formula,

$\Rightarrow P(x, y) = \Big(\dfrac{mx$2 + nx1}{m + n}, \dfrac{my2 + ny1}{m + n}\Big) \\[1em] = \Big(\dfrac{m(-3) + n(2)}{m + n}, \dfrac{m(6) + n(1)}{m + n}\Big) \\[1em] = \Big(\dfrac{-3m + 2n}{m + n}, \dfrac{6m + n}{m + n}\Big).

Since, point P lies on line x - 5y + 15 = 0, substituting values we get :

$\Rightarrow \dfrac{-3m + 2n}{m + n} - 5 \Big(\dfrac{6m + n}{m + n}\Big) + 15 = 0 \\[1em] \Rightarrow \dfrac{-3m + 2n - 5(6m + n) + 15(m + n)}{(m + n)} = 0 \\[1em] \Rightarrow -3m + 2n - 30m - 5n + 15m + 15n = 0 \\[1em] \Rightarrow -18m + 12n = 0 \\[1em] \Rightarrow 6(-3m + 2n) = 0 \\[1em] \Rightarrow -3m + 2n = 0 \\[1em] \Rightarrow 2n = 3m \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{3}.$

By section formula,

$\Rightarrow (x, y) = \Big(\dfrac{mx$2 + nx1}{m + n}, \dfrac{my2 + ny1}{m + n}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{2(-3) + 3(2)}{2 + 3}, \dfrac{2(6) + 3(1)}{2 + 3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{-6 + 6}{5}, \dfrac{12 + 3}{5}\Big) \\[1em] \Rightarrow (x, y) = 0, \dfrac{15}{5} \\[1em] \Rightarrow (x, y) = (0, 3).

Hence, the line x – 5y + 15 = 0 divides AB in the ratio 2 : 3 and the co-ordinates of their point of intersection are (0, 3).