In what ratio does the line x - y - 2 = 0 divide the line segment joining the points (3, -1) and (8, 9)? Also, find the coordinates of the point of division.

Let the point A be (3, -1) and point B be (8, 9) and let the line x - y - 2 = 0 divide the line segment AB in the ratio m₁ : m₂ at point P(x, y) then,

$x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{m1 \times 8 + m2 \times 3}{m1 + m2} \\[1em] = \dfrac{8m1 + 3m2}{m1 + m2} \qquad \text{….[Eq 1]} \\[1em] \text{and y } = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{m1 \times 9 + m2 \times -1}{m1 + m2} \\[1em] = \dfrac{9m1 - m2}{m1 + m2} \qquad \text{….[Eq 2]}.

Given, the point P(x, y) lies on the line x - y - 2 = 0.

$\therefore \dfrac{8m$1 + 3m2}{m1 + m2} - \dfrac{9m1 - m2}{m1 + m2} - 2 = 0 \\[1em] \Rightarrow \dfrac{8m1 + 3m2 - 9m1 + m2 - 2m1 - 2m2}{m1 + m2} = 0 \\[1em] \Rightarrow -3m1 + 2m2 = 0 \\[1em] \Rightarrow 3m1 = 2m2 \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{2}{3} \\[1em].

Putting value of m₁ : m₂ in Eq 1,

$\Rightarrow x = \dfrac{8 \times 2 + 3 \times 3}{2 + 3} \\[1em] = \dfrac{16 + 9}{5} \\[1em] = \dfrac{25}{5} \\[1em] = 5.$

Putting value of m₁ : m₂ in Eq 2,

$\Rightarrow y = \dfrac{9 \times 2 - 3}{2 + 3} \\[1em] = \dfrac{18 - 3}{5} \\[1em] = \dfrac{15}{5} \\[1em] = 3.$

Hence, the ratio is 2 : 3 and coordinates of P are (5, 3).