The line segment joining A(2, 3) and B(6, -5) is intercepted by the x-axis at the point k. Find the ratio in which k divides AB. Also, write the co-ordinates of the point k.

Since the point k lies on the x-axis, its y-coordinate must be 0. Let the coordinates of k be (x, 0).

Let k divide the line segment joining A(2, 3) and B(6, -5) in the ratio m₁:m₂

By section-formula,

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

Substituting values we get :

$\Rightarrow 0 = \Big(\dfrac{m$1(-5) + m2(3)}{m1 + m2}\Big) \\[1em] \Rightarrow -5m1 + 3m2 = 0 \\[1em] \Rightarrow 5m1 = 3m2 \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{3}{5} \\[1em] \Rightarrow m1 : m2 = 3 : 5.

By section-formula,

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

Substituting values we get :

$\Rightarrow x = \Big(\dfrac{3(6) + 5(2)}{3 + 5}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{18 + 10}{8}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{28}{8}\Big) \\[1em] \Rightarrow x = \dfrac{7}{2}$

Hence, ratio in which k divides AB = 3 : 5 and coordinates of the point k are $\Big(\dfrac{7}{2}, 0\Big)$ .