In what ratio is the segment joining the points A(6, 5) and B(-3, 2) divided by the y-axis? Find the point at which the y-axis cuts AB.

When a point lies on the y-axis, its x-coordinate is always 0. Let the point where the y-axis cuts AB be P(0, y).

Let ratio in which P divides AB be m₁ : m₂.

By section-formula,

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

Substituting values we get :

$\Rightarrow 0 = \Big(\dfrac{m$1(-3) + m2(6)}{m1 + m2}\Big) \\[1em] \Rightarrow 0 = -3m1 + 6m2 \\[1em] \Rightarrow 3m1 = 6m2 \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{6}{3} = 2.

Thus, m₁ : m₂ = 2 : 1.

By section-formula,

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

Substitute values we get:

$\Rightarrow y = \Big(\dfrac{2(2) + 1(5)}{2 + 1}\Big) \\[1em] = \Big(\dfrac{4 + 5}{3}\Big) \\[1em] = \Big(\dfrac{9}{3}\Big) \\[1em] = 3.$

P = (0, y) = (0, 3).

Hence, AB is divided in ratio 2 : 1 and point at which the y-axis cuts AB is (0, 3).