Given a line segment AB joining the points A(-4, 6) and B(8, -3). Find :

(i) the ratio in which AB is divided by the y-axis.

(ii) find the co-ordinates of the point of intersection.

(iii) the length of AB.

(i) Let the y-axis divide AB in the ratio m₁ : m₂.

By section-formula, the x-coordinate = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}\Big)

Since, the x-coordinate on y-axis is 0. Putting value in above formula we get :

$\Rightarrow 0 = \dfrac{m$1 \times 8 + m2 \times -4}{m1 + m2} \\[1em] \Rightarrow 8m1 - 4m2 = 0 \\[1em] \Rightarrow 8m1 = 4m2 \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{4}{8} = \dfrac{1}{2} \\[1em] \Rightarrow m1 : m2 = 1 : 2.

Hence, required ratio = 1 : 2.

(ii) The x-coordinate equals to zero on y-axis.

By section formula, the y-coordinate = $\Big(\dfrac{m$1y2 + m2y1}{m1 + m2}\Big)

Substituting value in above formula, we get :

$\Rightarrow y = \dfrac{1 \times -3 + 2 \times 6}{1 + 2} \\[1em] = \dfrac{-3 + 12}{3} \\[1em] = \dfrac{9}{3} \\[1em] = 3.$

Hence, the coordinates of the point of intersection are (0, 3).

(iii) By distance formula,

Distance between two points = $\sqrt{(y$2 - y1)^2 + (x2 - x1)^2}

Substituting values we get :

$AB = \sqrt{(-3 - 6)^2 + (8 - (-4))^2} \\[1em] = \sqrt{(-9)^2 + (12)^2} \\[1em] = \sqrt{81 + 144} \\[1em] = \sqrt{225} \\[1em] = 15 \text{ units}.$

Hence, AB = 15 units.