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) The coordinates 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} \\[1em] \Rightarrow m1 : m2 = 1 : 2

Hence, required ratio = $\dfrac{1}{2} : 1$ or 1 : 2.

(ii) By section formula, the y-coordinate is given by

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

Putting 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) Distance formula = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

By distance formula,

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

Hence, the length of AB = 15 units.