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

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

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

(iii) equation of perpendicular bisector 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 :

$\Rightarrow y = \dfrac{1 \times 3 + 2 \times 6}{1 + 2} \\[1em] = \dfrac{3 + 12}{3} \\[1em] = \dfrac{15}{3} \\[1em] = 5.$

Hence, the coordinates of the point of intersection are (0, 5).

(iii) By formula,

Mid-point = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Let M be the mid-point of AB.

M = $\Big(\dfrac{(-4) + 8}{2}, \dfrac{6 + 3}{2}\Big) = \Big(\dfrac{4}{2}, \dfrac{9}{2}\Big) = \Big(2, \dfrac{9}{2}\Big)$.

Slope of AB = $\dfrac{3 - 6}{8 - (-4)} = \dfrac{-3}{12} = -\dfrac{1}{4}$

We know that,

Product of slope of perpendicular lines = -1.

Let slope of perpendicular bisector be h.

∴ h × $-\dfrac{1}{4}$ = -1

⇒ h = 4.

By point-slope form,

Equation : y - y₁ = m(x - x₁)

⇒ y - $\dfrac{9}{2}$ = 4(x - 2)

⇒ $\dfrac{2y - 9}{2}$ = 4(x - 2)

⇒ 2y - 9 = 8(x - 2)

⇒ 2y - 9 = 8x - 16

⇒ 2y - 8x - 9 + 16 = 0

⇒ 2y - 8x + 7 = 0

⇒ 8x - 2y = 7.

Hence, equation of perpendicular bisector of AB is 8x - 2y = 7.