A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:

(i) the co-ordinates of A and B.

(ii) the equation of line through P and perpendicular to AB.

(i) Let’s assume the co-ordinates of point A, lying on x-axis be (x, 0) and the co-ordinates of point B (lying on y-axis) be (0, y).

Given,

P = (-4, -2) and AP : PB = 1 : 2

By section formula, we get

$\Rightarrow P = \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{1 \times 0 + 2 \times x}{1 + 2}, \dfrac{1 \times y + 2 \times 0}{1 + 2}\Big) \\[1em] \Rightarrow (-4, -2) = \Big(\dfrac{2x}{3}, \dfrac{y}{3}\Big) \\[1em] \Rightarrow -4 = \dfrac{2x}{3} \text{ and } -2 = \dfrac{y}{3} \\[1em] \Rightarrow x = \dfrac{-4 \times 3}{2} \text{ and } y = -6 \\[1em] \Rightarrow x = -6 \text{ and } y = -6.

Hence, the co-ordinates of A and B are (-6, 0) and (0, -6).

(ii) By formula,

Slope = $\dfrac{y$2 - y1}{x2 - x1}

Slope of AB = $\dfrac{-6 - 0}{0 - (-6)} = \dfrac{-6}{6}$ = -1.

Let slope of perpendicular line be m.

⇒ m × -1 = -1

⇒ -m = -1

⇒ m = 1.

Therefore, the required equation of the line passing through P and perpendicular to AB is given by

⇒ y – y₁ = m(x – x₁)

⇒ y - (-2) = 1[x - (-4)]

⇒ y + 2 = x + 4

⇒ y = x + 2.

Hence, the equation of line through P and perpendicular to AB is y = x + 2.