A line AB meets X-axis at A and Y-axis at B. P(4, -1) divides AB in the ratio 1 : 2.

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

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

(i) As A lies on x-axis let its co-ordinates be (a, 0) and B lies on y-axis so, co-ordinates = (0, b).

By section-formula,

$\Rightarrow P = \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] \Rightarrow (4, -1) = \Big(\dfrac{1 \times 0 + 2 \times a}{1 + 2}, \dfrac{1 \times b + 2 \times 0}{1 + 2}\Big) \\[1em] \Rightarrow (4, -1) = \Big(\dfrac{2a}{3}, \dfrac{b}{3}\Big) \\[1em] \Rightarrow \dfrac{2a}{3} = 4 \text{ and } \dfrac{b}{3} = -1 \\[1em] \Rightarrow a = \dfrac{12}{2} = 6 \text{ and } b = -3.

Hence, A = (6, 0) and B = (0, -3).

(ii) By formula,

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

Slope of AB = $\dfrac{-3 - 0}{0 - 6} = \dfrac{-3}{-6} = \dfrac{1}{2}$.

Let slope of perpendicular line be m₁.

Since, slope of product of perpendicular lines = -1.

⇒ m₁ × Slope of AB = -1

⇒ m₁ $\times \dfrac{1}{2} = -1$

⇒ m₁ = -2.

By point-slope from,

Equation of line passing through P and slope = -2 is :

⇒ y - y₁ = m(x - x₁)

⇒ y - (-1) = -2(x - 4)

⇒ y + 1 = -2(x - 4)

⇒ y + 1 = -2x + 8

⇒ 2x + y = 7.

Hence, equation of required line is 2x + y = 7.