A line through point P(4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.

Let co-ordinates of A be (a, 0) and B be (0, b).

Given,

BP = 2PA

$\dfrac{BP}{PA} = \dfrac{2}{1} \text{ or } \dfrac{PA}{BP} = \dfrac{1}{2}$.

∴ P divides AB in the ratio 1 : 2.

By section formula,

$\Rightarrow P = \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{1 \times 0 + 2 \times a}{1 + 2}, \dfrac{1 \times b + 2 \times 0}{1 + 2}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{0 + 2a}{3}, \dfrac{b + 0}{3}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{2a}{3}, \dfrac{b}{3}\Big) \\[1em] \Rightarrow 4 = \dfrac{2a}{3} \text{ and } 3 = \dfrac{b}{3} \\[1em] \Rightarrow a = \dfrac{4 \times 3}{2} \text{ and } b = 9 \\[1em] \Rightarrow a = 6 \text{ and } b = 9.

A = (a, 0) = (6, 0) and B = (0, b) = (0, 9).

Slope of AB = $\dfrac{9 - 0}{0 - 6} = -\dfrac{9}{6} = -\dfrac{3}{2}$.

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{3}{2}$(x - 6)

⇒ 2y = -3(x - 6)

⇒ 2y = -3x + 18

⇒ 3x + 2y = 18.

Hence, equation of AB is 3x + 2y = 18.