The line 3x - 4y - 12 = 0 meets x-axis at point A and y-axis at point B.

(i) Find the co-ordinates of the point P on line segment AB dividing AB in the ratio 2 : 1.

(ii) Find the equation of the line that passes through the point P and is perpendicular to AB.

(i) Given,

The line 3x - 4y - 12 = 0 meets x-axis at point A and y-axis at point B.

Substituting x = 0, we get :

⇒ 3.0 - 4y - 12 = 0

⇒ 4y = -12

⇒ y = -3.

B = (0, -3)

Substituting y = 0, we get :

⇒ 3x - 4.0 - 12 = 0

⇒ 3x = 12

⇒ x = $\dfrac{12}{3}$ = 4.

A = (4, 0)

By section formula,

Point of division = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get :

$\Rightarrow P = \Big(\dfrac{2 \times 0 + 1 \times 4}{2 + 1}, \dfrac{2 \times -3 + 1 \times 0}{2 + 1}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{4}{3}, \dfrac{-6}{3}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{4}{3}, -2\Big).$

Hence, coordinates of P = $\Big(\dfrac{4}{3}, -2\Big)$.

(ii) Slope of AB = $\dfrac{-3 - 0}{0 - 4} = \dfrac{-3}{-4} = \dfrac{3}{4}$.

We know that,

Product of slope of perpendicular lines = -1. Let slope of line perpendicular to AB be m.

$\Rightarrow m \times \dfrac{3}{4} = -1 \\[1em] \Rightarrow m = -\dfrac{4}{3}.$

By point slope form :

Equation of line :

y - y₁ = m(x - x₁)

Equation of line passing through P and perpendicular to AB is :

$\Rightarrow y - (-2) = -\dfrac{4}{3}\Big(x - \dfrac{4}{3}\Big) \\[1em] \Rightarrow y + 2 = -\dfrac{4}{3}\Big(x - \dfrac{4}{3}\Big) \\[1em] \Rightarrow 3(y + 2) = -4\Big(x - \dfrac{4}{3}\Big) \\[1em] \Rightarrow 3y + 6 = -4x + \dfrac{16}{3} \\[1em] \Rightarrow 3(3y + 6) = -12x + 16 \\[1em] \Rightarrow 9y + 18 = -12x + 16 \\[1em] \Rightarrow 12x + 9y + 2 = 0.$

Hence, equation of line passing through P and perpendicular to AB is 12x + 9y + 2 = 0.