AB is a line segment joining the points A(3, 2) and B(4, 1). Find :

(i) the ratio in which AB is divided by point P on x-axis

(ii) the co-ordinates of point P

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

(i) Let point P be (a, 0).

By section formula,

y = $\dfrac{m$1y2 + m2y1}{m1 + m2}

Let ratio in which P divides AB be k : 1, substituting values we get :

$\Rightarrow 0 = \dfrac{k \times 1 + 1 \times 2}{k + 1} \\[1em] \Rightarrow 0 = k + 2 \\[1em] \Rightarrow k = -2$

Since, k is negative it means the division is external.

∴ k : 1 = 2 : 1 (externally)

Hence, point P on x-axis divides the line segment AB in the ratio 2 : 1.

(ii) By section-formula,

x = $\dfrac{m$1x2 + m2x1}{m1 + m2}

Substituting values we get :

$\Rightarrow x = \dfrac{2 \times 4 + 1 \times 3}{2 + 1} \\[1em] = \dfrac{8 + 3}{3} \\[1em] = \dfrac{11}{3}.$

P = (a, 0) = $\Big(\dfrac{11}{3}, 0\Big).$

Hence, co-ordinates of P = $\Big(\dfrac{11}{3}, 0\Big)$.

(iii) By formula,

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

Slope of AB = $\dfrac{-1 - 2}{4 - 3} = \dfrac{-3}{1}$ = -3.

We know that,

Product of slope of perpendicular lines = -1.

∴ -3 × Slope of AB = -1

⇒ Slope of AB = $\dfrac{-1}{-3} = \dfrac{1}{3}$.

By point-slope form :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\dfrac{1}{3} \times \Big(x - \dfrac{11}{3}\Big)$

⇒ y = $\dfrac{1}{3} \times \Big(\dfrac{3x - 11}{3}\Big)$

⇒ y = $\dfrac{3x - 11}{9}$

⇒ 9y = 3x - 11

⇒ 3x - 9y - 11 = 0

Hence, required equation is 3x - 9y - 11 = 0.