AB is a line segment with A = (2, 4) and B = (6, 12). Point P lies on the line segment AB so that P = (3, x), then the ratio AP : PB is :

1. 3 : 2

2. 2 : 3

3. 3 : 1

4. 1 : 3

Let ratio in which P divides AB be k : 1.

By section-formula,

(x, y) = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get :

$\Rightarrow (3, x) = \Big(\dfrac{k \times 6 + 1 \times 2}{k + 1}, \dfrac{k \times 12 + 1 \times 4}{k + 1}\Big) \\[1em] \Rightarrow (3, x) = \Big(\dfrac{6k + 2}{k + 1}, \dfrac{12k + 4}{k + 1}\Big) \\[1em] \Rightarrow \dfrac{6k + 2}{k + 1} = 3 \\[1em] \Rightarrow 6k + 2 = 3(k + 1) \\[1em] \Rightarrow 6k + 2 = 3k + 3 \\[1em] \Rightarrow 6k - 3k = 3 - 2 \\[1em] \Rightarrow 3k = 1 \\[1em] \Rightarrow k = \dfrac{1}{3}.$

Substituting value of k in k : 1, we get :

⇒ $\dfrac{1}{3} : 1$

⇒ 1 : 3.

Hence, Option 4 is the correct option.