If 2PA = 3PB, find :

(i) co-ordinates of points A and B.

(ii) equation of line AB.

(i) From figure,

A lies on x-axis and B lies on y-axis.

Coordinates of A be (a, 0) and B be (0, b).

Given,

⇒ 2PA = 3PB

⇒ $\dfrac{PA}{PB} = \dfrac{3}{2}$

or P divides line segment AB in the ratio 3 : 2.

By section-formula,

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

Substituting values we get :

$\Rightarrow (2, 3) = \Big(\dfrac{3 \times 0 + 2 \times a}{3 + 2}, \dfrac{3 \times b + 2 \times 0}{3 + 2}\Big) \\[1em] \Rightarrow (2, 3) = \Big(\dfrac{2a}{5}, \dfrac{3b}{5}\Big) \\[1em] \Rightarrow \dfrac{2a}{5} = 2 \text{ and } \dfrac{3b}{5} = 3 \\[1em] \Rightarrow a = \dfrac{2 \times 5}{2} \text{ and } b = \dfrac{3 \times 5}{3} \\[1em] \Rightarrow a = 5 \text{ and } b = 5.$

A = (a, 0) = (5, 0) and B = (0, b) = (0, 5).

Hence, co-ordinates of A = (5, 0) and B = (0, 5).

(ii) By two point form,

Equation of line : y - y₁ = $\dfrac{y$2 - y1}{x2 - x1}(x - x₁)

Substituting values we get :

$\Rightarrow y - 0 = \dfrac{5 - 0}{0 - 5}(x - 5) \\[1em] \Rightarrow y = \dfrac{5}{-5}(x - 5) \\[1em] \Rightarrow y = -1(x - 5) \\[1em] \Rightarrow y = -x + 5 \\[1em] \Rightarrow x + y - 5 = 0.$

Hence, equation of AB is x + y - 5 = 0.