A and B are the two points on the x-axis and y-axis respectively. P(2, -3) is the mid-point of AB.

Find :

(i) the coordinates of A and B.

(ii) the slope of the line AB.

(iii) the equation of the line AB.

(i) Let the coordinates of A be (x, 0) and B be (0, y).

P(2, -3) is the mid-point of AB. So we have,

$\Rightarrow 2 = \dfrac{x + 0}{2} \text{ and } -3 = \dfrac{0 + y}{2} \\[1em] \Rightarrow 2 = \dfrac{x}{2} \text{ and } -3 = \dfrac{y}{2} \\[1em] \Rightarrow x = 4 \text{ and } y = -6.$

Hence. the coordinates of A are (4, 0) and B are (0, -6).

(ii) Slope of AB = $\dfrac{y$2 - y1}{x2 - x1}

Putting values we get slope,

$= \dfrac{(-6 - 0)}{(0 - 4)} \\[1em] = \dfrac{-6}{-4} \\[1em] = \dfrac{3}{2}.$

Hence, the slope of the line AB is $\dfrac{3}{2}.$

(iii) Equation of AB will be

⇒ y - y₁ = m(x - x₁)
⇒ y - 0 = $\dfrac{3}{2}$(x - 4)
⇒ 2y = 3x - 12
⇒ 3x - 2y = 12.

Hence, the equation of AB is 3x - 2y = 12.