A and B are two points on the x-axis and y-axis respectively.

(i) Write down the coordinates of A and B.

(ii) P is a point on AB such that AP : PB = 3 : 1. Using section formula, find the coordinates of point P.

(iii) Find the equation of a line passing through P and perpendicular to AB.

(i) From figure,

A = (4, 0) and B = (0, 4).

(ii) Let coordinates of P be (x, y).

By section formula,

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

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{3(0) + 1(4)}{3 + 1}, \dfrac{3(4) + 1(0)}{3 + 1}\Big) \\[1em] = \Big(\dfrac{0 + 4}{4}, \dfrac{12 + 0}{4}\Big) \\[1em] = \Big(\dfrac{4}{4}, \dfrac{12}{4}\Big) \\[1em] = (1,3).$

Hence, coordinates of P = (1, 3).

(iii) By formula,

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

Substituting values we get :

Slope of AB = $\dfrac{4 - 0}{0 - 4} = \dfrac{4}{-4} = -1$

We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of line perpendicular to AB = -1

⇒ -1 × Slope of line perpendicular to AB = -1

⇒ Slope of line perpendicular to AB = $\dfrac{-1}{-1} = 1$

Line passing through P and perpendicular to AB :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = 1(x - 1)

⇒ y - 3 = x - 1

⇒ y = x - 1 + 3

⇒ x - y + 2 = 0

Hence, required equation is x - y + 2 = 0.