P and Q are two points on the x-axis and y-axis respectively. M(3, 2) divides the line segment PQ in the ratio 2 : 3. Find :

(i) the co-ordinates of the points P and Q

(ii) the slope of line segment PQ

(iii) the equation of the line through point P and perpendicular to PQ.

(i) Let points P and Q be (x, 0) and (0, y) respectively.

Given,

M(3, 2) divides the line segment PQ in the ratio 2 : 3.

By section-formula,

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

Substituting values we get :

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

P = (x, 0) = (5, 0) and Q = (0, y) = (0, 5).

Hence, P = (5, 0) and Q = (0, 5).

(ii) By formula,

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

Substituting values we get :

Slope of PQ = $\dfrac{5 - 0}{0 - 5} = \dfrac{5}{-5}$ = -1.

Hence, slope of PQ = -1.

(iii) We know that,

Product of slope of perpendicular lines equal to -1.

Let slope of line perpendicular to PQ be m

∴ -1 × m = -1

⇒ m = 1.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 1(x - 5)

⇒ y = x - 5

⇒ x - y = 5.

Hence, equation of the line through point P and perpendicular to PQ is x - y = 5.