The line joining P(-4, 5) and Q(3, 2) intersects the y-axis at point R. PM and QN are perpendiculars from P and Q on the x-axis. Find :

(i) the ratio PR : RQ.

(ii) the co-ordinates of R.

(iii) the area of the quadrilateral PMNQ.

(i) Since, R lies on y-axis. Let its co-ordinates be (0, y).

Let R divide PQ in ratio m₁ : m₂.

By section formula,

$x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] \Rightarrow 0 = \dfrac{m1 \times 3 + m2 \times -4}{m1 + m2} \\[1em] \Rightarrow 3m1 - 4m2 = 0 \\[1em] \Rightarrow 3m1 = 4m2 \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{4}{3}.

m₁ : m₂ = 4 : 3.

Hence, PR : RQ = 4 : 3.

(ii) Substituting m₁ : m₂ = 4 : 3 in section formula we get,

$y = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{4 \times 2 + 3 \times 5}{4 + 3} \\[1em] = \dfrac{8 + 15}{7} \\[1em] = \dfrac{23}{7} \\[1em] = 3\dfrac{2}{7}

R = (0, y) = $\Big(0, 3\dfrac{2}{7}\Big)$.

Hence, co-ordinates of R = $\Big(0, 3\dfrac{2}{7}\Big)$.

(iii) From graph,

PMNQ is a trapezium and 1 block = 1 unit.

Area of trapezium = $\dfrac{1}{2} \times$ (Sum of || sides) × Distance between them

$= \dfrac{1}{2} \times (PM + QN) \times MN \\[1em] = \dfrac{1}{2} \times (5 + 2) \times 7 \\[1em] = \dfrac{1}{2} \times 7 \times 7 \\[1em] = \dfrac{1}{2} \times 49 \\[1em] = 24.5$

Hence, area of PMNQ = 24.5 sq. units.