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

(i) the ratio PR : RQ

(ii) the coordinates of R.

(iii) the area of quadrilateral PMNQ.

(i) The point R lies on the y-axis, so its x-coordinate is 0.

Let R(0, y) divide the line segment PQ in the ratio m₁ : m₂

By section-formula,

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

Substitute values we get,

$\Rightarrow 0 = \Big(\dfrac{m$1(3) + m2(-4)}{m1 + m2}\Big) \\[1em] \Rightarrow 0 = 3m1 - 4m2 \\[1em] \Rightarrow 3m1 = 4m2 \\[1em] \Rightarrow \dfrac{m1}{m2} = \dfrac{4}{3}.

Hence, the ratio PR : RQ = 4 : 3.

(ii) Given,

P(-4, 5) and Q(3, 2)

m₁ : m₂ = 4 : 3.

By section-formula,

y = $\Big(\dfrac{m$1y2 + m2y1}{m1 + m2}\Big)

Substitute values we get,

$\Rightarrow y = \Big(\dfrac{4(2) + 3(5)}{4 + 3}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{8 + 15}{7}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{23}{7}\Big).$

Hence, the coordinates of R = $\Big(0, \dfrac{23}{7}\Big)$.

(iii) From figure,

PQNM is a trapezium, with PM and QN being the parallel sides, as both are perpendicular to x-axis.

From figure,

PM = 5 units, MN = 7 units and QN = 2 units.

The area of a trapezoid is given by :

Area = $\dfrac{1}{2}$ × Sum of parallel sides × Distance between them

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

Hence, area of PQNM = 24.5 sq.units.