Question

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

(1) the ratio $PR:RQ$.

(2) the coordinate of $R$.

(3) the area of the quadrilateral $PMNQ$.

Answer 1

(i) Let, $R$ divides $PQ$ in the ratio $k:1$. Since, $R$ lies on $y$-axis, therefore, $R(0,y)$

Using Sectional formula,

$R(0,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Here, $m:n=k:1$

$\Rightarrow R(0,y)=(\frac{k\left(3\right)+1(-4)}{k+1},\frac{k\left(2\right)+1\left(5\right)}{k+1})$ ...(A)

On comparing the coordinates, we get,

$\Rightarrow 0=(\frac{k\left(3\right)+1(-4)}{k+1}$

$\Rightarrow 3k=4$

$\Rightarrow k=\frac{4}{3}$

Hence, $R$ divides $PQ$ in the ratio $4:3$.

Therefore, $PR:RQ=4:3$

(ii) On comparing the $y$ coordination from (A), we get,

$\Rightarrow y=\left(\frac{k\left(2\right)+1\left(5\right)}{k+1}\right)$

$\Rightarrow y=\left(\frac{\frac{4}{3}\left(2\right)+1\left(5\right)}{\frac{4}{3}+1}\right)$

$\Rightarrow y=\frac{8+15}{4+3}$

$\therefore y=\frac{23}{7}$

Hence, $R(0,\frac{23}{7})$

(iii) Here, $PM=5$ units, $QN=2$ units and $MN=4+3=7$ units

Area of $PMNQ=\frac{1}{2}\times (PM+QN)\times MN$

$=\frac{1}{2}\times (5+2)\times 7$

$=24.5$ sq. units