Question

Let PQRST be a pentagon in which the sides PQ and RS are parallel and sides TP and QR are parallel. If PQ : RS is 3 : 1 and TP : QR is 1 : 2 and diagonals PS and QT meet at M, then PM : MS equals

• A. 3 : 1
• B. 3 : 5
• C. 2 : 1
• D. 1 : 2

Answer 1

The correct option is B 3 : 5


Let position vectors of P, Q, R, S, T be $\overrightarrow{0},\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$
$\begin{array}{c}\overrightarrow{PQ}=3\overrightarrow{SR}\Rightarrow \overrightarrow{a}=3(\overrightarrow{b}-\overrightarrow{c})\\ \overrightarrow{QR}=2\overrightarrow{PT}\Rightarrow \overrightarrow{b}-\overrightarrow{a}=2\overrightarrow{d}\end{array}\}\Rightarrow \frac{\overrightarrow{a}+\overrightarrow{3c}}{3}=\overrightarrow{a}+\overrightarrow{2d}$
$\Rightarrow \overrightarrow{c}=\frac{2\overrightarrow{a}+6\overrightarrow{d}}{3}$
$\therefore M=\frac{\mu \overrightarrow{a}+\overrightarrow{d}}{\mu +1}=\frac{\lambda \overrightarrow{c}}{\lambda +1}=\lambda \frac{\left(\frac{2\overrightarrow{a}+6\overrightarrow{d}}{3}\right)}{\lambda +1}$
$\frac{\mu }{\mu +1}=\frac{2\lambda }{3(\lambda +1)}\&\frac{1}{\mu +1}=\frac{2\lambda }{(\lambda +1)}$
$\frac{2\lambda \mu }{(\lambda +1)}=\frac{2\lambda }{3(\lambda +1)}\Rightarrow -\lambda =\frac{3}{5}$