Question

Points X and Y are taken on the sides QR and RS, respectively of a parallelogram PQRS, so that QX = 4XR sand RY = 4YS. The line XY cuts the line PR at Z. Find the ratio PZ:ZR.

• A. $4:21$
• B. $3:4$
• C. $21:4$
• D. $4:3$

Answer 1

The correct option is C $21:4$

$A=OW-OX$

$B=OY-OZ$

$A/B=\frac{OW-OX}{OY-OZ}=\frac{\cos \beta -\cos \alpha }{\sin \alpha -\sin \beta }$

$A/B=\frac{\cos \alpha -\cos \beta }{\sin \beta -\sin \alpha }$

pv of $x=[\overrightarrow{q}+4(q+\overrightarrow{s}\left)\right]|5$

$=(5\overrightarrow{q}+4\overrightarrow{s})/5$

pv of $y=\frac{4\overrightarrow{s}+\overrightarrow{q}+\overrightarrow{s}}{5}=\frac{\overrightarrow{q}+5\overrightarrow{q}}{5}$

Let $=\frac{PZ}{ZR}=\frac{1}{\lambda }$ and $\frac{YZ}{ZX}=\mu$

Pv of P$=\frac{\overrightarrow{q}+\overrightarrow{s}}{\lambda +1}=\frac{\mu (q+\frac{4}{5}\overrightarrow{a})+(\frac{\overrightarrow{q}}{5}+\overrightarrow{s})}{\mu +1}$

$\frac{1}{\lambda +1}=\frac{\mu +1/5}{\mu +1}$ .....$\left(1\right)$

$\frac{\perp }{\lambda +1}=\frac{\frac{4\mu }{5}+1}{\mu +1}$ ...............$\left(2\right)$

$\mu =4$

$\lambda =4/21$

$\frac{PZ}{ZR}=\frac{21}{4}$

$\frac{PZ}{PR}=\frac{21}{25}$.