Question

Point $X$ and $Y$ are taken on the sides $QR$ and $RS$, respectively of a parallelogram $PQRS$, so that $QX=4XR$ and $\overrightarrow{RY}=4\overrightarrow{YS}$. The line $XY$ cuts the line $PR$ at $Z$. Prove that $\overrightarrow{PZ}=\left(\frac{21}{25}\right)\overrightarrow{PR}$.

Answer 1

Let point P be taken as origin & $\overrightarrow{q}\cdot \overrightarrow{r}$ are the position vectors of Q & S points respectively.

$\overrightarrow{PR}=\overrightarrow{q}+\overrightarrow{r}$

Position vector of $x=\frac{\overrightarrow{9}+4(\overrightarrow{q}+\overrightarrow{s})}{5}$

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

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

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

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

$\Rightarrow \frac{1}{\lambda +1}=\frac{\mu +\frac{1}{5}}{\mu +1}$ ........(i)

$\Rightarrow \frac{1}{\lambda +1}=\frac{\frac{4\mu }{5}+1}{\mu +1}$ ........(ii)

From eq (i) & eq (ii) we get

$\mu =4$ & $\lambda =\frac{4}{21}$

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

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