A line L intersects the three sides $mathrm{BC}, mathrm{CA} {text { and }} mathrm{AB}$ of a triangle ABC at $mathrm{P}, mathrm{Q}$ and R , respectively. Then $frac{mathrm{BP}}{mathrm{PC}} cdot frac{mathrm{CQ}}{mathrm{QA}} cdot frac{mathrm{AR}}{mathrm{RB}}=mathrm{k}$, where $mathrm{k}=$

A line intersects the three sides <img alt="\mathrm{BC, CA}" src="https://entrancecorner.oncodec

A line $\mathrm{L}$ intersects the three sides $\mathrm{BC, CA}$ and $\mathrm{AB}$ of a triangle $\mathrm{ABC}$ at $\mathrm{P,Q}$ and $\mathrm{R}$ , respectively. Then $\mathrm{\frac{B P}{P C} \cdot \frac{C Q}{Q A} \cdot \frac{A R}{R B}=k}$ , where $\mathrm{k=}$
Option: 1
$\mathrm{-1}$

Option: 2
$\mathrm{1}$

Option: 3
$\mathrm{2}$

Option: 4
$\mathrm{3}$

Answers (1)

Let $\mathrm{A\left ( x_{1} ,y_{1}\right ),B\left ( x_{2},y_{2} \right )}$ and $\mathrm{C\left ( x_{3} , y_{3}\right )}$ be the vertices of $\mathrm{\bigtriangleup ABC, }$ and let $\mathrm{\varphi _x+m y+n=0 }$ be equation of the line $\mathrm{L }$ .
If $\mathrm{P }$ divides $\mathrm{BC }$ in the ratio $\mathrm{ \lambda :1, }$ then the coordinates of $\mathrm{ P }$ are $\mathrm{ \left(\frac{\lambda \mathrm{x}_3+\mathrm{x}_2}{\lambda+1}, \frac{\lambda \mathrm{y}_3+\mathrm{y}_2}{\lambda+1}\right) }$
Also, as $\mathrm{ P }$ lies on $\mathrm{ L }$ , we have $\mathrm{ ........\left ( ii \right )}$
$\mathrm{ \varphi \left(\frac{\lambda \mathrm{x}_3+\mathrm{x}_2}{\lambda+1}\right)+\mathrm{m}\left(\frac{\lambda \mathrm{y}_3+\mathrm{y}_2}{\lambda+1}\right)+\mathrm{n}=0 }$
$\mathrm{ \Rightarrow \quad-\frac{\varphi \mathrm{x}_2+\mathrm{my}_2+\mathrm{n}}{\varphi \mathrm{x}_3+\mathrm{my}_3+\mathrm{n}}=\lambda=\frac{\mathrm{BP}}{\mathrm{PC}}\ .......\left ( i \right ) }$
Similarly, we obtain $\mathrm{ \frac{\mathrm{CQ}}{\mathrm{QA}}=-\frac{\varphi \mathrm{x}_3+\mathrm{my}_3+\mathrm{n}}{\varphi \mathrm{x}_1+\mathrm{my} \mathrm{y}_1+\mathrm{n}} }$
and $\mathrm{ \frac{A R}{R B}=-\frac{\varphi x_1+m y_1+n}{\varphi x_2+m y_2+n}\ \ \ \ ........\left ( iii \right )}$
Multiplying $\mathrm{\left ( i \right )}$ , $\mathrm{\left ( ii \right )}$ and $\mathrm{\left ( iii \right )}$ , we get the required result.