Question

A line $L$ intersects the three sides $BC,CA$ and $AB$ of a $\Delta ABC$ at $P,Q$ and $R$ respectively. Then, $\frac{BP}{PC}\cdot \frac{CQ}{QA}\cdot \frac{AR}{RB}$ is equal to

• A. 1
• B. 0
• C. -1
• D. None of these

Answer 1

Answer: (C)

Let $A\left(x_1,y_1\right),B\left(x_2,y_2\right)$ and $C\left(x_3,y_3\right)$ be the vertices of $\Delta ABC$ and let $lx+my+n=0$ be the equation of the line. If $P$ divides $BC$ in the ratio $\lambda :1$, then the coordinates of $P$ are

$\left(\frac{\lambda x_3+x_2}{\lambda +1},\frac{\lambda y_3+y_2}{\lambda +1}\right)$
Also, as $P$ lies on $L$, we have
$l\left(\frac{\lambda x_3+x_2}{\lambda +1}\right)+m\left(\frac{\lambda y_3+y_2}{\lambda +1}\right)+n=0$
$\Rightarrow -\frac{l{x}_2+m{y}_2+n}{l{x}_3+m{y}_3+n}=\lambda =\frac{BP}{PC}\dots$(i)
Similarly, we obtain
$\frac{CQ}{QA}=-\frac{l{x}_3+m{y}_3+n}{l{x}_1+m{y}_1+n}\dots$(ii)
and $\frac{AR}{RB}=-\frac{l{x}_1+m{y}_1+n}{l{x}_2+m{y}_2+n}\dots$(iii)
On multiplying Eqs. (i), (ii) and (iii), we get
$\frac{BP}{PC}\cdot \frac{CQ}{QA}\cdot \frac{AR}{RB}=-1$