Question

P, Q, R are the points of intersection of a line / with the sides BC, CA, AB of a$\Delta$ ABC respectively, then $\frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}$is

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

The correct option is B -1

Let $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ be the vertices of the $\Delta$ ABC. Let the equation of the line / be ax + by + c = 0.
Suppose line / divides BC at P in the ratio m: 1 ie, $\frac{BP}{PC}=\frac{m}{1}$ Therefore, coordinates of P are $(\frac{m{x}_3+x_2}{m+1},\frac{m{y}_3+y_2}{m+1})$
Since, P lies on ax + by + c = 0, therefore a $\left(\frac{m{x}_3+x_2}{m+1}\right)+b\left(\frac{m{y}_3+y_2}{m+1}\right)+c=0$
$\Rightarrow m(a{x}_3+b{y}_3+c)+(a{x}_2+b{y}_2+c)=0$
$\Rightarrow \frac{m}{1}=-\left(\frac{a{x}_2+b{y}_2+c}{a{x}_3+b{y}_3+c}\right)$
$\Rightarrow \frac{BP}{PC}=-\left(\frac{a{x}_2+b{y}_2+c}{a{x}_3+b{y}_3+c}\right)...........\left(i\right)$
$Similarly\frac{CQ}{QA}=-\left(\frac{a{x}_3+b{y}_3+c}{a{x}_1+b{y}_1+c}\right)...........\left(ii\right)$
$And\frac{AR}{RB}=-\left(\frac{a{x}_a+b{y}_1+c}{a{x}_2+b{y}_2+c}\right)...........\left(iii\right)$
Multiplying Eqs. (i), (ii) amd (iii), we get
$\frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=-1$