Straight Line Question 218

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}.\frac{CQ}{QA}.\frac{AR}{RB} $ is equal to

Options:

A) 1.

B) 0

C) -1

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

  • [c] Let $ A(x_1,y_1),B(x_2,y_2) $ and $ C(x_3,y_3) $ 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 $ ( \frac{\lambda x_3+x_2}{\lambda +1},\frac{\lambda y_3+y_2}{\lambda +1} ) $ Also, as P lies on L, we have $ l( \frac{\lambda x_3+x_2}{\lambda +1} )+m( \frac{\lambda y_3+y_2}{\lambda +1} )+n=0 $

$ \Rightarrow \frac{lx_2+my_2+n}{lx_3+my_3+n}=\lambda =\frac{BP}{PC} $ ? (i) Similarly, we obtain $ \frac{CQ}{QA}=-\frac{lx_3+my_3+n}{lx_1+my_1+n} $ …. (ii) and $ \frac{AR}{RB}=\frac{lx_1+my_1+n}{lx_2+my_2+n} $ ?. (iii) On multiplying Eqs. (i), (ii), and (iii), we get $ \frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=1 $



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