question_answer

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

A) 1

B) 0

C) -1

D) None of these

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

\[\left( \frac{\lambda {{x}_{3}}+{{x}_{2}}}{\lambda +1},\frac{\lambda {{y}_{3}}+{{y}_{2}}}{\lambda +1} \right)\]

Also, as P lies on L,

we have

\[l\left( \frac{\lambda {{x}_{3}}+{{x}_{2}}}{\lambda +1} \right)+m\left( \frac{\lambda {{y}_{3}}+{{y}_{2}}}{\lambda +1} \right)+n=0\]

\[\Rightarrow -\frac{l{{x}_{2}}+m{{y}_{2}}+n}{l{{x}_{3}}+m{{y}_{3}}+n}=\lambda =\frac{BP}{PC}\]            ? (i)

Similarly, we obtain

\[\frac{CQ}{QA}=-\frac{l{{x}_{3}}+m{{y}_{3}}+n}{l{{x}_{1}}+m{{y}_{1}}+n}\]                .... (ii)

and \[\frac{AR}{RB}=\frac{l{{x}_{1}}+m{{y}_{1}}+n}{l{{x}_{2}}+m{{y}_{2}}+n}\]                        ?. (iii)

On multiplying Eqs. (i), (ii) and (iii), we get

\[\frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=1\]