Question

$A \,B \,C \,D$ is a parallelogram and $P$ is a point on the segment $AD$ dividing it internally in the ratio $3: 1$. If the line $BP$ meets the diagonal $A \,C$ in $Q$, then $A Q: Q C$ equals

• A. 3 : 4
• B. 4 : 3
• C. 3 : 2
• D. 2 : 3

Answer 1

Answer: (A)

Since, $P$ divides $A D$ in the ratio $3: 1$, so $P$ will be $\frac{3 d }{4}$

Let $Q$ divides $B P$ in $1: \lambda$ and $A C$ in $1: \mu$.
$\therefore \frac{ b + d }{1+\mu}=\frac{\frac{3 d }{4}+\lambda b }{1+\lambda}$
$\Rightarrow b (1+\lambda)+(1+\lambda) d =\frac{3}{4}(1+\mu) d +\lambda(1+\mu) d$
On equating the vectors $b$ and $d$, we get
$1+\lambda=\frac{3}{4}(1+\mu)$ and $1+\lambda=\lambda(1+\mu)$
$ \Rightarrow \lambda =\frac{3}{4} \text { and } \mu=\frac{4}{3} $
$\therefore A Q: Q C =1: \mu $
$ =1: \frac{4}{3}=3: 4 $