Question

$ABCD$ 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 $AC$ in $Q$, then $AQ:QC$ equals

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

Answer 1

Answer: (A)

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

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