$ABCD$ is a parallelogram and $P$ is a point on the segment $\mathbf{AD}$ dividing it internally in the ratio $3:1$. If the line $\mathbf{BP}$ meets the diagonal $AC$ in $Q$, then $AQ:QC$ equals

• Correct answer
  A

  $3:4$
• B

  $4:3$
• C

  $3:2$
• D

  $2:3$

Solution

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 $1:\mu$. $$\begin{array}{rl}& \therefore \frac{\mathbf{b}+\mathbf{d}}{1+\mu }=\frac{\frac{3\mathbf{d}}{4}+\lambda \mathbf{b}}{1+\lambda }\\ & \Rightarrow \mathbf{b}(1+\lambda )+(1+\lambda )\mathbf{d}=\frac{3}{4}(1+\mu )\mathbf{d}+\lambda (1+\mu )\mathbf{d}\end{array}$$ On equating the vectors $\mathbf{b}$ and $\mathbf{d}$, we get $$\begin{array}{rlrl}& & 1+\lambda =\frac{3}{4}(1+\mu )& \text{ and }1+\lambda =\lambda (1+\mu )\\ \Rightarrow & \lambda & =\frac{3}{4}\text{ and }\mu =\frac{4}{3}& \\ \therefore & AQ:QC& =1:\mu & \\ & =& 1:\frac{4}{3}=3:4& \end{array}$$

Hence, the correct answer is option: 'A'.