Question

In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio $1:2$ and $Q$ is the mid-point of $AC$.

If $AB+2AD+BC-2DC=kPQ$, then $k$ is equal to?

• A. $-6$
• B. $-4$
• C. $6$
• D. $4$

Answer 1

The correct option is A

$-6$

Explanation for the correct option.

$\begin{array}{rcl}AB+2AD+BC-2DC& =& kPQ\\ \Rightarrow AB+BC+2AD-2DC& =& kPQ\\ & \Rightarrow & AC+2\left(AC+CD\right)-2DC=kPQ\\ & \Rightarrow & AC+2\left(AC-DC\right)-2DC=kPQ\\ \Rightarrow 3AC-4DC& =& kPQ\\ \Rightarrow 3\left(2QC\right)-4\left(\frac{3}{2}PC\right)& =& kPQ\left[\mathrm{As}\mathrm{per}\mathrm{the}\mathrm{given}\mathrm{information}\right]\\ \Rightarrow 6QC-6PC& =& kPQ\\ \Rightarrow 6\left(QC+CP\right)& =& kPQ\\ \Rightarrow 6QP& =& kPQ\\ \Rightarrow -6PQ& =& kPQ\\ \therefore k& =& -6\end{array}$

Hence, option A is correct.