Question

ABCD is parallelogram. If L and M are the middle points of BC and CD, then $\overline{AL}+\overline{AM}$ equals?

• A. $\frac{1}{2}\overline{AC}$
• B. $\frac{3}{2}\overline{AC}$
• C. $\overline{AC}$
• D. $\frac{2}{3}\overline{AC}$

Answer 1

The correct option is B $\frac{3}{2}\overline{AC}$

Take $A$ as origin.

Let $\overrightarrow{a},\overrightarrow{b}$ be the position vectors of the points $B,D$ respectively.

Then, $\overrightarrow{AB}=\overrightarrow{a}$ and $\overrightarrow{AD}=\overrightarrow{b}$

By parallelogram law of addition of vectors,

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}$

$\Rightarrow$ $\overrightarrow{AC}=\overrightarrow{a}+\overrightarrow{b}$

$\therefore$ The position vector of $C$ is $\overrightarrow{a}+\overrightarrow{b}$.

Since, $L,M$ are mid-points of $BC,$ and $DC$ respectively.

The position vector of $L=\frac{\overrightarrow{a}+(\overrightarrow{a}+\overrightarrow{b})}{2}=\frac{2\overrightarrow{a}+\overrightarrow{b}}{2}$

The position vector of $M=\frac{\overrightarrow{b}+(\overrightarrow{a}+\overrightarrow{b})}{2}=\frac{\overrightarrow{a}+2\overrightarrow{b}}{2}$

$\overrightarrow{AL}=\frac{2\overrightarrow{a}+\overrightarrow{b}}{2}=\overrightarrow{a}+\frac{1}{2}\overrightarrow{b}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}$

$\overrightarrow{AM}=\frac{\overrightarrow{a}+2\overrightarrow{b}}{2}=\frac{1}{2}\overrightarrow{a}+\overrightarrow{b}=\frac{1}{2}\overrightarrow{AB}+\overrightarrow{AD}$

$\Rightarrow$ $\overrightarrow{AL}+\overrightarrow{AM}=(\overrightarrow{a}+\frac{1}{2}\overrightarrow{b})+(\frac{1}{2}\overrightarrow{a}+\overrightarrow{b})$

$=\frac{3}{2}(\overrightarrow{a}+\overrightarrow{b})$

$\Rightarrow$ $\overrightarrow{AL}+\overrightarrow{AM}=\frac{3}{2}\overrightarrow{AC}$