Single Choice

$$ABCD$$ is a quadrilateral, $$E$$ is the point of intersection of the line joining the midpoints of the opposite sides. If $$O$$ is any point and $$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = \vec{x OE},$$ then $$x$$ is equal to

A$$3$$
B$$9$$
C$$7$$
D$$4$$
Correct Answer

Solution

Let $$\vec{OA} = \vec{a}, \vec{OB} = \vec{b}, \vec{OC} = \vec{c}$$ and $$\vec{OD} = \vec{d}$$.

Therefore,
$$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = \vec{a} + \vec{b} + \vec{c} + \vec{d}$$.
The midpoint $$P$$ of $$AB$$, is $$\displaystyle \dfrac{\vec{a} + \vec{b}}{2}$$.
The position vector of the midpoint $$Q$$ of $${CD}$$, is $$\displaystyle \dfrac{\vec{c} + \vec{d}}{2}$$.

Therefore, the position vector of the midpoint of $${PQ} $$ is $$\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$$.

Similarly, the position vector of the midpoint of $$RS$$ is $$\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$$, i.e., $$\vec{OE} =\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4} \Rightarrow x = 4$$

Hence, option '$$D$$' is correct.


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