MCQ
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ and $\vec{\text{d}}$ are the position vector of points A, B, C, D such that no three of them are collinear and $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$, then ABCD is a,
  • A
    Rhombus.
  • B
    Rectangle.
  • C
    Square.
  • Parallelogram.

Answer

Correct option: D.
Parallelogram.
Given:

$\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$

$\Rightarrow\vec{\text{c}}-\vec{\text{d}}=\vec{\text{b}}-\vec{\text{a}}$

$\Rightarrow\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$

And $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$

$\Rightarrow\vec{\text{c}}-\vec{\text{b}}=\vec{\text{d}}-\vec{\text{a}}$

$\Rightarrow\overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}$

Also, since $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$

$\Rightarrow\frac{1}2\big(\vec{\text{a}}+\vec{\text{c}}\big)=\frac{1}2\big(\vec{\text{b}}+\vec{\text{d}}\big)$

So, position vector of mid-point of BD = position vector of mid-point of AC.

Hence diagonals bisect each other.

The given ABCD is a parallelogram.

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