Question

If $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ and $\overrightarrow{d}$ are the position vectors of points A, B, C, D such that no three of them are collinear and $\overrightarrow{a}+\overrightarrow{c}=\overrightarrow{b}+\overrightarrow{d},$ then ABCD is a
(a) rhombus
(b) rectangle
(c) square
(d) parallelogram

Answer 1

Given:
$$\begin{gathered} \overrightarrow{a}+\overrightarrow{c}=\overrightarrow{b}+\overrightarrow{d} \\ \Rightarrow \overrightarrow{c}-\overrightarrow{d}=\overrightarrow{b}-\overrightarrow{a} \\ \Rightarrow \overrightarrow{AB}=\overrightarrow{DC} \\ And\overrightarrow{a}+\overrightarrow{c}=\overrightarrow{b}+\overrightarrow{d} \\ \Rightarrow \overrightarrow{c}-\overrightarrow{b}=\overrightarrow{d}-\overrightarrow{a} \\ \Rightarrow \overrightarrow{AD}=\overrightarrow{BC} \\ Also,\sin ce\overrightarrow{a}+\overrightarrow{c}=\overrightarrow{b}+\overrightarrow{d} \\ \Rightarrow \frac{1}{2}(\overrightarrow{a}+\overrightarrow{c)}=\frac{1}{2}(\overrightarrow{b}+\overrightarrow{d}) \\ so,positionvectorofmidpointofBD=positionvectorofmidpointofAC. \\ hencediagonalsbisecteachother. \\ thegivenABCDisaparalle\log ram. \end{gathered}$$
Option (d).