Let PQRS be a quadrilateral. If M and N are m | vedclass

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(C) Let the position vectors of points $P, Q, R, S, M, N$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{m}, \vec{n}$ respectively.Since $M$ is the midp

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$2 \overrightarrow{MN}$

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(C) Let the position vectors of points $P, Q, R, S, M, N$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{m}, \vec{n}$ respectively.Since $M$ is the midpoint of $PQ$,we have $\vec{m} = \frac{\vec{p} + \vec{q}}{2}$,which implies $\vec{p} + \vec{q} = 2\vec{m}$.Since $N$ is the midpoint of $RS$,we have $\vec{n} = \frac{\vec{r} + \vec{s}}{2}$,which implies $\vec{r} + \vec{s} = 2\vec{n}$.Now,consider the expression $\overrightarrow{PS} + \overrightarrow{QR} = (\vec{s} - \vec{p}) + (\vec{r} - \vec{q})$.Rearranging the terms,we get $(\vec{s} + \vec{r}) - (\vec{p} + \vec{q})$.Substituting the values derived from the midpoints,we get $2\vec{n} - 2\vec{m}$.This simplifies to $2(\vec{n} - \vec{m}) = 2\overrightarrow{MN}$.

Let $PQRS$ be a quadrilateral. If $M$ and $N$ are midpoints of the sides $PQ$ and $RS$ respectively,then $\overrightarrow{PS} + \overrightarrow{QR} =$

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