In a quadrilateral $A B C D, M$ and $N$ are the mid-points of the sides $A B$ and $C D$ respectively. If…
In a quadrilateral $A B C D, M$ and $N$ are the mid-points of the sides $A B$ and $C D$ respectively. If $\overline{A D}+\overline{B C}=t \overline{M N}$, then $t=$
4
2
$\frac{1}{2}$
$\frac{3}{2}$
Solution
$(\mathrm{C})$
Let $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{m}, \bar{n}$ be the position vectors of $A, B, C$, $\mathrm{D}, \mathrm{M}, \mathrm{N}$ respectively. $\mathrm{M}$ and $\mathrm{N}$ are the midpoints of $\mathrm{AB}$ and CD respectively.
$\overline{\mathrm{m}}=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}$, and $\overline{\mathrm{n}}=\frac{\overline{\mathrm{c}}+\overline{\mathrm{d}}}{2} \Rightarrow \overline{\mathrm{a}}+\overline{\mathrm{b}}=2 \overline{\mathrm{m}}$ and $\overline{\mathrm{c}}+\overline{\mathrm{d}}=2 \overline{\mathrm{n}}$
We have $\overline{\mathrm{AD}}+\overline{\mathrm{BC}}=\mathrm{t}(\overline{\mathrm{MN}})$
$(\bar{d}-\bar{a})+(\bar{c}-\bar{b})=t(\bar{n}-\bar{m})$
$\overline{\mathrm{d}}-\overline{\mathrm{a}}+\overline{\mathrm{c}}-\overline{\mathrm{b}}=\mathrm{t}(\overline{\mathrm{n}}-\mathrm{m})$
$(\overline{\mathrm{d}}+\overline{\mathrm{c}})-(\overline{\mathrm{a}}+\overline{\mathrm{b}})=\mathrm{t}(\overline{\mathrm{n}}-\overline{\mathrm{m}})$
$2 \bar{n}-2 \bar{m}=t(\bar{n}-\bar{m}) \Rightarrow t=2$
