Question

Let $P,Q,R,S$ be points on the plane with position vectors $-2\hat{i}-\hat{j},4\hat{i},3\hat{i}+3\hat{j},-3\hat{i}+2\hat{j}$ respectively.The quadrilateral $PQRS$ must be a

• A. parallelogram, which is neither a rhombus nor a rectangle
• B. square
• C. rectangle, but not a square
• D. rhombus, but not a square

Answer 1

The correct option is A parallelogram, which is neither a rhombus nor a rectangle

$\overrightarrow{P}=-2\hat{i}-\hat{j},\overrightarrow{Q}=4\hat{i},\overrightarrow{R}=3\hat{i}+3\hat{j},\overrightarrow{S}=-3\hat{i}+2\hat{j}$
Midpoint of diagonal $PR$ is $=\frac{-2\hat{i}-\hat{j}+3\hat{i}+3\hat{j}}{2}$
$=\frac{\hat{i}+2\hat{j}}{2}=\frac{\hat{i}}{2}+\hat{j}$
Midpoint of diagonal $QS$ is
$=\frac{4\hat{i}-3\hat{i}+2\hat{j}}{2}$
$=\frac{\hat{i}+2\hat{j}}{2}$
$=\frac{\hat{i}}{2}+\hat{j}$
It is a parallelogram since diagonal bisect each other
$\overrightarrow{PR}=\overrightarrow{OR}-\overrightarrow{OP}$
$=3\hat{i}+3\hat{j}-(-2\hat{i}-\hat{j})$
$=3\hat{i}+3\hat{j}+2\hat{i}+\hat{j}=5\hat{i}+4\hat{j}$
$\overrightarrow{QS}=\overrightarrow{OS}-\overrightarrow{OQ}$
$=-3\hat{i}+2\hat{j}-\left(4\hat{i}\right)$
$=-7\hat{i}+2\hat{j}$
$\left|\overrightarrow{PR}\right|=\sqrt{25+16}=\sqrt{41}$
$\left|\overrightarrow{QS}\right|=\sqrt{49+4}=\sqrt{53}$
It is not a rectangle since diagonals are not equal
$\overrightarrow{PR}.\overrightarrow{QS}=(5\hat{i}+4\hat{j}).(-7\hat{i}+2\hat{j})$
$=-35+8=-27\ne 0$
It is not a rhombus since diagonals are not perpendicular