Question

Let P, Q, R and S be the points on the plane with position vectors (-2i - j), 4i, (3i + 3j) and (-3i + 2j) respectively. The quadilateral 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

Answer: (A)

Parallelogram, which is neither a rhombus nor a rectangle

Consider the problem

As we given

$\left(-2\hat{i}-\hat{j}\right),\left(4\hat{i}\right),\left(3\hat{i}+3\hat{j}\right),\left(-3i+2\hat{j}\right)$

On converting above vectors into Cartesian coordinates

Then , consider,

$P\left(-2,-1\right),Q\left(4,0\right),R\left(3,3\right),S\left(-3,2\right)$

By distance formula

$=\sqrt{(x_1-x_2{)}^2+(y_1+y_2{)}^2}$

$d\left(PQ\right)=\sqrt{(4+2{)}^2+(0+1{)}^2}=\sqrt{37}$

Similarly,

$d\left(QR\right)=\sqrt{10}$

$d\left(RS\right)=\sqrt{37}$

$d\left(SP\right)=\sqrt{10}$

Therefore,

$d\left(PQ\right)=d\left(RS\right)$ and $d\left(QR\right)=d\left(SP\right)$

Hence,

$PQRS$ is parallelogram

Now,

Slope $=\frac{y_2-y_1}{x_2-x_1}$

Therefore, let slope is $S$

then ,

$S\left(PQ\right)=\frac{0-(-1)}{4-(-2)}=\frac{0+1}{4+2}=\frac{1}{6}$

Similarly,

$S\left(QR\right)=\frac{3}{-1}=-3$

$S\left(RS\right)=\frac{-1}{-6}=\frac{1}{6}$

$S\left(PS\right)=-3$

$S\left(PQ\right)=S\left(RS\right)$ this implies $PQ\parallel RS$

And

$S\left(QR\right)=S\left(SP\right)$ this implies $QR\parallel SP$

$S\left(PQ\right)\ne S\left(QR\right)$ i.e. they aren't parallel

$S\left(PQ\right)\times S\left(QR\right)\ne -1$ i.e. they aren't perpendicular.

Therefore

$PQRS$ is a parallelogram but it is nor a rhombus and not rectangle.

Hence option $A$ is the correct answer.