Question

The area of a square whose vertices are $(0,-2),(3,1)(0,4)$ and $(-3,1)$ is

• A. $\sqrt{18}$ sq. units
• B. $18$ sq. units
• C. $15$ sq. units
• D. $\sqrt{15}$ sq. units

Answer 1

The correct option is B $18$ sq. units

Let the parallelogram be $ABCD$ with $A=(x_1,y_1)=(0,-2),B=(x_2,y_2)=(3,1),C=(x_3,y_3)=(0,4)$ and $D=(x_4,y_4)=(-3,1)$

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

$=\sqrt{({3-0)}^2+({1+2)}^2}$

$=\sqrt{18}$

$\therefore ar\left(ABCD\right)={AB}^2={\left(\sqrt{18}\right)}^2=18$ sq. units