Question

The quadrilateral whose vertices are $A(-2,5),B(4,-1),C(9,1),D(3,7)$ is a parallelogram

• A. True
• B. False

Answer 1

The correct option is A True

Now, we are going to find distance between $AB,BC,CD$ and $DA.$

$AB=\sqrt{(4+2{)}^2+(-1-5{)}^2}=\sqrt{36+36}=\sqrt{72}$

$BC=\sqrt{(9-4)+(1+1{)}^2}=\sqrt{25+4}=\sqrt{29}$

$CD=\sqrt{(3-9{)}^2+(7-1{)}^2}=\sqrt{36+36}=\sqrt{72}$

$DA=\sqrt{(-2-3{)}^2+(5-7{)}^2}=\sqrt{25+4}=\sqrt{29}$

Here we can see, $AB=CD$ and $BC=DA$

We know that in parallelogram opposites sides are equal.

$\therefore$ $ABCD$ is a parallelogram.