Question

Prove that the points $(-4,-1),(-2,-4),(4,0)$ and $(2,3)$ are the vertices of a rectangle.

Answer 1

hit $A(-4,-1),B(-2,4),C(4,0)$ and $D(2,3)$ be the given points (circles)

Now, $AD=\sqrt{(2+4{)}^2+(3+1{)}^2}$

$=\sqrt{52}$

$BC=\sqrt{(4+2{)}^2+(0+4{)}^2}$

$=\sqrt{52}$

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

$=\sqrt{4+9}$

$=\sqrt{13}$

$CD=\sqrt{(2-4{)}^2+(3-0{)}^2}$

$=\sqrt{13}$

$\Rightarrow AD=BC$ and $AB=CD$

Hence $ABCD$ is a parallelogram

Now, $AC=\sqrt{(4+4{)}^2+(0+1{)}^2}$

$=\sqrt{65}$

$BD=\sqrt{(2+2{)}^2+(3+4{)}^2}=\sqrt{65}$

Clearly $A{C}^2=65=A{B}^2+B{C}^2$

$=(\sqrt{52}{)}^2+(\sqrt{13}{)}^2=65$

and $(BD{)}^2=B{C}^2+C{D}^2$

$65=52+13=65$

Hence, $ABCD$ is a rectangle

$\Rightarrow$ The points $(-4,-1,)(4,0)$ and $(2,3)$ are which of rectangle.