Question

Show that the following points $(3,-2),(3,2),(-1,2)and(-1,-2)$ taken in order are vertices of a square.

Answer 1

Let the vertices be taken as A(3, -2), B(3, 2), C(-1, 2) and D(-1, -2).
$A{B}^2=(3-3{)}^2+(2+2{)}^2=4^2=16$
$B{C}^2=(3+1{)}^2+(2-2{)}^2=4^2=16$
$C{D}^2=(-1+1{)}^2+(2+2{)}^2=4^2=16$
$D{A}^2=(-1-3{)}^2+(-2+2{)}^2=(-4{)}^2=16$
$A{C}^2=(3+1{)}^2+(-2-2{)}^2=4^2+(-4{)}^2=16+16=32$
$B{D}^2=(3+1{)}^2+(2+2{)}^2=4^2+4^2=16+16=32$
$AB=BC=CD=DA=\sqrt{16}=4$. (That is, all the sides are equal)
$AC=BD=\sqrt{32}=4\sqrt{2}$. (That is, the diagonals are equal.)
Hence, the points A, B, C and D form a square