Question

Show that the following points are the vertices of a square:

A(0, –2), B(3, 1), C(0, 4) and D(–3, 1)

Answer 1

The given points are  A (0,-2), B(3,1), C(0,4) and D(-3,1)

`AB = sqrt ((3-0)^2 +(1+2)^2) = sqrt((3)^2+(3)^2) = sqrt(9+9) = sqrt(18) = 3sqrt(2)   units`

`BC = sqrt ((0-3)^2 +(4-1)^2) = sqrt((-3)^2 +(3)^2) = sqrt(9+9) = sqrt(18) = 3 sqrt(2)  units`

`CD = sqrt((-3-0)^2 + (1-4)^2)  = sqrt((-3)^2 +(-3)^2 ) = sqrt(9+9) = sqrt(18) = 3 sqrt(2)  units`

`DA = sqrt((-3-0)^2 +(1+2)^2) = sqrt((-3)^2 +(3)^2) = sqrt(9+9) = sqrt(18) = 3 sqrt(2)  units`

Therefore, `AB = BC = CD = DA = 3 sqrt(2)  units`

Also , 

 `AC= sqrt((0-0)^2 + (4+2)^2) = sqrt((0)^2 +(6)^2 ) = sqrt(36) = 6  units`

`BD = sqrt((-3-3)^2 +(1-1)^2) = sqrt((-6)^2 +(0)^2) = sqrt(36) =6  units`

Thus, diagonal AC = diagonal BD 

Therefore, the given points from a square.