Show that the coplanar points `(0,4,1),(2,3,-1),(4,5,0) and 92,6,2)` are the vertices of a square.

Options

A. Rhombus
B. Rectangle
C. Square
D. Parallelogram

Answer

C

Explanation

Let the coordinates of A,B,C and D are (0,4,1), (2,3,-1)(4,5,0) and (2,6,2) respectively.
We find its sides and diagonals as below
`AB=sqrt((2-0)^2+(3-4)^2+(-1-1)^2)`
`=sqrt(4+1+4)=sqrt9=3`
`BC=sqrt((4-2)^2+(5-3)^2+(0+1)^2)`
`=sqrt(4+4+1)=sqrt9=3`
`CD=sqrt((2-4)^2+((6-5)^2+(2-0)^2)`
`=sqrt(4+1+4)=sqrt9=3`
`DA=sqrt((0-2)^2+(5-4)^2+(0-1)^2)`
`=sqrt(4+4+1)=sqrt9=3`
`AC=sqrt((4-0)^2+(5-4)^2+(0-1)^2)`
`=sqrt(16+1+1)=sqrt18`
and `BD=sqrt((2-2)^2+(6-3)^2+(2+1)^2)`
`=sqrt(9+9)=sqrt18`
Since, AB=BC=CD=DA, sides are equal and AC = BD, diagonals are also equal.
Hence, A, B, C and D are the vertices of a square.