Question

show that the points 3, 20, 5 - 3, 2 and 0 - 1 are the vertex of a square

Answer 1

let A(3,2) B(0,5) C(-3,2) D(0,-1)
USING DISTANCE FORMULA,
AB = √(3-0)² + (2-5)²
AB = √(3)² + (-3)²
AB = √9+9
AB = √18 UNITS
BC = √(0+3)² + (5-2)²
BC = √(3)² + (3)²
BC = √9+9 = √18 UNITS
CD = √(-3-0)² + (2+1)²
CD = √(-3)² + (3)²
CD = √9+9 = √18 UNITS
AD = √(3-0)² + (2+1)²
AD = √(3)² + (3)²
AD = √9+9 = √18 UNITS
SO, AB=BC=CD=AD
NOW AC = √(3+3)² + (2+2)² 
AC = √(6)² + (4)²
AC = √36+16 = √52 UNITS
BD = √(5+1)² = √(6)² = √36 UNITS
DIAGONALS ARE NOT EQUAL.
SO IT IS NOT A SQUARE.