Let us consider the points as A(0,0) ; B(3,4) ; C(0,8) ; D(-3,4)
The distance b/w AB = √(3-0)²+(4-0)² = √25 = 5 
The distance b/w BC = √(0-3)²+(8-4)² = √9+16 = 5
The distance b/w CD = √(-3-0)²+(4-8)² = √9+16 = 5
The distance b/w  DA = √(-3-0)²+(4-0)² = √25 = 5
The distance b/w AC =√(0-0)²+(8-0)² = 8
The distance b/w BD = √(-3-3)²+(4-4)² = 6
As the distance of all sides are equal and diagonals are not equal.
Therefore the given points represents Rhombus
1 5 1
Indirectly U r asking to prove that this points represent a rhombus or not
As we all know that rhombus has all its sides equal.So lets name the points as A(0,0),B(3,4),C(0,8),D(-3,4)
By taking out the difference from this points, we get,
we can see AB=CD=BC=AD This mean that the points are the edges of a rhombus.
3 3 3
good enough
you have shown only all sides are equal.we can't decide whether the given points form a rhombus.Becuase all sides equal fo both rhombus and square.If want to whether it is a square or rhombus you need find distance between AC and BD.If you get AC=BD it is a square. If you get AC ≠D it is a rhombus.