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Show that the following points taken in order form the vertices of a rhombus.

(0, 0), (3, 4), (0, 8) and ( - 3, 4)

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(0, 0), (3, 4), (0, 8) and ( - 3, 4)

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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

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,

AB=(0-3)+(0-4)=1

BC=(0-3)+(8-4)=1

CD=(-3-0)+(4-8)=1

AD=(-3-0)+(4-0)=1

we can see AB=CD=BC=AD This mean that the points are the edges of a rhombus.