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

(0, 0), (3, 4), (0, 8) and (−3, 4)

Formula used:

(0, 0), (3, 4), (0, 8) and (–3, 4)

Let the vertices be taken as A (0, 0), B (3, 4), C (0, 8) and D (–3, 4).

Distance of AB

⇒ AB = √ ((3 – 0)² + ((4 – 0)²)

⇒ AB = √ ((3)² + (4)²)

⇒ AB = √ (9 + 16)

⇒ AB = √ 25

⇒ AB = 5

Distance of BC

⇒ BC= √ ((0 – 3)² + (8 – 4)²)

⇒ BC = √ ((–3)² + (4)²)

⇒ BC = √ (9 + 16)

⇒ BC = √ 25

⇒ BC = 5

Distance of CD

⇒ CD = √ ((–3 – 0)² + (4 – 8)²)

⇒ CD = √ ((–3)² + (–4)²)

⇒ CD = √ (9 + 16)

⇒ CD = √25

⇒ CD = 5

Distance of AD

⇒ AD = √ ((–3 – 0)² + (4 – 0)²)

⇒ AD = √ ((–3)² + (4)²)

⇒ AD = √ (9 + 16)

⇒ AD = √ 25

⇒ AD = 5

Distance of AC

⇒ AC = √ ((0 – 0)² + (8 – 0)²)

⇒ AC = √ ((0)² + (8)²)

⇒ AC = √ (64)

⇒ AC = 8

Distance of BD

⇒ BD = √ ((–3 – 3)² + (4 – 4)²)

⇒ BD = √ ((–6)² + (0)²)

⇒ BD = √ (36 +0)

⇒ BD = √ 36

⇒ BD = 6

AB = BC = CD = DA = 5 (That is, all the sides are equal.)

AC ≠ BD (That is, the diagonals are not equal.)

Hence the points A, B, C and D form a rhombus.