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

(1, 0), (5, 3), (2, 7) and (−2, 4)

Formula used:

(1, 0), (5, 3), (2, 7) and (–2, 4)

Let the vertices be taken as A (1, 0), B (5, 3), C (2, 7) and D (–2, 4).

Distance of AB

⇒ AB = √ ((5 – 1)² + (3 – 0)²)

⇒ AB = √ ((4)² + (3)²)

⇒ AB = √ (16 + 9)

⇒ AB = √ 25

⇒ AB = 5

Distance of BC

⇒ BC= √ ((2 – 5)² + (7 – 3)²)

⇒ BC = √ ((3)² + (4)²)

⇒ BC = √ (9 + 16)

⇒ BC = √ 25

⇒ BC = 5

Distance of CD

⇒ CD = √ ((–2 – 2)² + (4 – 7)²)

⇒ CD = √ ((–4)² + (–3)²)

⇒ CD = √ (16 + 9)

⇒ CD = √25

⇒ CD = 5

Distance of AD

⇒ AD = √ ((–2 – 1)² + (4 – 0)²)

⇒ AD = √ ((–3)² +(4)²)

⇒ AD = √ (9 + 16)

⇒ AD = √ 25

⇒ AD = 5

Distance of AC

⇒ AC = √ ((2 – 1)² + (7 – 0)²)

⇒ AC = √ ((1)² + (7)²)

⇒ AC = √ (1 + 49)

⇒ AC = √ 50

Distance of BD

⇒ BD = √ ((–2 – 5)² + (4 – 3)²)

⇒ BD = √ ((–7)² + (1)²)

⇒ BD = √ (49 + 1)

⇒ BD = √ 50

AB = BC = CD = DA = 10 (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.