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

(15, 20), (−3, 12), (−11, −6) and (7, 2)

Formula used:

(15, 20), (–3, 12), (–11, –6) and (7, 2)

Let the vertices be taken as A (15, 20), B (–3, 12), C (–11, –6) and D (7, 2).

Distance of AB

⇒ AB = √ ((–3 – 15)² + (12 – 20)²)

⇒ AB = √ ((–18)² + (–8)²)

⇒ AB = √ (324 + 64)

⇒ AB = √ 388

Distance of BC

⇒ BC= √ ((–11 –(–3))² + (–6 – 12)²)

⇒ BC = √ (–11 + 3)² + (–6 – 12)²)

⇒ BC = √ ((–8)² + (–18)²)

⇒ BC = √ (64 + 324)

⇒ BC = √ 388

Distance of CD

⇒ CD = √ ((7 – (–11))² + (2 – (–6))²)

⇒ CD = √ ((7 + 11)² + (2 + 6)²)

⇒ CD = √ ((18)² + (8)²)

⇒ CD = √ (324 + 64)

⇒ CD = √388

Distance of AD

⇒ AD = √ ((7 – 15))² + (2 – 20)²)

⇒ AD = √ ((–8)² +(–18)²)

⇒ AD = √ (64 + 324)

⇒ AD = √ 388

Distance of AC

⇒ AC = √ ((–11 – 15)² + (–6 – 20)²)

⇒ AC = √ ((–26)² +(–26)²)

⇒ AC = √ (676 + 676)

⇒ AC = √ 1352

Distance of BD

⇒ BD = √ ((7 – (–3))² + (2 – 12)²)

⇒ BD = √ ((7 + 3))² + (2 – 12)²)

⇒ BD = √ ((10)² + (–10)²)

⇒ BD = √ (100 + 100)

⇒ BD = √ 200

AB = BC = CD = DA = √388 (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.