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

(9, 5), (6, 0), (−2, −3) and (1, 2)

Formula used:

(9,5), (6, 0), (–2, –3) and (1, 2)

Let A, B, C and D represent the points (9, 5), (6, 0), (–2, –3) and (1, 2)

Distance of AB

⇒ AB = √ ((6 – 9))² + (0 – 5)²)

⇒ AB = √ ((–3)² + (5)²)

⇒ AB = √ (9 + 25)

⇒ AB = √ 34

Distance of BC

⇒ BC= √ ((–2 – 6)² + (–3 – 0)²)

⇒ BC = √ ((–8)² + (–3)²)

⇒ BC = √ (64 + 9)

⇒ BC = √73

Distance of CD

⇒ CD = √ ((1 – (–2))² + (2 – (–3))²)

⇒ CD = √ ((1 + 2)² + (2 + 3))²)

⇒ CD = √ ((3)² + (5)²)

⇒ CD = √ (9 + 25)

⇒ CD = √36

Distance of AD

⇒ AD = √ ((1 – 9))² + (2 – 5)²)

⇒ AD = √ ((–8)² + (–3)²)

⇒ AD = √ (64 + 9)

⇒ AD = √ 73

So, AB = CD = √36 and BC = AD = √73

i.e., The opposite sides are equal. Hence ABCD is a parallelogram.