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

(0, 0), (7, 3), (10, 6) and (3, 3)

Formula used:

(0,0) (7, 3), (10, 6) and (3, 3)

Let A, B, C and D represent the points (0, 0), (7, 3), (10, 6) and (3, 3)

Distance of AB

⇒ AB = √ ((7 – 0))² + (3 – 0)²)

⇒ AB = √ ((7)² + (3)²)

⇒ AB = √ (49 + 9)

⇒ AB = √ 58

Distance of BC

⇒ BC= √ ((10 – 7)² + (6 – 3)²)

⇒ BC = √ ((3)² + (3)²)

⇒ BC = √ (9 + 9)

⇒ BC = √18

Distance of CD

⇒ CD = √ ((3 – 10)² + (3 – 6)²)

⇒ CD = √ ((–7)² + (–3)²)

⇒ CD = √ (49 + 9)

⇒ CD = √58

Distance of AD

⇒ AD = √ ((3 – 0))² + (3 – 0)²)

⇒ AD = √ ((3)² + (3)²)

⇒ AD = √ (9 + 9)

⇒ AD = √18

So, AB = CD = √58 and BC = AD = √18

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