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

(−2, 5), (7, 1), (−2, −4) and (7, 0)

Formula used:

(–2, 5), (7, 1), (–2, –4) and (7, 0)

Let A, B, C and D represent the points (–2, 5), (7, 1), (–2, –4) and (7, 0)

Distance of AB

⇒ AB = √ ((7 – (–2)))² + (1 – 5)²)

⇒ AB = √ ((7 + 2))² + (1 – 5)²)

⇒ AB = √ ((9)² + (–4)²)

⇒ AB = √ (81 + 16)

⇒ AB = √ 97

Distance of BC

⇒ BC= √ ((–2 – 7)² + (–4 – 1)²)

⇒ BC = √ ((–9)² + (–5)²)

⇒ BC = √ (81 + 25)

⇒ BC = √106

Distance of CD

⇒ CD = √ ((7 – (–2))² + (0 – (–4))²)

⇒ CD = √ ((7 + 2)² + (0 + 4))²)

⇒ CD = √ ((9)² + (4)²)

⇒ CD = √ (81 + 16)

⇒ CD = √97

Distance of AD

⇒ AD = √ ((7 – (–2))² + (0 – 5)²)

⇒ AD = √ ((7 + 2)² + (0 – 5)²)

⇒ AD = √ ((9)² + (–5)²)

⇒ AD = √ (81 + 25)

⇒ AD = √ 106

So, AB = CD = √97 and BC = AD = √106

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