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

(−7, –5), (−4, 3), (5, 6) and (2, −2)

Formula used:

(–7, –5), (–4, 3), (5, 6) and (2, –2)

Let A, B, C and D represent the points (–7, –5), (–4, 3), (5, 6) and (2, –2)

Distance of AB

⇒ AB = √ ((–4 – (–7)))² + (3 – (–5))²)

⇒ AB = √ ((–4 + 7))² + (3 + 5)²)

⇒ AB = √ ((3)² + (8)²)

⇒ AB = √ (9 + 64)

⇒ AB = √73

Distance of BC

⇒ BC= √ ((5 – (–4))² + (6 – 3)²)

⇒ BC= √ ((5 + 4))² + (6 – 3)²)

⇒ BC = √ ((9)² + (3)²)

⇒ BC = √ (81 + 9)

⇒ BC = √ 90

Distance of CD

⇒ CD = √ ((2 – 5)² + (–2 – 6)²)

⇒ CD = √ ((–3)² + (–8)²)

⇒ CD = √ (9 + 64)

⇒ CD = √73

Distance of AD

⇒ AD = √ ((2 – (–7)))² + (–2 – (–5))²)

⇒ AD = √ ((2 + 7))² + (–2 + 5)²)

⇒ AD = √ ((9)² + (3)²)

⇒ AD = √ (81 + 9)

⇒ AD = √ 90

So, AB = CD = √73 and BC = AD = √90

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