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

(3, −5), (−5, −4), (7, 10) and (15, 9)

Formula used:

(3, –5), (–5, –4), (7, 10) and (15, 9)

Let A, B, C and D represent the points (3, –5), (–5, –4), (7, 10) and (15, 9)

Distance of AB

⇒ AB = √ ((–5 – 3)² + ((–4 – (–5))²)

⇒ AB = √ ((–5 – 3))² + (–4 + 5)²)

⇒ AB = √ ((–8)² + (1)²)

⇒ AB = √ (64 + 1)

⇒ AB = √ 65

Distance of BC

⇒ BC= √ ((7 – (–5))² + (10 – (–4))²)

⇒ BC= √ ((7 + 5)² + (10 + 4)²)

⇒ BC = √ ((12)² + (14)²)

⇒ BC = √ (144 + 196)

⇒ BC = √ 340

Distance of CD

⇒ CD = √ ((15 – 7)² + (9 – 10)²)

⇒ CD = √ ((8)² + (–1)²)

⇒ CD = √ (64 + 1)

⇒ CD = √65

Distance of AD

⇒ AD = √ ((15 – 3)² + (9 – (–5))²)

⇒ AD = √ ((15 – 3)² + (9 + 5)²)

⇒ AD = √ ((12)² + (14)²)

⇒ AD = √ (144 + 196)

⇒ AD = √ 340

So, AB = CD = √65 and BC = AD = √340

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