Show that the following points form a right–angled triangle.

(2, −3), (−6, −7) and (−8, −3)

Formula used:

(2, –3), (–6, –7) and (–8, –3)

Let the points be A (2, –3), B (–6, –7) and C (–8, –3)

Distance of AB

⇒ AB = √ ((–6 – 2)² + (–7 – (–3))²)

⇒ AB = √ ((–6 – 2)² + (–7 + 3)²)

⇒ AB = √ ((–8)² + (–4)²)

⇒ AB = √ (64 + 16)

⇒ AB = √ 80

Distance of BC

⇒ B C= √ ((–8 – (–6))² + (–3 – (–7))²)

⇒ BC = √ ((–8 + 6)² + (–3 + 7)²)

⇒ BC = √ ((–2)² + (4)²)

⇒ BC = √ (4 + 16)

⇒ BC = √ 20

Distance of AC

⇒ AC = √ ((–8 – 2)² + (–3 – (–3))²)

⇒ AC = √ ((–8 – 2)² + (–3 + 3)²)

⇒ AC = √ ((–10)² + (0)²)

⇒ AC = √ (100 + 0)

⇒ AC = √ 100

i.e. AB² + BC²

= (√80)² + (√20)²

= 80 + 20

= 100 = (AC)²

Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.