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

(−11, 13), (−3, −1) and (4, 3)

Formula used:

(–11, 13), (–3, –1) and (4, 3)

Let the points be A (–11, 13), B (–3, –1) and C (4, 3)

Distance of AB

⇒ AB = √ ((–3 – (–11))² + (–1 – 13)²)

⇒ AB = √ ((–3 + 11)² + (–1 – 13)²)

⇒ AB = √ ((8)² + (–14)²)

⇒ AB = √ (64 + 196)

⇒ AB = √260

Distance of BC

⇒ B C= √ ((4 – (–3))² + (3 – (–1))²)

⇒ BC = √ ((4 + 3)² + (3 + 1)²)

⇒ BC = √ ((7)² + (4)²)

⇒ BC = √ (49 + 16)

⇒ BC = √ 65

Distance of AC

⇒ AC = √ ((4 – (–11))² + (3 – 13))²)

⇒ AC = √ ((4 + 11)² + (3 – 13)²)

⇒ AC = √ ((15)² + (–10)²)

⇒ AC = √ (225 + 100)

⇒ AC = √ 325

i.e. AB² + BC²

= (√260)² + (√65)²

= 260 + 65

= 325 = (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.