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

(0, 0), (a, 0) and (0, b)

Formula used:

(0, 0), (a, 0) and (0, b)

Let the points be A (0, 0), B (a, 0) and C (0, b)

Distance of AB

⇒ AB = √ ((a – 0)² + (0 – 0)²)

⇒ AB = √ ((a)² + (0)²

⇒ AB = √ a²

Distance of BC

⇒ BC = √ ((0 – a)² + (b – 0)²)

⇒ BC = √ ((–a)² + (b)²

⇒ BC = √ a² + b²

Distance of AC

⇒ AC = √ ((0 – 0)² + (b – 0)²)

⇒ AC = √ ((0)² + (b)²

⇒ AC = √ b²

i.e. AB² + AC²

= (√a²)² + (√b²)²

= a² + b² = BC²

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.