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

(10, 0), (18, 0) and (10, 15)

Formula used:

(10, 0), (18, 0) and (10, 15)

Let the points be A (10, 15), B (10, 0) and C (18, 0)

Distance of AB

⇒ AB = √ ((10 – 10))² + (0 – 15)²)

⇒ AB = √ ((0)² + (–15)²)

⇒ AB = √ (0 + 225)

⇒ AB = √225

Distance of BC

⇒ B C= √ ((18 – 10)² + (0 – 0)²)

⇒ BC = √ ((8)² + (0)²)

⇒ BC = √ (64 + 0)

⇒ BC = √ 64

Distance of AC

⇒ AC = √ ((18 – 10)² + (0 – 15))²)

⇒ AC = √ ((8)² + (–15)²)

⇒ AC = √ (64 + 225)

⇒ AC = √289

i.e. AB² + BC²

= (√225)² + (√64)²

= 225 + 64

= 289 = (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.