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

(5, 9), (5, 16) and (29, 9)

Formula used:

(5, 9), (5, 16) and (29, 9)

Let the points be A (5, 16), B (5, 9) and C (29, 9)

Distance of AB

⇒ AB = √ ((5 – 5)² + (9 – 16)²)

⇒ AB = √ ((0)² + (–7)²)

⇒ AB = √ (0 + 49)

⇒ AB = √49

Distance of BC

⇒ B C= √ ((29 – 5)² + (9 – 9)²)

⇒ BC = √ ((24)² + (0)²)

⇒ BC = √ (576 + 0)

⇒ BC = √576

Distance of AC

⇒ AC = √ ((29 – 5)² + (9 – 16))²)

⇒ AC = √ ((24)² + (–7)²)

⇒ AC = √ (576 + 49)

⇒ AC = √ 625

i.e. AB² + BC²

= (√49)² + (√576)²

= 49 + 576

= 625 = (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.