Show that the following points are collinear.

(3, 7), (6, 5) and (15, −1)

Formula used:

(3, 7), (6, 5) and (15, –1)

Let the points be A (15, –1), B (6, 5) and C (3, 7)

Distance of AB

⇒ AB = √ (6 – 15)² + (5 – (–1))²

⇒ AB = √ (–9)² + (6)²

⇒ AB = √ (81 + 36)

⇒ AB = √ 117 = √ 3 × 3 × 13

⇒ AB = 3√13

Distance of BC

⇒ BC = √ (3 – 6)² + (7 – 5)²

⇒ BC= √ (3)² + (2)²

⇒ BC = √ (9 + 4)

⇒ BC= √ 13

Distance of AC

⇒ AC = √ (3 – 15)² + (7 – (–1))²

⇒ AC = √ (3 – 15)² + (7 + 1)²

⇒ AC= √ (–12)² + (8)²

⇒ AC = √ (144 + 64)

⇒ AC= √ 208 = √ 4 × 4 × 13

⇒ AC = 4√13

i.e. AB + BC = AC

⇒ 3√13 + √13 = 4√13

∴ A, B and C are collinear