Show that the following points are collinear.

(4, 1), (5, −2) and (6, −5)

Formula used:

(4, 1), (5, –2) and (6, –5)

Let A (4, 1), B (5, –2) and C (6, –5)

Distance of AB

⇒ AB =√ ((5 – 4)² + (–2 – 1)²)

⇒ AB = √ (1)² + (–3)²

⇒ AB = √ (1 + 9)

⇒ AB = √10

Distance of BC

⇒ BC =√ ((6 – 5)² + (–5 – (–2))²)

⇒ BC = √ (6 – 5)² + (–5 + 2)²)

⇒ BC = √ (1)² + (–3)²

⇒ BC = √ (1 + 9)

⇒ BC = √ 10

Distance of AC

⇒ AC =√ ((6 – 4)² + (–5 – 1)²)

⇒ AC = √ (2)² + (–6)²

⇒ AC = √ (4 + 36)

⇒ AC = √20 =

i.e. AB + BC = AC

⇒ √10 + √10 = √20

Squaring both sides

⇒ (√10)² + (√10)² = (√20)²

⇒ 10 + 10 = 20

∴ A, B and C are collinear.