Answer :

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.


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