Q. 2 E3.5( 2 Votes )

# Show that the fol

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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