# Using distance formula prove that the following points are collinear : A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

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Using distance formula prove that the following points are collinear : A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

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Given: A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

To prove: Points A, B and C are collinear

Formula used:

Points A, B and C are collinear if AB + BC = AC or AB + AC = BC or AC + BC = AB

The distance between any two points (a, b, c) and (m, n, o) is given by,

$\sqrt{(a-m)^2+(b-n)^2+(c-0)^2}$

Therefore,

Distance between A(4, -3, -1) and B(5, -7, 6) is AB,

$\sqrt{(4-5)^2+(-3-(-7)^2+(-1-6)^2}$

$\sqrt{(-1)^2+(-4)^2+(-7)^2}$

$\sqrt{1+16+49}$

$\sqrt{66}$

Distance between B(5, -7, 6) and C(3, 1, -8) is BC,

$\sqrt{(5-3)^2+(-7-1)^2+(6-(-8))^2}$

$\sqrt{(-2)^2+(-8)^2+(14)^2}$

$\sqrt{4+64+196}$

$\sqrt{264}$

$2\sqrt{66}$

Distance between A(4, -3, -1) and C(3, 1, -8) is AC,

$\sqrt{(4-3)^2+(-3-1)^2+(-1-(-8))^2}$

$\sqrt{(1)^2+(-4)^2+(7)^2}$

$\sqrt{1+16+49}$

$\sqrt{66}$

Clearly,

AB + AC

$\sqrt{66}+\sqrt{66}$

$2\sqrt{66}$

= BC

Hence, A, B and C are collinear

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