# Using distance formula prove that the following points are collinear : A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

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Using distance formula prove that the following points are collinear :

A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

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Given: A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

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 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(3, -5, 1) and B(-1, 0, 8) is AB,

$\sqrt{(3-(-1))^2+(-5-0)^2+(1-8)^2}$

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

$\sqrt{16+25+49}$

$\sqrt{90}$

$3\sqrt{10}$

Distance between B(-1, 0, 8) and C(7, -10, -6) is BC,

$\sqrt{(-1-7)^2+(0-(-10))^2+(8-(-6))^2}$

$\sqrt{(-8)^2+(10)^2+(14)^2}$

$\sqrt{64+100+196}$

$\sqrt{360}$

$6\sqrt{10}$

Distance between A(3, -5, 1) and C(7, -10, -6) is AC,

$\sqrt{(3-7)^2+(-5-(-10))^2+(-1-(-6))^2}$

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

$\sqrt{16+25+49}$

$\sqrt{90}$

$3\sqrt{10}$

Clearly,

AB + AC

$3\sqrt{10}+3\sqrt{10}$

$6\sqrt{10}$

= BC

Hence, A, B and C are collinear

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