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
