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Question
Using distance formula prove that the following points are collinear:
A(3, –5, 1), B(–1, 0, 8) and C(7, –10, –6)
Solution
AB =\[\sqrt{\left( - 1 - 3 \right)^2 + \left( 0 + 5 \right)^2 + \left( 8 - 1 \right)^2}\]
\[= \sqrt{\left( - 4 \right)^2 + \left( 5 \right)^2 + \left( 7 \right)^2}\]
\[ = \sqrt{16 + 25 + 49}\]
\[ = \sqrt{90}\]
\[ = 3\sqrt{10}\]
BC =\[\sqrt{\left( 7 + 1 \right)^2 + \left( - 10 - 0 \right)^2 + \left( - 6 - 8 \right)^2}\]
\[= \sqrt{\left( 8 \right)^2 + \left( - 10 \right)^2 + \left( - 14 \right)^2}\]
\[ = \sqrt{64 + 100 + 196}\]
\[ = \sqrt{360}\]
\[ = 6\sqrt{10}\]
AC =\[\sqrt{\left( 7 - 3 \right)^2 + \left( - 10 + 5 \right)^2 + \left( - 6 - 1 \right)^2}\]
\[= \sqrt{\left( 4 \right)^2 + \left( - 5 \right)^2 + \left( - 7 \right)^2}\]
\[ = \sqrt{16 + 25 + 49}\]
\[ = \sqrt{90}\]
\[ = 3\sqrt{10}\]
\[Here, AB + AC = 3\sqrt{10} + 3\sqrt{10}\]
\[ = 6\sqrt{10}\]
\[ = BC\]
Hence, the points are collinear.
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