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Question
Using distance formula prove that the following points are collinear:
P(0, 7, –7), Q(1, 4, –5) and R(–1, 10, –9)
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Solution
PQ =\[\sqrt{\left( 1 - 0 \right)^2 + \left( 4 - 7 \right)^2 + \left( - 5 + 7 \right)^2}\]
\[= \sqrt{\left( 1 \right)^2 + \left( - 3 \right)^2 + \left( 2 \right)^2}\]
\[ = \sqrt{1 + 9 + 4}\]
\[ = \sqrt{14}\]
QR =\[\sqrt{\left( - 1 - 1 \right)^2 + \left( 10 - 4 \right)^2 + \left( - 9 + 5 \right)^2}\]
\[= \sqrt{\left( - 2 \right)^2 + \left( 6 \right)^2 + \left( - 4 \right)^2}\]
\[ = \sqrt{4 + 36 + 16}\]
\[ = \sqrt{56}\]
\[ = 2\sqrt{14}\]
PR\[= \sqrt{\left( - 1 - 0 \right)^2 + \left( 10 - 7 \right)^2 + \left( - 9 + 7 \right)^2}\]
\[= \sqrt{\left( - 1 \right)^2 + \left( 3 \right)^2 + \left( - 2 \right)^2}\]
\[ = \sqrt{1 + 9 + 4}\]
\[ = \sqrt{14}\]
\[\text{ Here }, PQ + PR = \sqrt{14} + \sqrt{14}\]
\[ = 2\sqrt{14}\]
\[ = QR\]
Hence, the points are collinear.
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