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Question
Verify the following:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
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Solution
Let A(\[-\]1, 2, 1) , B(1, \[-\]7, 8), D(2, \[-\]3, 4) be the vertices of quadrilateral \[\square ABCD\]
\[ = \sqrt{4 + 16 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]
\[BC = \sqrt{\left( 4 - 1 \right)^2 + \left( - 7 + 2 \right)^2 + \left( 8 - 5 \right)^2}\]
\[ = \sqrt{9 + 25 + 9}\]
\[ = \sqrt{43}\]
\[CD = \sqrt{\left( 2 - 4 \right)^2 + \left( - 3 + 7 \right)^2 + \left( 4 - 8 \right)^2}\]
\[ = \sqrt{4 + 16 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]
\[DA = \sqrt{\left( - 1 - 2 \right)^2 + \left( 2 + 3 \right)^2 + \left( 1 - 4 \right)^2}\]
\[ = \sqrt{9 + 25 + 9}\]
\[ = \sqrt{43}\]
\[ \therefore AB = CD \text{ and } BC = DA\]
Since, each pair of opposite sides are equal.
Thus, quadrilateral \[\square ABCD\]is a parallelogram.
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