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Question
Show that the points A(1, 3, 4), B(–1, 6, 10), C(–7, 4, 7) and D(–5, 1, 1) are the vertices of a rhombus.
Solution
Let A(1,3,4) , B(\[-\]1,6,10) , C(\[-\]7,4,7) and D (\[-\]5,1,1) be the vertices of quadrilateral \[\square ABCD\]
\[AB = \sqrt{\left( - 1 - 1 \right)^2 + \left( 6 - 3 \right)^2 + \left( 10 - 4 \right)^2}\]
\[ = \sqrt{4 + 9 + 36} \]
\[ = \sqrt{49}\]
\[ = 7\]
\[BC = \sqrt{\left( - 7 + 1 \right)^2 + \left( 4 - 6 \right)^2 + \left( 7 - 10 \right)^2}\]
\[ = \sqrt{36 + 4 + 9}\]
\[ = \sqrt{49}\]
\[ = 7\]
\[CD = \sqrt{\left( - 5 + 7 \right)^2 + \left( 1 - 4 \right)^2 + \left( 1 - 7 \right)^2}\]
\[ = \sqrt{4 + 9 + 36}\]
\[ = \sqrt{49}\]
\[ = 7\]
\[DA = \sqrt{\left( 1 + 5 \right)^2 + \left( 3 - 1 \right)^2 + \left( 4 - 1 \right)^2}\]
\[ = \sqrt{36 + 4 + 9}\]
\[ = \sqrt{49}\]
\[ = 7\]
\[ \therefore AB = BC = CD = DA\]
Hence, ABCD is a rhombus.
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