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Question
Verify the following:
(0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
Solution
Let A(0, 7, \[-\]10) , B(1, 6, ) , C(4, 9, \[-\]6) be the vertices of \[\bigtriangleup ABC\]Then,
AB = \[\sqrt{\left( 1 - 0 \right)^2 + \left( 6 - 7 \right)^2 + \left( - 6 + 10 \right)^2}\]
\[= \sqrt{1^2 + \left( - 1 \right)^2 + 4^2}\]
\[ = \sqrt{1 + 1 + 16}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]
BC = \[\sqrt{\left( 4 - 1 \right)^2 + \left( 9 - 6 \right)^2 + \left( - 6 + 6 \right)^2}\]
\[= \sqrt{3^2 + 3^2 + 0}\]
\[ = \sqrt{9 + 9}\]
\[ = \sqrt{18}\]
\[ = 3\sqrt{2}\]
CA= \[\sqrt{\left( 0 - 4 \right)^2 + \left( 7 - 9 \right)^2 + \left( - 10 + 6 \right)^2}\]
\[= \sqrt{16 + 4 + 16}\]
\[ = \sqrt{36}\]
\[ = 6\]
Clearly, AB = BC
Thus, the given points are the vertices of an isosceles triangle.
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