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Question
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
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Solution
Let A(a,b,c) , B(b,c,a) and C(c,a,b) be the vertices of \[\bigtriangleup ABC\] Then,AB =\[\sqrt{\left( b - a \right)^2 + \left( c - b \right)^2 + \left( a - c \right)^2}\]
\[= \sqrt{b^2 - 2ab + a^2 + c^2 - 2bc + b^2 + a^2 - 2ca + c^2}\]
\[ = \sqrt{2 a^2 + 2 b^2 + 2 c^2 - 2ab - 2bc - 2ca}\]
\[ = \sqrt{2\left( a^2 + b^2 + c^2 - ab - bc - ca \right)}\]
BC =\[\sqrt{\left( c - b \right)^2 + \left( a - c \right)^2 + \left( b - a \right)^2}\]
\[= \sqrt{c^2 - 2bc + b^2 + a^2 - 2ca + c^2 + b^2 - 2ab + a^2}\]
\[ = \sqrt{2 a^2 + 2 b^2 + 2 c^2 - 2ab - 2bc - 2ca}\]
\[ = \sqrt{2\left( a^2 + b^2 + c^2 - ab - bc - ca \right)}\]
CA =\[\sqrt{\left( a - c \right)^2 + \left( b - a \right)^2 + \left( c - b \right)^2}\]
\[= \sqrt{a^2 - 2ca + c^2 + b^2 - 2ab + a^2 + c^2 - 2bc + b^2}\]
\[ = \sqrt{2 a^2 + 2 b^2 + 2 c^2 - 2ab - 2bc - 2ca}\]
\[ = \sqrt{2\left( a^2 + b^2 + c^2 - ab - bc - ca \right)}\]
\[\therefore\]AB = BC = CA
Therefore,\[\bigtriangleup ABC\] is an equilateral triangle.
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