# Show that the Points (A, B, C), (B, C, A) and (C, A, B) Are the Vertices of an Equilateral Triangle. - Mathematics

Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.

#### 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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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 28 Introduction to three dimensional coordinate geometry
Exercise 28.2 | Q 18 | Page 10