Question

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

Answer 1

Here, A(a, b, c), B(b, c, a), C(c, a, b)

AB = $\sqrt{(b-a{)}^2+(c-b{)}^2+\left)\right(a-c{)}^2}$

=$\sqrt{b^2-2ab+a^2+c^2-2bc+b^2+a^2-2ac+c^2}$

=$\sqrt{2a^2+2b^2+2c^2-2ab-2bc-2ca}$

=$\sqrt{2(a^2+b^2+c^2-ab-bc-ca)}$

=$\sqrt{b^2+c^2-2bc+c^2+a^2-2ac+a^2+b^2-2ab}$

BC=$\sqrt{(c-b{)}^2+(a-c{)}^2+(b-a{)}^2}$

=$\sqrt{c^2-2bc+b^2+a^2}-2ac+c^2+b^2-2ab+a^2$

=$\sqrt{2a^2+2b^2+2c^2-2ab}-2bc-2ca$

=$\sqrt{2(a^2+b^2+c^2-ab-bc-ca)}$

CA=$\sqrt{(a-c{)}^2+(b-a{)}^2+(c-b{)}^2}$

=$\sqrt{a^2+c^2-2ac+b^2+a^2-2ab}+b^2+c^2-2bc$

CA=$\sqrt{2a^2+2b^2+2c^2-2ab}-2bc-2ca$

Since, AB = BC = CA, so

$△$ABC is an equilateral triangle.