Question

Show that the points $(a,a),(-a,-a)$ and $(-a\sqrt{3},a\sqrt{3})$ form an equilateral triangle.

Answer 1

Let the points be represented by $A(a,a),B(-a,-a)$ and $C(-a\sqrt{3},a\sqrt{3})$. Using the distance formula
$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$, we have
$AB=\sqrt{(a+a{)}^2+(a+a{)}^2}$
$=\sqrt{(2a{)}^2+(2a{)}^2}=\sqrt{4a^2+4a^2}=\sqrt{8a^2}=2\sqrt{2}a$
$BC=\sqrt{(-a\sqrt{3}+a{)}^2+(a\sqrt{3}+a{)}^2}=\sqrt{3a^2+a^2-2a^2\sqrt{3}+3a^2+a^2+2a^2\sqrt{3}}$
$=\sqrt{8a^2}=\sqrt{4\times 2a^2}=2\sqrt{2}a$
$CA=\sqrt{(a+a\sqrt{3}{)}^2+(a-a\sqrt{3}{)}^2}=\sqrt{a^2+2a^2\sqrt{3}+3a^2+a^2-2a^2\sqrt{3}+3a^2}$
$={\sqrt{8a}}^2=2\sqrt{2}a$
$\therefore AB=BC=CA=2\sqrt{2}a$.
Since all the sides are equal the points form an equilateral triangle.