Question

Show that the points $(a,a),(-a,-a)$ and $(-\sqrt{3a},\sqrt{3a})$ are the vertices of an equilateral triangle. Also, find its area.

Answer 1

Let $A(a,a),B(-a,-a)andC(-\sqrt{3}a,\sqrt{3}a)$ be the given points. Then, we have
$AB=\sqrt{{(-a-a)}^2+{(-a-a)}^2}=\sqrt{{4a}^2+{4a}^2}=2\sqrt{2}a$
$BC=\sqrt{{(-\sqrt{3}a+a)}^2+{(\sqrt{3}a+a)}^2}$
$BC=\sqrt{a^2{(1-\sqrt{3})}^2+a^2{(\sqrt{3}+1)}^2}$
$BC=a\sqrt{{(1-\sqrt{3})}^2+{(1+\sqrt{3})}^2}$
$BC=a\sqrt{1+3-2\sqrt{3}+1+3+2\sqrt{3}}=a\sqrt{8}=2\sqrt{2}a$
$AC=\sqrt{{(-\sqrt{3}a-a)}^2+{(\sqrt{3}a-a)}^2}$
$AC=\sqrt{a^2{(\sqrt{3}+1)}^2+a^2{(\sqrt{3}-1)}^2}$
$AC=a\sqrt{{(\sqrt{3}+1)}^2+{\left(\sqrt{3}\right)-1)}^2}$
$AC=a\sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}=a\sqrt{8}=2\sqrt{2}a$
Clearly, we have
AB=BC=AC
Hence, the triangle ABC formed by the given points is an equilateral triangle.
Now,
$Areaof△ABC=\frac{\sqrt{3}}{4}{\left(Side\right)}^2$
$\Rightarrow Areaof△ABC=\frac{\sqrt{3}}{4}\times {AB}^2$
$\Rightarrow Areaof△ABC=\frac{\sqrt{3}}{4}\times {\left(2\sqrt{2a}\right)}^{2}sq.units=2\sqrt{3}{a}^{2}sq.units$