Question

Show that the points A(a, a), B(−a, −a) and $\mathrm{C}(-a\sqrt{3},a\sqrt{3})$ are the vertices of an equilateral triangle.

Answer 1

The given points are $\mathrm{A}\left(\mathrm{a},\mathrm{a}\right),\mathrm{B}\left(-\mathrm{a},-\mathrm{a}\right)\mathrm{and}\mathrm{C}\left(-\mathrm{a}\sqrt{3},\mathrm{a}\sqrt{3}\right)$.

$\mathrm{AB}=\sqrt{{\left[a-\left(-a\right)\right]}^2+{\left[a-\left(-a\right)\right]}^2}=\sqrt{{\left(2a\right)}^2+{\left(2a\right)}^2}=\sqrt{4a^2+4a^2}=\sqrt{8a^2}=2\sqrt{2}a$ units
$\mathrm{BC}=\sqrt{{\left[-a-\left(-a\sqrt{3}\right)\right]}^2+{\left[-a-\left(a\sqrt{3}\right)\right]}^2}$
$=\sqrt{{\left[a\left(\sqrt{3}-1\right)\right]}^2+{\left[a\left(\sqrt{3}+1\right)\right]}^2}$
$=\sqrt{a^2\left(3+1-2\sqrt{3}\right)+a^2\left(3+1+2\sqrt{3}\right)}$
$=\sqrt{a^2\left(4-2\sqrt{3}\right)+a^2\left(4+2\sqrt{3}\right)}=\sqrt{a^2\left(4-2\sqrt{3}+4+2\sqrt{3}\right)}=\sqrt{8a^2}=2\sqrt{2}a$ units
$CA=\sqrt{{\left[a-\left(-a\sqrt{3}\right)\right]}^2+{\left[a-a\sqrt{3}\right]}^2}$
$=\sqrt{{\left[a\left(1+\sqrt{3}\right)\right]}^2+{\left[a\left(1-\sqrt{3}\right)\right]}^2}$
$=\sqrt{a^2\left(1+3+2\sqrt{3}\right)+a^2\left(1+3-2\sqrt{3}\right)}$
$=\sqrt{a^2\left(4+2\sqrt{3}\right)+a^2\left(4-2\sqrt{3}\right)}=\sqrt{a^2\left(4+2\sqrt{3}+4-2\sqrt{3}\right)}=\sqrt{8a^2}=2\sqrt{2}a$ units
∴ AB = BC = CA = $2\sqrt{2}a$ units
Hence, ΔABC is an equilateral triangle and each of its sides is $2\sqrt{2}a$ units.