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# If the coordinates of the vertices of an equilateral triangle with sides of length 'a' are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then ${\begin{vmatrix}x_1 & y_1 &1\\x_2 & y_2 &1\\x_3 & y_3 & 1\end{vmatrix}}^2=\frac{3a^4}{4}.$

Toolbox:
• Area of a triangle ABC is $\Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 &y_2 & 1\\x_3 & y_3 &1\end{vmatrix}$
• Area of an equilateral triangle is $\frac{\sqrt 3}{4}a^2$
Given the sides of the equilateral triangle is 'a'
The area of the triangle $\Delta ABC$ is given by
$\Delta=1/2\begin{vmatrix}x_1 & y_1 &1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
$2\Delta=\begin{vmatrix}x_1 & y_1 &1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}$
Now squaring on both sides,we get
$4{\Delta}^2={\begin{vmatrix}x_1 & y_1 &1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}}^2$-------(1)
We know the area of the equilateral triangle with each side equal to a is $\frac{\sqrt 3}{4}a^2$
Therefore $\Delta=\frac{\sqrt 3}{4}a^2$
Now squaring on both sides,we get
${\Delta}^2=\frac{3}{16}a^4$
Multiply by 4 on both sides
$4{\Delta}^2=\frac{3.4}{16}a^4$
${4\Delta}^2=\frac{3}{4}a^4$--------(2)
From equ(1)&equ(2) we get,
${\begin{vmatrix}x_1 & y_1 &1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}}^2=\frac{3a^4}{4}$
edited Apr 3, 2013