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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 \[{\begin{vmatrix}x_1 & y_1 &1\\x_2 & y_2 &1\\x_3 & y_3 & 1\end{vmatrix}}^2=\frac{3a^4}{4}.\]

1 Answer

  • 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
Multiply by 4 on both sides
From equ(1)&equ(2) we get,
${\begin{vmatrix}x_1 & y_1 &1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}}^2=\frac{3a^4}{4}$
answered Mar 16, 2013 by vijayalakshmi_ramakrishnans
edited Apr 3, 2013 by sreemathi.v

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