Question

If the coordinates of the vertices of an equilateral triangle with sides of length 'a' are $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$, then
${\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2=\frac{3a^4}{4}$

Answer 1

Since, we know that area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is given by
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$
$\Rightarrow {\Delta }^2=\frac{1}{4}{\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2$ ....(i)
We know that, area of an equilateral triangle with side a,
$\Delta =\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)a^2$
$\Rightarrow {\Delta }^2=\frac{3}{16}{a}^4$ ...... (ii)
From Eqs. (i) and (ii), $\frac{3}{16}{a}^4=\frac{1}{4}{\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2$
$\Rightarrow {\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2=\frac{3}{4}{a}^4$
Hence proved.