Question

If $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are vertices of an equilateral triangle whose each side is equal to $a$, the prove that ${\left[\begin{array}{ccc}{x}_1& y_1& 2\\ x_2& y_2& 2\\ x_3& y_3& 2\end{array}\right]}^2={3a}^4$.

Answer 1

The area of triangle with vertices

$\left(x_1{y}_1\right)(x_2,y_2)$ and $(x_3,y_3)$ is

$△=\frac{1}{2}$ $\left[\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right]...\left(1\right)$

The area of equilateral triangle with side 'a' is $\frac{\sqrt{3}}{4}{a}^2...\left(2\right)$

We have to prove ${\left[\begin{array}{ccc}{x}_1& y_1& 2\\ x_2& y_2& 2\\ x_3& y_3& 2\end{array}\right]}^2={3a}^4$

From (1) & (2)

$\frac{1}{2}$ $\left[\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right]$ $=\frac{\sqrt{3}}{4}{a}^2$

${△}^2=\frac{1}{4}$ ${\left[\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right]}^2$ $=\frac{3}{16}{a}^4$

$\frac{1}{16}$ ${\left[\begin{array}{ccc}{x}_1& y_1& 2\\ x_2& y_2& 2\\ x_3& y_3& 2\end{array}\right]}^2=\frac{3}{16}{a}^4$

${\left[\begin{array}{ccc}{x}_1& y_1& 2\\ x_2& y_2& 2\\ x_3& y_3& 2\end{array}\right]}^2={3a}^4$.