Question

If the co-ordinates of the vertices of an equilateral triangle with sides of length $‘a^{\prime }$ are $\left(x_{1,}{y}_1\right),\left(x_{2,}{y}_2\right),\left(x_{3,}{y}_3\right),$
Then ${\left|\begin{array}{c}{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

Let $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$ be vertices of a triangle then area is $\Delta =\frac{1}{2}\left|\begin{array}{c}{X}_1{Y}_{1}1\\ X_2{Y}_{2}1\\ X_3{Y}_{3}1\end{array}\right|$
squaring both sides
${\Delta }^2=\frac{1}{4}{\left|\begin{array}{c}{X}_1{Y}_{1}1\\ X_2{Y}_{2}1\\ X_3{Y}_{3}1\end{array}\right|}^2$..........(1)
Also if side of an equilateral triangle is $a,$ then its area $=\sqrt{\frac{3}{4}{a}^2}$ …(2)
From equation (1) and (2) we can say

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

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

Hence proved.