Question

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

• A. $3a^4$
• B. $12a^4$
• C. $9a^4$
• D. $16a^4$

Answer 1

Answer: (B)

Let $\Delta$ be the area of triangle $ABC$. Then,

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

$\Rightarrow 8\Delta =4\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\left|\begin{array}{ccc}{x}_1& y_1& 4\\ x_2& y_2& 4\\ x_3& y_3& 4\end{array}\right|$

$\Rightarrow 64{\Delta }^2={\left|\begin{array}{ccc}{x}_1& y_1& 4\\ x_2& y_2& 4\\ x_3& y_3& 4\end{array}\right|}^2\dots \left(i\right)$

But, the area of an equilateral triangle with each side $a$ is $\frac{\sqrt{3}}{4}{a}^2$.

$\therefore \Delta =\frac{\sqrt{3}}{4}{a}^2\Rightarrow 16{\Delta }^2=3a^4\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

${\left|\begin{array}{ccc}{x}_1& y_1& 4\\ x_2& y_2& 4\\ x_3& y_3& 4\end{array}\right|}^2=12a^4$