Question

An equilateral triangle has each of its sides of length $6$ cm. If $(x_1,y_1);(x_2,y_2)$ & $(x_3,y_3)$ are its vertices, then the value of the determinant ${\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2$ is equal to:

• A. $-972$
• B. $-486$
• C. $486$
• D. $972$

Answer 1

Answer: (D)

$972$

The area of a triangle having $(x_1,y_1);(x_2,y_2)$ & $(x_3,y_3)$ as its vertices is,

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

Now, ${\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2=4A^2$.

The given triangle is equilateral and each side is of $6$ cm.

Then area $\left(A\right)=\frac{\sqrt{3}}{4}\times 6^2$.

Then $4A^2=4\times \frac{3}{4^2}\times 36\times 36=3\times 9\times 36=972$.