An equilateral triangle has each of its sides | vedclass

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(D) The area $A$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), 	ext{ and } (x_3, y_3)$ is given by $A = \frac{1}{2} \left| \begin{array}{ccc}

Vedclass · Accepted Answer

$972$

Vedclass · Answer

(D) The area $A$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), 	ext{ and } (x_3, y_3)$ is given by $A = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{array} ight|$.Therefore,$\left| \begin{array}{ccc} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{array} ight|^2 = (2A)^2 = 4A^2$.For an equilateral triangle with side length $s = 6 	ext{ cm}$,the area $A$ is $\frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} 	imes 6^2 = \frac{\sqrt{3}}{4} 	imes 36 = 9\sqrt{3}$.Thus,$4A^2 = 4 	imes (9\sqrt{3})^2 = 4 	imes (81 	imes 3) = 4 	imes 243 = 972$.

An equilateral triangle has each of its sides of length $6 \text{ cm}$. If $(x_1, y_1), (x_2, y_2), \text{ and } (x_3, y_3)$ 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:

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