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 determinant (in nearest integer value), $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$ is equal to a then find $\frac{a}{31}.$

Answer 1

We know that area of equilateral triangle is given by

$\Delta =\frac{\sqrt{3}}{4}{a}^2$

Given $a=6$

$\Delta =15.588$ sq.units
We know that area of triangle with vertices $(x_1,y_1);(x_2,y_2)$ and $(x_3,y_3)$ is

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

$\Rightarrow \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=31.17$
And $\frac{a}{31}=1$