 # If the co-ordinates of the vertices of

Question:

If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are (x1y1),

$\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|^{2}=\frac{3 a^{4}}{4} .$

(x2y2), (x3y3), then

Solution:

We know that, the area of a triangle with vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by

$\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|$

Also, we know the area of an equilateral triangle with side a is given by

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

Hence,

$\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=\frac{\sqrt{3}}{4} a^{2}$

On squaring both the sides, we get

$\Rightarrow \Delta^{2}=\frac{1}{4}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|^{2}=\frac{3}{16} a^{4}$

or $\quad\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=\frac{3 a^{4}}{4}$