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  • If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are (x1, y1), (x2, y2), (x3, y3), then

If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are \left (x_1, y_1 \right ), \left (x_2, y_2 \right ), \left (x_3, y_3 \right ), then \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}

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Area of a triangle with the given vertices will be:


\\\Delta=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|\\$ Given: Length of the sides of the equilateral triangle = a \\Thus, the area $=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$ \\$\therefore \frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$\\ Square both sides; \\$\Rightarrow\left(\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|\right)^{2}=\left(\frac{\sqrt{3}}{4} \mathrm{a}^{2}\right)^{2}$

\\\Rightarrow \frac{1}{4}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|^{2}=\frac{3}{16} \mathrm{a}^{4}$

\\\begin{aligned} &\Rightarrow\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3} \end{array}\right|^{2}=\frac{3}{4} \mathrm{a}^{4}\\ &\text { Hence Proved } \end{aligned}

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