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`I have a terms written below

$((x_1-x_2)(y_2-y_3)-(x_2-x_3)(y_1-y_2))^2$

Also i am given vertices of equilateral triangle with side $a$ and vertices as $(x_1,y_1),(x_2,y_2),(x_3,y_3)$. I am to write above term in terms of $a$.I tried using distance formula for equilateral triangle and then to eliminate $x_i$ in original term , but it didnot work out.

Any suggestions

Thanks

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    $\begingroup$ Your question is not enough clear. $\endgroup$ – Kanwaljit Singh Feb 5 '17 at 12:25
  • $\begingroup$ It is not a question nor an equation. $\endgroup$ – mathreadler Feb 5 '17 at 12:26
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Area of triangle with vertices $(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})$ is given by

$\triangle=\dfrac{1}{2}\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}$

$=\dfrac{1}{2}\begin{vmatrix}x_{1}-x_{2}&y_{1}-y_{2}&0\\x_{2}-x_{3}&y_{2}-y_{3}&0\\x_{3}&y_{3}&1\end{vmatrix}=\dfrac{1}{2}\left[(x_{1}-x_{2})(y_{2}-y_{3})-(x_{2}-x_{3})(y_{1}-y_{2})\right]$

Thus,$\left((x_{1}-x_{2})(y_{2}-y_{3})-(x_{2}-x_{3})(y_{1}-y_{2})\right)^2=4\times$ $(\triangle)^2=4\times\left(\dfrac{\sqrt{3}}{4}a^2\right)^2=\dfrac{3}{4}a^4$

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  • $\begingroup$ oh that was so simple..thanks a lot $\endgroup$ – Sophie Clad Feb 6 '17 at 1:09
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You can try Napolean's theorem as the theorem states that if three equilateral triangles are drawn on the legs of any triangle (either all drawn inwards or outwards) and the centers of these triangles are connected, the result is another equilateral triangle Here is the formula below with the distances of a point from the three corners of an equilateral triangle, a, b, and c, the length of a side s is given by 3(a^4+b^4+c^4+s^4)=(a^2+b^2+c^2+s^2)^2

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