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