Let ABC be an equilateral triangle in which coordinates of B and C are (5,0) and (-5,0) respectively . Find the coordinates of the oint A.

By symmetry,the x coordinate is obviously 0.
Let the y coordinate be y
Thus,coordinate of A =(0,y)
Now,distance BC =10
Therefore,AB =AC=BC =10  (equilateral triangle)
$$\begin{gathered} \mathrm{Now}, \mathrm{AB}\hspace{0.17em}=10 \\ \Rightarrow \sqrt{(0-5{)}^2+(\mathrm{y}-0{)}^2}=10 \\ \Rightarrow 25 + {\mathrm{y}}^2=100 \\ \Rightarrow {\mathrm{y}}^2=100-25 \\ \Rightarrow {\mathrm{y}}^2=75 \\ \Rightarrow \mathrm{y}=\pm 5\sqrt{3} \end{gathered}$$


That,is A can be both in +y and -y axis.
Coordinate of A = $(0,\pm 5\sqrt{3})$