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.

Let the y coordinate be y

Thus,coordinate of A =(0,y)

Now,distance BC =10

Therefore,AB =AC=BC =10 (equilateral triangle)

$\mathrm{Now},\mathrm{AB}\hspace{0.17em}=10\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{(0-5{)}^{2}+(\mathrm{y}-0{)}^{2}}=10\phantom{\rule{0ex}{0ex}}\Rightarrow 25+{\mathrm{y}}^{2}=100\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{y}}^{2}=100-25\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{y}}^{2}=75\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{y}=\pm 5\sqrt{3}$

That,is A can be both in +y and -y axis.

Coordinate of A = $(0,\pm 5\sqrt{3})$

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