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)
Now, AB=10(0-5)2+(y-0)2=1025 + y2=100 y2=100-25 y2=75y=±53

That,is A can be both in +y and -y axis.
Coordinate of A = (0,±53)

  • 11

this soo easycan't you do it the answer is2 root 5

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answer ..... i also know but how?

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