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2016-01-01T20:20:47+05:30
Let A(x,y), B(-4,0), C(4,0)
Distance between BC = 8 (Calculate it)
Distance between AB = √ (-4-x)^2 + (0-y)^2
Distance between AC = √ (4-x)^2 + (0-y)^2
Given that the triangle is equilateral. So, AB=BC= AC

AB=AC
√(-4-x)^2+ (0-y)^2 = √ (4-x)^2+ (0-y)^2

(-4-x)^2+ (0-y)^2 = (4-x)^2+ (0-y)^2
(-4-x)^2= (4-x)^2
x^2+8x+16= x^2-8x+16
8x+8x= 16-16
x= 0 ..........(1)

Again, AC = BC

√(4-x)^2+ (0-y)^2 = 8
(4-x)^2+ (0-y)^2= 64
(4-0)^2+ (0-y)^2 { substituting 1}= 64
4^2 + y^2 = 64
y^2 = 64-16=38
y = +√38 or -√38

Therefore, x=0 and y = +√38 or -√38
Third vertex is (0, √38) or (0,-√38)
1 5 1
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