ABC is an equilateral triangle with coordinate of and c as b=(-3,0) c=(3,0).Find the coordinates of the vertex A(assuming A lies on the positive side of y axis
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As in equilateral triangle ABC
AB=BC=AC
First
AB=AC
(x+3)^2+y^2=(x-3)^2+y^2
x^2+6x+9+y^2=x^2-6x+9+y^2
6x=-6x
12x=0
x=0
Secondly
AB=BC
(0+3)^2+y^2=(-3-3)^2
9+y^2=36
y^2=25
y=5 {ignoring -ve value}
The the point A is (0,5)
AB=BC=AC
First
AB=AC
(x+3)^2+y^2=(x-3)^2+y^2
x^2+6x+9+y^2=x^2-6x+9+y^2
6x=-6x
12x=0
x=0
Secondly
AB=BC
(0+3)^2+y^2=(-3-3)^2
9+y^2=36
y^2=25
y=5 {ignoring -ve value}
The the point A is (0,5)
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