C(x;y)
Ca sa fie echilateral AB=AC=BC
[tex]AB= \sqrt{ (xB-xA)^{2} + (yB-yA)^{2} } [/tex]
AB=4=AC=BC(mai jos ca sa scap de radical am ridicat la patrat)
deci
[tex]AC= \sqrt{ (xC-xA)^{2} + (yC-yA)^{2} } [/tex]
[tex]AC= \sqrt{ (x-0)^{2} + (y-1)^{2} }=4 [/tex]
[tex] x^{2}+ y^{2}-2*y+1=16 [/tex]
[tex]BC= \sqrt{ (xC-xb)^{2} + (yC-yB)^{2} } [/tex]
[tex]BC= \sqrt{ (x-4)^{2} + (y-1)^{2} }=4 [/tex]
[tex] x^{2} -8*x+16+ y^{2}-2*y+1=16 [/tex]
AC=BC
[tex] x^{2} + y^{2} -2*y+1= x^{2} -8*x+16+ y^{2} -2*y+1[/tex]
Din ambii termeni dispare [tex]x^{2}+y^{2} -2*y+1[/tex]
-8*x+16=0
8*x=16
x=2
