Question

Fie A(0,1) si B(4,1). Aflati coordonatele punctului C cu proprietatea ca tringhiulABC este echilateral.

Answer 1

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