Me ajudem???? O triangulo equilatero ABC é tal que A(0,2)B(2,0) Obtenha o vertice C.

d² = (Ax - Bx)² + (Ay - By)² = (0 - 2)² + (2 - 0)² = 8

distância AC =

d² = (Ax - Cx)² + (Ay - Cy)² = (0 - x)² + (2 - y)² = x² + y² - 4y + 4 = 8

d² = (Bx - Cx)² + (By - Cy)² = (2 - x)² + (0 - y)² = x² - 4x + 4 + y² = 8

x² + y² - 4y + 4 = x² - 4x + 4 + y² = 8

-4y + 4x = 0

x = y

x² - 4x + 4 + y² = 8

x² - 4x + 4 + x² = 8

2x² - 4x - 4 = 0

x² - 2x - 2 = 0

FORMULA DE BASKHARA:

ax² + bx + c = 0

a = 1

b = -2

c = -2

∆² = b² - 4ac = (-2)² - 4.1.(-2) = 4 + 8 = 12

∆ = 2√3

x' = -b/2a + ∆/2a = 2/2 + 2√3/2 = 1 + √3

x" = -b/2a - ∆/2a = 2/2 - 2√3/2 = 1 - √3

C'(1+√3; 1+√3) e C"(1-√3; 1-√3)