* Demostrar que el triangulo ABC con vértice

A = (3, 3), B = (-3, -3), C =(-3√3, 3√3) es equilatero.

PAPB = √[(-3-3)² + (-3-33)²]

PAPB = √36+36

PAPB = 8.48

PAPC = √[(-3√3 -3)² + (3 √3-3)²]

PAPC = √67.17 + 4.82

PAPC = 8.48

PBPC = √[(-3√3 – (-3))² + (3√3 – (-3))²]

PBPC = 8.48