Bonjour,
4)a)
AB = |zB - zA| = |-5 + 3i| = √[(-5)² + 3²] = √34
AC = |zC - zA| = |-2 - 2i| = √[(-2)² + (-2)²] = √8
BC = |zC - zB| = |3 - 5i| =√[(-3)² + (-5)²] = √34
b) On constate AB = BC
Donc ABC est isocèle en B
5) AB² = |zB - zA|² = |8 - i| = 64 + 1 = 65
AC² = |zC - zA|² = |2 - 4i|² = 4 + 16 = 20
BC² = |zC - zB|² = |-6 -3i| = 36 + 9 = 45
⇒ AB² = AC² + BC²
⇒ ABC rectangle en C
