Réponses

2013-04-03T18:15:03+02:00

exercice 2

OC = (13;0)+ (0;5) = (13;5)

CD = (8;-5) et CE = (-13;8)

(8;-5) différent de k(-13;8) donc C,D, E non alignés

 

exercice 3 soitM (x;y)

1.Il faut AM = BC => (x+3;y-2) = (3;-7) donc x+3 = 3 => x = 0

                                                                      y-2 = -7 => y -5

    M(0;-5)

2.Il faut CM = k.CE  => (x-5;y+3) = k(5;2) => x-5 = 5k => x = 5 + 5k et x = 0 donc k = -1

                                                                       y+3 = 2k=> y = 2k - 3 => y = -5

M(0;-5)

3.Il faut que B soit le milieu de CM donc

(x+5)/2 = 2 => x = -1

(y-3)/2 = 4 => y = 11

M(-1;11)