Réponse:

1) AM ( x-3; y+4) BC(2+5; -1-0) BC(7;0)

donc M : x-3=7 soit x=10

y+4=0 y=-4 M(10; -4)

2) ABMC est un parallelogramme <=> AB=CM

AB(-5-3; 0+4) AB(-8; 4)

CM(x-2; y+1)

AB =CM <=>

x-2 =-8 x=-6

y+1=4 y=3 M(6; 3)

3) 2AM+3MB=0

2AM+3(MA+AB) =0

-AM+3AB=0

AM=3AB

3AB(3*-8; 3*4) 3AB(-24; 12)

AM( x-3; y+4)

x-3=-24 x=-21

y+4=12 y=8 M(-21; 8)