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)
