Question

A(4,-3)
B(2,8)
C(-6,9)
rA=
rB=
rC=
rG(triungiul ABC)
AB->
|AB->|
|AC->|
BC->
|BC->|
M? (mijl AB)
N(mijlAC)
PM(P mijl BC)
rG(triunghiul MNP) ​

Answer 1

Răspuns:

AB= AO+OB=OB-OA=rB-rA=-3i-j-j=-3i-2j

AC=rC-rA=6i+5j-j=6i+4j=-2(-3i-2j)=-2AB

deci exista -2∈R, asa fel incat AC=-2AB⇔A,B,C coliniare