Răspuns:
A,B,C - coliniare <=> AB coliniar cu BC
[tex]AB=(xb-xa,yb-ya)=((m+2)-m;m-1) =(2;m-1)\\BC=(xc-xb,yc-yb)=((m+1)-(m+2);3-m)=(-1;3-m)[/tex]
Ca AB sa fie coliniar cu BC=> exista un numar n, astfel incat AB=n*BC=>
[tex]n=\frac{AB}{BC} <=>\frac{2}{-1}=\frac{m-1}{3-m}=>2(3-m)=-1(m-1)=6-2m=-m+1=>-m+2m+1-6=0=>m-5=0=>m=5[/tex]
=>[tex]A(5,1),B(7,5),C(6,3) -[/tex]coliniare
