OA = (1,1),OB = (2,3),OC = (m+1,m-1)
A、B、C能构成三角形=> A,B,C 不在同一直线上
AB = OB-OA
= (1,2)
AC = OC -OA
= (m,m-2)
A,B,C 不在同一直线上 =>
AB ≠kAC (where k is a constant)
(1,2)≠(m,m-2)
=> m ≠ (m-2)/2
=>2m≠ m-2
=> m ≠ -2
for m≠ -2 A、B、C能构成三角形