(1+2m)x+(1-m)y-3-3m=0
x+2mx+y-my-3-3m=0
x+y+(2x-y-3)m-3=0
恒過定點,則與m的取值無關(guān),所以可得含m項的系數(shù)為0,可得：
2x-y-3=0 ····················1
此時原方程可化為：
x+y-3=0·····················2
聯(lián)立1、2兩式得：
x=2,y=1
綜上可得直線恒過定點(2,1).