已知直线的方程为（2+m）x+（1-2m）y+4-3m=0.有一定点（-1,-2）

直线(2+m)x+(1-2m)y+4-3m=0过定点M(-1,-2)
x=0时,(1-2m)y+4-3m=0 => y=-(4-3m)/(1-2m)
即直线与y轴的交点为A[0,-(4-3m)/(1-2m)]
y=0时,(2+m)x+4-3m=0 => x=-(4-3m)/(2+m)
即直线与x轴的交点为B[-(4-3m)/(2+m),0]
欲使AB被定点M平分,则有
-1*2=-(4-3m)/(2+m)+0,
解得m=0
∴直线方程为：2x+y+4=0