由{■(y=kx+1"," @2x+y"-" 8=0"," )┤得B的横坐标xB=7/(k+2).

　　∵点M平分线段AB,∴7/(3k"-" 1)+7/(k+2)=0,解得k=-1/4.故所求的直线方程为x+4y-4=0.

解法二设所求直线与l1,l2分别交于A,B两点,且设A(3m-10,m),B(a,8-2a).

　　∵M为线段AB的中点,∴{■((3m"-" 10+a)/2=0"," @(m+8"-" 2a)/2=1"," )┤

　　解得{■(a=4"," @m=2"," )┤∴A(-4,2),B(4,0),

　　∴直线AB即所求直线的方程为x+4y-4=0.