设过L1:3X+4Y-2=0与L2:2X+Y+2=0交点的直线方程为3X+4Y-2+m(2X+Y+2)=0,(3+2m)x+(4+m)y+(2m-2)=0该直线与X-2Y+3=0 平行(3+2m)/1=(4+m)/(-2)≠(2m+2)/3左边;-6-4m=4+m, m=-2代入右边(4-2)/(-2)≠(-4+2)/3成立所求的直线方程-x+2y-6=0, x-2y+6=0故得所求的直线方程: x-2y+6=0。