设所求直线为y=kx+b,而另两条直线的交点是(1,2),2=k+b;按照点到直线距离公式有l3k-1+b|=|5k-1+b|,求|2k+1|=|4k+1l,k=-1/3;k=0;对应b=7/3;b=2,故所求为y=-x/3+7/3或y=2