已知两直线L1：ax-by+4=0和L2:(a-1)x+y+b=0.若L1 // L2且坐标原点到两直线的距离相等,求a,b的值?

若L1 // L2,则有：a/b=(a-1)/(-1)
a/b=1-a
a/(1-a)=b
坐标原点到两直线的距离相等,则有：
4/√(a^2+b^2)=|b|/√[(a-1)^2+1]
16/(a^2+b^2)=b^2/[(a-1)^2+1]
16[(a-1)^2+1)]=a^2/(a-1)^2*[a^2+a^2/(a-1)^2]
16[(a-1)^2+1]=a^2/(a-1)^2*a^2/(a-1)^2*[(a-1)^2+1)]
16=a^4/(a-1)^4
a/(a-1)=(+/-)2
(1)a=2.b=2/(1-2)=-2
(2)a=2/3,b=2/3/(1-2/3)=-2