两直线l1:ax-by+b=0和l2(a-1)x+y+b=0,若l1‖l2且l1与l2的距离为根号2/2,求a,b的值
问题描述：
两直线l1:ax-by+b=0和l2(a-1)x+y+b=0,若l1‖l2且l1与l2的距离为根号2/2,求a,b的值
答
l1过A(0,1)l1 // l2,a ：a-1 = -b :1,a = b - ab,a = b/(1+b)A(0,1)与l2的距离为d = |1 + b|/√[(a-1)^2 + 1] = √2/2(1+b)^2/√[(a-1)^2 + 1] = 1/22(1+b)^2 = [b/(1+b) -1]^2 +1b = -1+√2 (a = (2-√2)/2) 或b = ...