已知两直线L1:ax-by+4=0,L2:(a-1)x+y+b=0,求分别满足下列条件的a,b (1)直线L1过点(-3,-1),并且直线L1与L2垂直 L1过点(-3,-1),则-3a+b+4=0,得b=3a-4,L1:ax-(3a-4)y+4=0直线L1与直线L2垂直,则a(a-1)-(3a-4)*1=0,得a

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已知两直线L1:ax-by+4=0,L2:(a-1)x+y+b=0,求分别满足下列条件的a,b (1)直线L1过点(-3,-1),并且直线L1与L2垂直 L1过点(-3,-1),则-3a+b+4=0,得b=3a-4,L1:ax-(3a-4)y+4=0直线L1与直线L2垂直,则a(a-1)-(3a-4)*1=0,得a
已知两直线L1:ax-by+4=0,L2:(a-1)x+y+b=0,求分别满足下列条件的a,b
(1)直线L1过点(-3,-1),并且直线L1与L2垂直

L1过点(-3,-1),
则-3a+b+4=0,
得b=3a-4,
L1:ax-(3a-4)y+4=0
直线L1与直线L2垂直,
则a(a-1)-(3a-4)*1=0,
得a=2,
此时b=2
a(a-1)-(3a-4)*1=0,这个式子怎么来的?详解

已知两直线L1:ax-by+4=0,L2:(a-1)x+y+b=0,求分别满足下列条件的a,b (1)直线L1过点(-3,-1),并且直线L1与L2垂直 L1过点(-3,-1),则-3a+b+4=0,得b=3a-4,L1:ax-(3a-4)y+4=0直线L1与直线L2垂直,则a(a-1)-(3a-4)*1=0,得a
l1与l2垂直啊,所以两条直线的斜率互为负倒数,化简一下就得到你问的那个公式了

俩直线垂直,有个公式是关于一般式的 x y前对应相前的系数乘积相等的 a1a2=b1b2

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