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](/uploads/image/z/8568373-13-3.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%A4%E7%9B%B4%E7%BA%BFL1%EF%BC%9Aax-by%2B4%3D0%2CL2%EF%BC%9A%EF%BC%88a-1%EF%BC%89x%2By%2Bb%3D0%2C%E6%B1%82%E5%88%86%E5%88%AB%E6%BB%A1%E8%B6%B3%E4%B8%8B%E5%88%97%E6%9D%A1%E4%BB%B6%E7%9A%84a%2Cb++%281%29%E7%9B%B4%E7%BA%BFL1%E8%BF%87%E7%82%B9%EF%BC%88-3%2C-1%EF%BC%89%2C%E5%B9%B6%E4%B8%94%E7%9B%B4%E7%BA%BFL1%E4%B8%8EL2%E5%9E%82%E7%9B%B4++L1%E8%BF%87%E7%82%B9%28-3%2C-1%29%2C%E5%88%99-3a%2Bb%2B4%3D0%2C%E5%BE%97b%3D3a-4%2CL1%3Aax-%283a-4%29y%2B4%3D0%E7%9B%B4%E7%BA%BFL1%E4%B8%8E%E7%9B%B4%E7%BA%BFL2%E5%9E%82%E7%9B%B4%2C%E5%88%99a%28a-1%29-%283a-4%29%2A1%3D0%2C%E5%BE%97a)
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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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