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的值.
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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的值.
(1)直线l1过点(-3,-1),并且直线l1与l2垂直;
(2)直线l1与直线l2平行,并且坐标原点到l1,l2的距离相等.
(1)直线l1过点(-3,-1),并且直线l1与l2垂直;
(2)直线l1与直线l2平行,并且坐标原点到l1,l2的距离相等.
![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的值.](/uploads/image/z/16061872-40-2.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%A4%E7%9B%B4%E7%BA%BFl1%EF%BC%9Aax-by%2B4%3D0%EF%BC%8Cl2%EF%BC%9A%EF%BC%88a-1%EF%BC%89x%2By%2Bb%3D0%EF%BC%8E%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%EF%BC%8Cb%E7%9A%84%E5%80%BC%EF%BC%8E)
(1)∵l1⊥l2,
∴a(a-1)+(-b)•1=0,即a2-a-b=0①
又点(-3,-1)在l1上,
∴-3a+b+4=0②
由①②得a=2,b=2.
(2)∵l1∥l2,∴
a
b=1-a,∴b=
a
1−a,
故l1和l2的方程可分别表示为:
(a-1)x+y+
4(a−1)
a=0,(a-1)x+y+
a
1−a=0,
又原点到l1与l2的距离相等.
∴4|
a−1
a|=|
a
1−a|,∴a=2或a=
2
3,
∴a=2,b=-2或a=
2
3,b=2.
∴a(a-1)+(-b)•1=0,即a2-a-b=0①
又点(-3,-1)在l1上,
∴-3a+b+4=0②
由①②得a=2,b=2.
(2)∵l1∥l2,∴
a
b=1-a,∴b=
a
1−a,
故l1和l2的方程可分别表示为:
(a-1)x+y+
4(a−1)
a=0,(a-1)x+y+
a
1−a=0,
又原点到l1与l2的距离相等.
∴4|
a−1
a|=|
a
1−a|,∴a=2或a=
2
3,
∴a=2,b=-2或a=
2
3,b=2.
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