Question

已知直线l₁:ax+2y+6=0和直线l₂:x+(a-1)y+a²-1=0,

（1）试判断l₁与l₂是否平行；

（2）l₁⊥l₂时，求a的值.

Answer 1

（1）a=-1时，l₁∥l₂（2）a=

解析:

（1）方法一  当a=1时，l₁:x+2y+6=0,

l₂:x=0,l₁不平行于l₂;

当a=0时，l₁:y=-3,

l₂:x-y-1=0,l₁不平行于l₂;                                                                                                                         

当a≠1且a≠0时，两直线可化为

l₁:y=--3,l₂:y=-(a+1),

l₁∥l_2，解得a=-1,                                                                                                         

综上可知，a=-1时，l₁∥l₂，否则l₁与l₂不平行.                                                                                              

方法二  由A₁B₂-A₂B₁=0，得a（a-1）-1×2=0，

由A₁C₂-A₂C₁≠0，得a(a²-1)-1×6≠0,                                                                                                           

∴l₁∥l₂                                                                                                                              

a=-1,                                                                                                                               

故当a=-1时，l₁∥l₂，否则l₁与l₂不平行.                                                                                     

（2）方法一  当a=1时，l₁:x+2y+6=0,l₂:x=0,

l₁与l₂不垂直，故a=1不成立.                                                                                                                               当a≠1时，l₁:y=-x-3,

l₂:y=-(a+1),                                                                                                                                                由·=-1a=.                                                                                                                        

方法二  由A₁A₂+B₁B₂=0，得a+2(a-1)=0a=.