Question

已知直線l₁:ax+2y+6=0和直線l₂:x+(a-1)y+a²-1=0,
（1）試判斷l(xiāng)₁與l₂是否平行；
（2）l₁⊥l₂時，求a的值.

Answer 1

（1）a=-1時，l₁∥l₂（2）a=

（1）方法一 當(dāng)a=1時，l₁:x+2y+6=0,
l₂:x=0,l₁不平行于l₂;
當(dāng)a=0時，l₁:y=-3,
l₂:x-y-1=0,l₁不平行于l₂;                                                     
當(dāng)a≠1且a≠0時，兩直線可化為
l₁:y=--3,l₂:y=-(a+1),
l₁∥l_2，解得a="-1,                                               "  
綜上可知，a=-1時，l₁∥l₂，否則l₁與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,                                              
∴l(xiāng)₁∥l₂                                                               
a="-1,                   "                                   
故當(dāng)a=-1時，l₁∥l₂，否則l₁與l₂不平行.                                       
（2）方法一 當(dāng)a=1時，l₁:x+2y+6=0,l₂:x=0,
l₁與l₂不垂直，故a=1不成立.                                                   當(dāng)a≠1時，l₁:y=-x-3,
l₂:y=-(a+1),                                                             由·=-1a=.                                                       
方法二 由A₁A₂+B₁B₂=0，得a+2(a-1)=0a=.