Question

6、若直線l₁：ax+2y+6=0與直線l₂：x+（a-1）y+（a²-1）=0平行且不重合，則a的值是

-1

Answer 1

分析：已知兩條直線：l₁：A₁x+B₁y+C₁=0與l₂：A₂x+B₂y+C₂=0．l₁∥l₂?$\left\{\begin{array}{l}{A}_{1}{B}_{2}-{A}_{2}{B}_{1}=0\\{A}_{1}{C}_{2}-{A}_{2}{C}_{1}≠0\end{array}\right.$，根據(jù)直線l₁：ax+2y+6=0與直線l₂：x+（a-1）y+（a²-1）=0的方程，代入構(gòu)造方程即可得到答案．

解答：解：若直線l₁：ax+2y+6=0與直線l₂：x+（a-1）y+（a²-1）=0平行
則a（a-1）-2=0，即a²-a-2=0
解得：a=2，或a=-1
又∵a=2時，l₁：x+y+3=0與l₂：x+y+3=0重合
故a=-1
故答案為：-1

點評：兩條直線：l₁：A₁x+B₁y+C₁=0與l₂：A₂x+B₂y+C₂=0．l₁∥l₂?$\left\{\begin{array}{l}{A}_{1}{B}_{2}-{A}_{2}{B}_{1}=0\\{A}_{1}{C}_{2}-{A}_{2}{C}_{1}≠0\end{array}\right.$或$\left\{\begin{array}{l}{A}_{1}{B}_{2}-{A}_{2}{B}_{1}=0\\{B}_{1}{C}_{2}-{B}_{2}{C}_{1}≠0\end{array}\right.$