Question

12．不論m取何實(shí)數(shù)，直線（m+2）x-（m+1）y+m+1=0恒過(guò)定點(diǎn)（0，1）．

Answer 1

分析 由直線（m+2）x-（m+1）y+m+1=0變形為m（x-y+1）+（2x-y+1）=0，令$\left\{\begin{array}{l}{x-y+1=0}\\{2x-y+1=0}\end{array}\right.$解得即可．

解答 解：由直線（m+2）x-（m+1）y+m+1=0變形為m（x-y+1）+（2x-y+1）=0，
令$\left\{\begin{array}{l}{x-y+1=0}\\{2x-y+1=0}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=0}\\{y=1}\end{array}\right.$，
∴該直線過(guò)定點(diǎn)（0，1），
故答案為（0，1）．

點(diǎn)評(píng) 本題考查了直線系過(guò)定點(diǎn)問(wèn)題，屬于基礎(chǔ)題．