Question

11．画出方程（x+y-1）$\sqrt{x-y-2}$=0所表示的曲线．

Answer 1

分析 由原方程可得$\left\{\begin{array}{l}{x+y-1=0}\\{x-y-2≥0}\end{array}\right.$或x-y-2=0，进一步求出$\left\{\begin{array}{l}{x+y-1=0}\\{x-y-2≥0}\end{array}\right.$的轨迹得答案．

解答 解：由（x+y-1）$\sqrt{x-y-2}$=0，
得$\left\{\begin{array}{l}{x+y-1=0}\\{x-y-2＞0}\end{array}\right.$或x-y-2=0．
由$\left\{\begin{array}{l}{x+y-1=0}\\{x-y-2＞0}\end{array}\right.$，得$\left\{\begin{array}{l}{x+y-1=0}\\{x＞\frac{3}{2}}\end{array}\right.$，∴$\left\{\begin{array}{l}{x+y-1=0}\\{x-y-2＞0}\end{array}\right.$表示无端点的射线x+y-1=0（x＞$\frac{3}{2}$）．
∴方程（x+y-1）$\sqrt{x-y-2}$=0的曲线是射线x+y-1=0（x＞$\frac{3}{2}$）和直线x-y-2=0．
∴曲线是直线x-y-2=0和直线x+y-1=0在直线x-y-2=0下方的射线．

点评 本题考查曲线的方程和方程的曲线概念，关键是注意根式有意义，是中档题．