Question

11．畫出方程（x+y-1）$\sqrt{x-y-2}$=0所表示的曲線．

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，進(jìn)一步求出$\left\{\begin{array}{l}{x+y-1=0}\\{x-y-2≥0}\end{array}\right.$的軌跡得答案．

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$\left\{\begin{array}{l}{x+y-1=0}\\{x＞\frac{3}{2}}\end{array}\right.$，∴$\left\{\begin{array}{l}{x+y-1=0}\\{x-y-2＞0}\end{array}\right.$表示無(wú)端點(diǎn)的射線x+y-1=0（x＞$\frac{3}{2}$）．
∴方程（x+y-1）$\sqrt{x-y-2}$=0的曲線是射線x+y-1=0（x＞$\frac{3}{2}$）和直線x-y-2=0．
∴曲線是直線x-y-2=0和直線x+y-1=0在直線x-y-2=0下方的射線．

點(diǎn)評(píng) 本題考查曲線的方程和方程的曲線概念，關(guān)鍵是注意根式有意義，是中檔題．