初中数学七年级题库5.9:平面直角坐标系中点与象限的关系例题解析
解析:因为,点A(x-1,2-x)在第一象限
所以,\(\displaystyle \left\{ \begin{array}{l}x-1>0\\2-x>0\end{array} \right.\)
求得,\(\displaystyle \left\{ \begin{array}{l}x>1\\x<2\end{array} \right.\)
所以,求得x的取值范围是1<x<2
解析:(1)、因为点P(3a-8,a-1)在y轴上,所以3a-8=0,求得\(\displaystyle a=\frac{8}{3}\)
则:3a-8=0,\(\displaystyle a-1=\frac{5}{3}\)
所以点P的坐标为\(\displaystyle \left( {0,\frac{5}{3}} \right)\)
(2)、因为点P在第二象限,所以\(\displaystyle \left\{ \begin{array}{l}3a-8<0\\a-1>0\end{array} \right.\)
求得:\(\displaystyle \left\{ \begin{array}{l}a<2\frac{2}{3}\\a>1\end{array} \right.\)
又因为:a为整数,则a=2
所以P点坐标为(-2,1)
