解：?$(1)$?令?$y=0，$?則?$0=\frac 12x+2，$?解得?$x=-4，$?∴點(diǎn)?$A(-4，$??$0)$?
令?$x=0，$?則?$y=2，$?∴點(diǎn)?$B(0，$??$2)$?
?$ (2)$?設(shè)點(diǎn)?$P(x，$??$y)$?
?$S_{\triangle BOP}=\frac 12×2×|x|=|x|，$?
?$S_{\triangle COP}=\frac 12×2×y=y$?
由?$|x|=y$?及?$y=\frac 12x+2$?
解得?$\begin {cases}x=4\\y =4\end {cases}$?或?$\begin {cases}x=-\frac 43\\y =\frac 43\end {cases}$?
∴點(diǎn)?$P $?的坐標(biāo)為?$(4，$??$4)$?或?$(-\frac 43，$??$\frac 43)$?
