解：?$(1)$?設(shè)一次函數(shù)表達(dá)式為?$y = kx + b$?
將點(diǎn)?$(1，$??$ -1)$?和?$(2，$??$ 1)$?代入得?$\begin {cases}-1 = k + b\\1 = 2k + b\end {cases}，$?解得?$\begin {cases}{k = 2}\\{b = -3}\end {cases}$?
∴表達(dá)式為?$y = 2x - 3$?
?$(2)$?令?$y = 0，$?則?$2x - 3 = 0，$?解得?$x = \frac 32，$?∴點(diǎn)?$A(\frac 32，$??$ 0)；$?
令?$x = 0，$?則?$y = -3，$?∴點(diǎn)?$B(0，$??$ -3)，$?
?$\triangle AOB$?的面積為?$\frac 12×|\frac 32|×|-3| = \frac 12×\frac 32×3 = \frac 94$?
