解:??$(3)$??旋轉(zhuǎn)后??$(3,$????$-1)$??
設(shè)直線表達(dá)式為??$y=kx+b$??
將點(diǎn)的坐標(biāo)代入表達(dá)式得
??$\begin {cases}{1=-k+b } \\{-1=3k+b} \end {cases}$??
解得??$k=-\frac {1}{2},$????$b=\frac {1}{2}$??
過(guò)點(diǎn)??$B$??的直線表達(dá)式為??$y=-\frac {1}{2}x+\frac {1}{2}$??