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