$解:①若點(diǎn)P移動(dòng)了$
$則b≠-1,P'(a-b+1,b)$
$由題有\(zhòng)begin{cases}{ a+2=b }\ \\ { a-b+1=-b } \end{cases}$
$或\begin{cases}{ a+2=-b }\ \\ { a-b+1=b } \end{cases} $
$解得\begin{cases}{ a=-1 }\ \\ {\ b=1} \end{cases}或\begin{cases}{ a=-\frac {5}{3} }\ \\ { b=-\frac {1}{3} } \end{cases}$
$∴P(1,1)或(\frac {1}{3},-\frac {1}{3})$
$②當(dāng)P沒(méi)有移動(dòng)時(shí),b=-1$
$∴a+2=±1$
$∴P(-1,-1)或(1,-1)$
$綜上,P(-1,-1)或(1,-1)或(1,1)或(\frac {1}{3},-\frac {1}{3})$
