Simone如图，直线$AB:y=\frac{{\sqrt{3}}}{3}x+b$，其中$B\left(-1,0\right)$，点$A$横坐标为$4$，点$C\left(3,0\right)$，直线$FG$垂直平分线段$BC$.
的有关信息介绍如下：
$(1)$把点$B\left(-1,0\right)$代入直线$AB:y=\frac{{\sqrt{3}}}{3}x+b$中，得$\frac{{\sqrt{3}}}{3}\times \left(-1\right)+b=0$，$\therefore b=\frac{{\sqrt{3}}}{3}$，$\therefore AB$的函数关系式：$y=\frac{{\sqrt{3}}}{3}x+\frac{{\sqrt{3}}}{3}$，当$x=4$时$y=\frac{5}{3}\sqrt{3}$，$\therefore $点$A$的坐标为$(4$，$\frac{5}{3}\sqrt{3})$，设$AC$的函数关系式为：$y=mx+n$，把$A(4$，$\frac{5}{3}\sqrt{3})$，$C\left(3,0\right)$两点坐标代入，得$\left\{\begin{array}{l}{4k+b=\frac{5}{3}\sqrt{3}}\\{3k+b=0}\end{array}\right.$，$\therefore k=\frac{5}{3}\sqrt{3}$，$b=5\sqrt{3}$，$\therefore $直线$AC$的函数表达式为：$y=\frac{5}{3}\sqrt{3}x-5\sqrt{3}$；$(2)$过点$A$作$AF\bot x$轴于点$F$，由点$A(4$，$\frac{5}{3}\sqrt{3})$，$B\left(-1,0\right)$得，$\tan \angle ABF=\frac{AF}{BF}=\frac{\frac{5}{3}\sqrt{3}}{5}=\frac{\sqrt{3}}{3}$，$\therefore \angle ABF=30^{\circ}$，由翻折得$\angle C'BC=60^{\circ}$，$BC=BC'=4$，在$Rt\triangle BGC'$中，$BG=2$，$GC'=2\sqrt{3}$，$\therefore $点$C'$的坐标为$(1$，$2\sqrt{3})$，在$Rt\triangle BDG$中，$\angle DBG=30^{\circ}$，$\therefore DG=\frac{BG}{\sqrt{3}}=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}$，$\therefore $点$D$坐标为$(1$，$\frac{2}{3}\sqrt{3})$；$(3)\because $直线$FG$垂直平分线段$BC$，在$FG$上取（2）中的$C'$，$\therefore BC=BC'$，$QB=QC$，由（2）知$\angle CBC'=60^{\circ}$，$\angle CBD=\angle C'BD$，$\therefore \triangle BCC'$是等边三角形，$AB$垂直平分$CC'$，①点$P$在$x$轴上方，如图$2$，$\because \triangle BCC'$和$\triangle PQC$都是等边三角形，$\therefore CC'=CB$，$CP=CQ$，$\angle PCQ=\angle BCC'=60^{\circ}$，$\therefore \angle PCC'=\angle QCB$，$\therefore \triangle PCC'$≌$\triangle QCB\left(SAS\right)$，$\therefore PC'=QB$，$\because QB=QC$，$CP=CQ$，$\therefore PC=PC'$，$\because BC=BC'$，$\therefore PB$垂直平分$CC'$，$\therefore $点$P$与点$A$重合，$\therefore PA$的函数关系式就是$AB$的函数关系式：$y=\frac{{\sqrt{3}}}{3}x+\frac{{\sqrt{3}}}{3}$；②点$P$在$x$轴下方，如图$3$，$\because \triangle BCC'$和$\triangle PQC$都是等边三角形，$\therefore CC'=CB$，$CP=CQ$，$\angle PCQ=\angle BCC'=60^{\circ}$，$\therefore \angle PCCB\angle QCBC'$，$\therefore \triangle PCB$≌$\triangle QCC'\left(SAS\right)$，$\therefore \angle CBP=\angle CC'Q=30^{\circ}$，在$Rt\triangle BHO$中，$BO=1$，$\therefore HO=\tan \angle OBH\cdot BO=\frac{\sqrt{3}}{3}$，$\therefore $点$H$的坐标为$(0$，$-\frac{\sqrt{3}}{3})$，设$BP$的函数关系式为$y=mx+n$，把点$B\left(-1,0\right)$，$H(0$，$-\frac{\sqrt{3}}{3})$代入得，$\left\{\begin{array}{l}{-k+b=0}\\{b=-\frac{\sqrt{3}}{3}}\end{array}\right.$，解得$k=-\frac{\sqrt{3}}{3}$，$b=-\frac{\sqrt{3}}{3}$，$\therefore PA$的函数关系式$y=-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}$，综上所述$PA$的函数关系式$y=\frac{\sqrt{3}}{3}x+\frac{\sqrt{3}}{3}$或$y=-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}$.
$(1)$把点$B\left(-1,0\right)$代入直线$AB:y=\frac{{\sqrt{3}}}{3}x+b$中，得$\frac{{\sqrt{3}}}{3}\times \left(-1\right)+b=0$，$\therefore b=\frac{{\sqrt{3}}}{3}$，$\therefore AB$的函数关系式：$y=\frac{{\sqrt{3}}}{3}x+\frac{{\sqrt{3}}}{3}$，当$x=4$时$y=\frac{5}{3}\sqrt{3}$，$\therefore $点$A$的坐标为$(4$，$\frac{5}{3}\sqrt{3})$，设$AC$的函数关系式为：$y=mx+n$，把$A(4$，$\frac{5}{3}\sqrt{3})$，$C\left(3,0\right)$两点坐标代入，得$\left\{\begin{array}{l}{4k+b=\frac{5}{3}\sqrt{3}}\\{3k+b=0}\end{array}\right.$，$\therefore k=\frac{5}{3}\sqrt{3}$，$b=5\sqrt{3}$，$\therefore $直线$AC$的函数表达式为：$y=\frac{5}{3}\sqrt{3}x-5\sqrt{3}$；$(2)$过点$A$作$AF\bot x$轴于点$F$，由点$A(4$，$\frac{5}{3}\sqrt{3})$，$B\left(-1,0\right)$得，$\tan \angle ABF=\frac{AF}{BF}=\frac{\frac{5}{3}\sqrt{3}}{5}=\frac{\sqrt{3}}{3}$，$\therefore \angle ABF=30^{\circ}$，由翻折得$\angle C'BC=60^{\circ}$，$BC=BC'=4$，在$Rt\triangle BGC'$中，$BG=2$，$GC'=2\sqrt{3}$，$\therefore $点$C'$的坐标为$(1$，$2\sqrt{3})$，在$Rt\triangle BDG$中，$\angle DBG=30^{\circ}$，$\therefore DG=\frac{BG}{\sqrt{3}}=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}$，$\therefore $点$D$坐标为$(1$，$\frac{2}{3}\sqrt{3})$；$(3)\because $直线$FG$垂直平分线段$BC$，在$FG$上取（2）中的$C'$，$\therefore BC=BC'$，$QB=QC$，由（2）知$\angle CBC'=60^{\circ}$，$\angle CBD=\angle C'BD$，$\therefore \triangle BCC'$是等边三角形，$AB$垂直平分$CC'$，①点$P$在$x$轴上方，如图$2$，$\because \triangle BCC'$和$\triangle PQC$都是等边三角形，$\therefore CC'=CB$，$CP=CQ$，$\angle PCQ=\angle BCC'=60^{\circ}$，$\therefore \angle PCC'=\angle QCB$，$\therefore \triangle PCC'$≌$\triangle QCB\left(SAS\right)$，$\therefore PC'=QB$，$\because QB=QC$，$CP=CQ$，$\therefore PC=PC'$，$\because BC=BC'$，$\therefore PB$垂直平分$CC'$，$\therefore $点$P$与点$A$重合，$\therefore PA$的函数关系式就是$AB$的函数关系式：$y=\frac{{\sqrt{3}}}{3}x+\frac{{\sqrt{3}}}{3}$；②点$P$在$x$轴下方，如图$3$，$\because \triangle BCC'$和$\triangle PQC$都是等边三角形，$\therefore CC'=CB$，$CP=CQ$，$\angle PCQ=\angle BCC'=60^{\circ}$，$\therefore \angle PCCB\angle QCBC'$，$\therefore \triangle PCB$≌$\triangle QCC'\left(SAS\right)$，$\therefore \angle CBP=\angle CC'Q=30^{\circ}$，在$Rt\triangle BHO$中，$BO=1$，$\therefore HO=\tan \angle OBH\cdot BO=\frac{\sqrt{3}}{3}$，$\therefore $点$H$的坐标为$(0$，$-\frac{\sqrt{3}}{3})$，设$BP$的函数关系式为$y=mx+n$，把点$B\left(-1,0\right)$，$H(0$，$-\frac{\sqrt{3}}{3})$代入得，$\left\{\begin{array}{l}{-k+b=0}\\{b=-\frac{\sqrt{3}}{3}}\end{array}\right.$，解得$k=-\frac{\sqrt{3}}{3}$，$b=-\frac{\sqrt{3}}{3}$，$\therefore PA$的函数关系式$y=-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}$，综上所述$PA$的函数关系式$y=\frac{\sqrt{3}}{3}x+\frac{\sqrt{3}}{3}$或$y=-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}$.
$(1)$把点$B\left(-1,0\right)$代入直线$AB:y=\frac{{\sqrt{3}}}{3}x+b$中，得$\frac{{\sqrt{3}}}{3}\times \left(-1\right)+b=0$，$\therefore b=\frac{{\sqrt{3}}}{3}$，$\therefore AB$的函数关系式：$y=\frac{{\sqrt{3}}}{3}x+\frac{{\sqrt{3}}}{3}$，当$x=4$时$y=\frac{5}{3}\sqrt{3}$，$\therefore $点$A$的坐标为$(4$，$\frac{5}{3}\sqrt{3})$，设$AC$的函数关系式为：$y=mx+n$，把$A(4$，$\frac{5}{3}\sqrt{3})$，$C\left(3,0\right)$两点坐标代入，得$\left\{\begin{array}{l}{4k+b=\frac{5}{3}\sqrt{3}}\\{3k+b=0}\end{array}\right.$，$\therefore k=\frac{5}{3}\sqrt{3}$，$b=5\sqrt{3}$，$\therefore $直线$AC$的函数表达式为：$y=\frac{5}{3}\sqrt{3}x-5\sqrt{3}$；$(2)$过点$A$作$AF\bot x$轴于点$F$，由点$A(4$，$\frac{5}{3}\sqrt{3})$，$B\left(-1,0\right)$得，$\tan \angle ABF=\frac{AF}{BF}=\frac{\frac{5}{3}\sqrt{3}}{5}=\frac{\sqrt{3}}{3}$，$\therefore \angle ABF=30^{\circ}$，由翻折得$\angle C'BC=60^{\circ}$，$BC=BC'=4$，在$Rt\triangle BGC'$中，$BG=2$，$GC'=2\sqrt{3}$，$\therefore $点$C'$的坐标为$(1$，$2\sqrt{3})$，在$Rt\triangle BDG$中，$\angle DBG=30^{\circ}$，$\therefore DG=\frac{BG}{\sqrt{3}}=\frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}$，$\therefore $点$D$坐标为$(1$，$\frac{2}{3}\sqrt{3})$；$(3)\because $直线$FG$垂直平分线段$BC$，在$FG$上取（2）中的$C'$，$\therefore BC=BC'$，$QB=QC$，由（2）知$\angle CBC'=60^{\circ}$，$\angle CBD=\angle C'BD$，$\therefore \triangle BCC'$是等边三角形，$AB$垂直平分$CC'$，①点$P$在$x$轴上方，如图$2$，$\because \triangle BCC'$和$\triangle PQC$都是等边三角形，$\therefore CC'=CB$，$CP=CQ$，$\angle PCQ=\angle BCC'=60^{\circ}$，$\therefore \angle PCC'=\angle QCB$，$\therefore \triangle PCC'$≌$\triangle QCB\left(SAS\right)$，$\therefore PC'=QB$，$\because QB=QC$，$CP=CQ$，$\therefore PC=PC'$，$\because BC=BC'$，$\therefore PB$垂直平分$CC'$，$\therefore $点$P$与点$A$重合，$\therefore PA$的函数关系式就是$AB$的函数关系式：$y=\frac{{\sqrt{3}}}{3}x+\frac{{\sqrt{3}}}{3}$；②点$P$在$x$轴下方，如图$3$，$\because \triangle BCC'$和$\triangle PQC$都是等边三角形，$\therefore CC'=CB$，$CP=CQ$，$\angle PCQ=\angle BCC'=60^{\circ}$，$\therefore \angle PCCB\angle QCBC'$，$\therefore \triangle PCB$≌$\triangle QCC'\left(SAS\right)$，$\therefore \angle CBP=\angle CC'Q=30^{\circ}$，在$Rt\triangle BHO$中，$BO=1$，$\therefore HO=\tan \angle OBH\cdot BO=\frac{\sqrt{3}}{3}$，$\therefore $点$H$的坐标为$(0$，$-\frac{\sqrt{3}}{3})$，设$BP$的函数关系式为$y=mx+n$，把点$B\left(-1,0\right)$，$H(0$，$-\frac{\sqrt{3}}{3})$代入得，$\left\{\begin{array}{l}{-k+b=0}\\{b=-\frac{\sqrt{3}}{3}}\end{array}\right.$，解得$k=-\frac{\sqrt{3}}{3}$，$b=-\frac{\sqrt{3}}{3}$，$\therefore PA$的函数关系式$y=-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}$，综上所述$PA$的函数关系式$y=\frac{\sqrt{3}}{3}x+\frac{\sqrt{3}}{3}$或$y=-\frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}$.
