解：?$(1)$?建立如圖所示的平面直角坐標(biāo)系
∵邊?$AB$?上的高為?$4，$?∴點(diǎn)?$C$?的縱坐標(biāo)為?$±4$?
?$①$?當(dāng)?$AB$?為底邊時，則點(diǎn)?$C$?的橫坐標(biāo)為?$AB$?的中點(diǎn)的橫坐標(biāo)
∴?$C(-1，$??$4)$?或?$C'(-1，$??$-4)$?
?$②$?當(dāng)?$AB$?為腰時
當(dāng)?$AC_{1}=AB=6，$?過點(diǎn)?$C_{1}$?作?$C_{1}E⊥x$?軸于點(diǎn)?$E$?
則?$C_{1}E=4，$?∴?$AE=\sqrt {AC_{1}^2-C_{1}E^2}=\sqrt {20}$?
∴?$OE=AE-AO=\sqrt {20}-4$?
∴?$C_{1}(\sqrt {20}-4，$??$4)，$??$C_{2}(\sqrt {20}-4，$??$-4)$?
當(dāng)?$BC_{3}=AB=6$?時，過點(diǎn)?$C_{3}$?作?$C_{3}D⊥x$?軸于點(diǎn)?$D$?
則?$C_{3}D=4，$?∴?$BD=\sqrt {BC_{3}^2-C_{3}D^2}=\sqrt {20}$?
∴?$OD=BD-OB=\sqrt {20}-2$?
∴?$C_{3}(2-\sqrt {20}，$??$4)，$??$C_{4}(2-\sqrt {20}，$??$-4)$?
綜上，點(diǎn)?$C$?的坐標(biāo)為?$(-1，$??$4)$?或?$(-1，$??$-4)$?或?$(\sqrt {20}-4，$??$4)$?
或?$(\sqrt {20}-4，$??$-4)$?或?$(2-\sqrt {20}，$??$4)$?或?$(2-\sqrt {20}，$??$-4)$?
?$(2)S_{△ABC}=\frac 12AB· 4=\frac 12×6×4=12$?
