\(\left|y_G\right|=\frac{1}{3}\Rightarrow\left[{}\begin{matrix}y_G=\frac{1}{3}\\y_G=-\frac{1}{3}\end{matrix}\right.\)
TH1: \(y_G=\frac{1}{3}\Rightarrow y_C=3y_G-\left(y_A+y_B\right)=1\)
Gọi \(C\left(x;1\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AC}=\left(x+2;1\right)\\\overrightarrow{BC}=\left(x-2;1\right)\end{matrix}\right.\)
\(AC\perp BC\Rightarrow\overrightarrow{AC}.\overrightarrow{BC}=0\Rightarrow\left(x-2\right)\left(x+2\right)+1=0\Rightarrow x=\pm\sqrt{3}\)
\(\Rightarrow\left[{}\begin{matrix}C\left(\sqrt{3};1\right)\\C\left(-\sqrt{3};1\right)\end{matrix}\right.\)
TH2: \(y_G=-\frac{1}{3}\Rightarrow y_C=-1\)
\(C\left(x;-1\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AC}=\left(x+2;-1\right)\\\overrightarrow{BC}=\left(x-2;-1\right)\end{matrix}\right.\) \(\Rightarrow\left(x-2\right)\left(x+2\right)+1=0\Rightarrow x=\pm\sqrt{3}\) \(\Rightarrow\left[{}\begin{matrix}C\left(\sqrt{3};-1\right)\\C\left(-\sqrt{3};-1\right)\end{matrix}\right.\)
