Gọi \(A\left(a;b\right)\)
Theo tính chất trọng tâm: \(\overrightarrow{AM}=2\overrightarrow{GM}\)
\(\Rightarrow\left(1-a;-1-b\right)=2\left(\frac{1}{3};-1\right)\)\(\Rightarrow\left\{{}\begin{matrix}1-a=\frac{2}{3}\\-1-b=-2\end{matrix}\right.\) \(\Rightarrow A\left(\frac{1}{3};1\right)\)
Ta có \(\overrightarrow{GM}=\left(\frac{1}{3};-1\right)\), mà ABC vuông cân nên \(AM\perp BC\Rightarrow\) đường thẳng BC nhận \(\overrightarrow{n_{BC}}=\left(1;-3\right)\) là 1 vtpt
Phương trình BC: \(1\left(x-1\right)-3\left(y+1\right)=0\Rightarrow x-3y-4=0\)
Do \(B\in BC\Rightarrow B\left(3b+4;b\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AM}=\left(\frac{2}{3};-2\right)\\\overrightarrow{MB}=\left(3b+3;b+1\right)\end{matrix}\right.\)
Theo tính chất tam giác vuông cân: \(AM=MB\Rightarrow AM^2=MB^2\)
\(\Rightarrow\left(\frac{2}{3}\right)^2+\left(-2\right)^2=\left(3b+3\right)^2+\left(b+1\right)^2\)
\(\Rightarrow10\left(b+1\right)^2=\frac{40}{9}\Rightarrow\left(b+1\right)^2=\frac{4}{9}\Rightarrow\left[{}\begin{matrix}b=-\frac{1}{3}\\b=-\frac{5}{3}\end{matrix}\right.\)
- Với \(b=-\frac{1}{3}\Rightarrow B\left(3;-\frac{1}{3}\right)\Rightarrow\left\{{}\begin{matrix}x_C=2x_M-x_B=-1\\y_C=2y_M-y_B=-\frac{5}{3}\end{matrix}\right.\) \(\Rightarrow C\left(-1;-\frac{5}{3}\right)\)
- Với \(b=-\frac{5}{3}\Rightarrow B\left(-1;-\frac{5}{3}\right)\Rightarrow C\left(3;-\frac{1}{3}\right)\)
