Do \(AH\perp BC\Rightarrow\) đường thẳng AH nhận \(\overrightarrow{n_{AH}}=\left(2;-1\right)\) là 1 vtpt
\(\Rightarrow\) pt AH: \(2\left(x+\frac{17}{5}\right)-1\left(y-\frac{21}{5}\right)=0\Leftrightarrow2x-y+11=0\)
Do A thuộc AH, gọi \(A\left(a;2a+11\right)\)
Gọi M là trung điểm BC \(\Rightarrow\overrightarrow{AG}=\frac{2}{3}\overrightarrow{AM}\Rightarrow\left\{{}\begin{matrix}x_M=\frac{3x_G-x_A}{2}=\frac{-6-a}{2}\\y_M=\frac{3y_G-y_A}{2}=-a-1\end{matrix}\right.\)
Do \(M\in BC\Rightarrow x_M+2y_M-5=0\)
\(\Leftrightarrow\frac{-6-a}{2}+2\left(-a-1\right)-5=0\Rightarrow a=-4\Rightarrow\left\{{}\begin{matrix}A\left(-4;3\right)\\M\left(-1;3\right)\end{matrix}\right.\)
\(\Rightarrow AH=\sqrt{\left(-\frac{17}{5}+4\right)^2+\left(\frac{21}{5}-3\right)^2}=\frac{3\sqrt{5}}{5}\)
\(\Rightarrow BC=\frac{2.S_{ABC}}{AH}=2\sqrt{5}\)
Gọi \(B\left(5-2b;b\right)\Rightarrow BM=\frac{BC}{2}=\sqrt{5}\)
\(\Rightarrow\sqrt{\left(2b-6\right)^2+\left(3-b\right)^2}=\sqrt{5}\) \(\Rightarrow b^2-6b+8=0\Rightarrow\left[{}\begin{matrix}b=2\\b=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}B\left(1;2\right)\\C\left(-3;4\right)\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}B\left(-3;4\right)\\C\left(1;2\right)\end{matrix}\right.\)
