\(\overrightarrow{BH}=\left(2;-2\right)\Rightarrow\) đường thẳng AC có 1 vtpt \(\overrightarrow{n_{AC}}=\left(1;-1\right)\)
\(\Rightarrow\) phương trình AC:
\(1\left(x-4\right)-1\left(y-3\right)-0\Leftrightarrow x-y-1=0\)
Gọi \(A\left(a;a-1\right)\Rightarrow\left\{{}\begin{matrix}x_C=2x_M-x_A=8-a\\y_C=2y_M-y_A=7-a\end{matrix}\right.\) \(\Rightarrow C\left(8-a;7-a\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AH}=\left(2-a;4-a\right)\\\overrightarrow{BC}=\left(8-a;2-a\right)\end{matrix}\right.\) mà \(AH\perp BC\Rightarrow\overrightarrow{AH}.\overrightarrow{BC}=0\)
\(\Rightarrow\left(2-a\right)\left(8-a\right)+\left(4-a\right)\left(2-a\right)=0\)
\(\Leftrightarrow\left(2-a\right)\left(12-2a\right)=0\Rightarrow\left[{}\begin{matrix}a=2\\a=6\end{matrix}\right.\)
- Với \(a=2\Rightarrow\left\{{}\begin{matrix}A\left(2;1\right)\\C\left(6;5\right)\end{matrix}\right.\) \(\Rightarrow x_A< x_C\) (loại)
- Với \(a=6\Rightarrow\left\{{}\begin{matrix}A\left(6;5\right)\\C\left(2;1\right)\end{matrix}\right.\) (thỏa mãn) \(\Rightarrow\overrightarrow{BA}=\left(6;0\right)\)
\(\Rightarrow\) đường thẳng AB nhận \(\overrightarrow{n_{AB}}=\left(0;1\right)\) là 1 vtpt
\(\Rightarrow\) phương trình AB: \(y-5=0\)
