\(\overrightarrow{AB}=\left(0;3\right);\overrightarrow{AC}=\left(11;2\right);\overrightarrow{BC}=\left(11;-1\right)\)
\(H\left(x;y\right)\)
\(\overrightarrow{AH}=\left(x+2;y-6\right);\overrightarrow{BH}=\left(x+2;y-9\right)\)
H là trực tâm \(\Delta ABC\) \(\Rightarrow\left\{{}\begin{matrix}AH\perp BC\\BH\perp AC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{AH}.\overrightarrow{BC}=0\\\overrightarrow{BH}.\overrightarrow{AC}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11.\left(x+2\right)-y+6=0\\11\left(x+2\right)+2\left(y-9\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11x-y=-28\\11x+2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{20}{11}\\y=8\end{matrix}\right.\)
Vậy \(H\left(-\dfrac{20}{11};8\right)\)
Gọi \(H\left(x;y\right)\) là tọa độ trực tâm của tam giác ABC
\(\overrightarrow{AH}\left(x+2;y-6\right)\perp\overrightarrow{BC}\left(11;-1\right)\\ \Rightarrow11\left(x+2\right)-1\left(y-6\right)=0\\ \Rightarrow11x-y=-28_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{ }}}}}}}}}}}}}}}\left(1\right)\)
\(\overrightarrow{BH}\left(x+2;y-9\right)\perp\overrightarrow{AC}\left(11;2\right)\\ \Rightarrow11\left(x+2\right)+2\left(y-9\right)=0\\ \Rightarrow11x+2y=-4_{_{_{_{_{_{_{_{_{_{ }}}}}}}}}}\left(2\right)\)
Từ (1) và (2), ta có hệ phương trình \(\left\{{}\begin{matrix}11x-y=-28\\11x+2y=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{20}{11}\\y=8\end{matrix}\right.\)
Vậy tọa độ trực tâm H cần tìm là: \(H\left(-\dfrac{20}{11};8\right)\)