Đáp án: H(0;1)
Giải thích các bước giải:
$\begin{array}{l}
H\left( {x;y} \right)\\
\overrightarrow {AB} = \left( {1;2} \right);\overrightarrow {CH} = \left( {x + 2;y - 2} \right)\\
\overrightarrow {AC} = \left( { - 2;1} \right);\overrightarrow {BH} = \left( {x - 1;y - 3} \right)\\
Do\,AB \bot CH;AC \bot BH\\
\Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AB} .\overrightarrow {CH} = 0\\
\overrightarrow {AC} .\overrightarrow {BH} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + 2 + 2\left( {y - 2} \right) = 0\\
- 2x + 2 + y - 3 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + 2y = 2\\
- 2x + y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.\\
\Rightarrow H\left( {0;1} \right)
\end{array}$
