Đáp án:
$\begin{array}{l}
a)\\
\left\{ \begin{array}{l}
{x_G} = \frac{{{x_A} + {x_B} + {x_C}}}{3} = \frac{5}{3}\\
{y_G} = \frac{{{y_A} + {y_B} + {y_C}}}{3} = \frac{{10}}{3}
\end{array} \right. \Rightarrow G\left( {\frac{5}{3};\frac{{10}}{3}} \right)
\end{array}$
$\begin{array}{l}
H\left( {x;y} \right) \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AH} = \left( {x - 5;y - 4} \right)\\
\overrightarrow {BH} = \left( {x - 2;y - 7} \right)\\
\overrightarrow {BC} = \left( { - 4; - 8} \right)\\
\overrightarrow {AC} = \left( { - 7; - 5} \right)
\end{array} \right.\\
Do:\left\{ \begin{array}{l}
AH \bot BC\\
BH \bot AC
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AH} .\overrightarrow {BC} = 0\\
\overrightarrow {BH} .\overrightarrow {AC} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 4\left( {x - 5} \right) - 8\left( {y - 4} \right) = 0\\
- 7\left( {x - 2} \right) - 5\left( {y - 7} \right) = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \frac{{11}}{3}\\
y = \frac{{14}}{3}
\end{array} \right. \Rightarrow H\left( {\frac{{11}}{3};\frac{{14}}{3}} \right)\\
I\left( {x;y} \right)\\
\Rightarrow \left\{ \begin{array}{l}
A{I^2} = B{I^2}\\
A{I^2} = C{I^2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x - 5} \right)^2} + {\left( {y - 4} \right)^2} = {\left( {x - 2} \right)^2} + {\left( {y - 7} \right)^2}\\
{\left( {x - 5} \right)^2} + {\left( {y - 4} \right)^2} = {\left( {x + 2} \right)^2} + {\left( {y + 1} \right)^2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \frac{2}{3}\\
y = \frac{8}{3}
\end{array} \right. \Rightarrow I\left( {\frac{2}{3};\frac{8}{3}} \right)
\end{array}$
b)
$\begin{array}{l}
\overrightarrow {IH} = \left( {3;2} \right);\overrightarrow {IG} = \left( {1;\frac{2}{3}} \right)\\
\Rightarrow \overrightarrow {IH} = \frac{3}{2}.\overrightarrow {IG}
\end{array}$
Vậy I,H,K thẳng hàng
