Đáp án:
$\begin{array}{l}
1)I\left( {x;y} \right)\\
\Rightarrow IA = IB = IC\\
\Rightarrow \left\{ \begin{array}{l}
I{A^2} = I{B^2}\\
I{A^2} = I{C^2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {x - 2} \right)^2} + {y^2}\\
{\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {x + 3} \right)^2} + {\left( {y - 1} \right)^2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
6x - 4y = - 1\\
- 4x - 2y = 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = - \frac{{11}}{{14}}\\
y = - \frac{{13}}{{14}}
\end{array} \right. \Rightarrow I\left( { - \frac{{11}}{{14}}; - \frac{{13}}{{14}}} \right)\\
2)\overrightarrow {BC} = \left( { - 5;1} \right)\\
M\left( {x:y} \right) \Rightarrow \overrightarrow {BM} = \left( {x - 2;y} \right)\\
MB = \frac{1}{3}BC\\
\Rightarrow \left\{ \begin{array}{l}
\overrightarrow {BM} = \frac{1}{3}\overrightarrow {BC} \\
\overrightarrow {BM} = - \frac{1}{3}\overrightarrow {BC}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3\left( {x - 2} \right) = - 5\\
3y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
3\left( {x - 2} \right) = 5\\
3y = - 1
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = \frac{{11}}{3}\\
y = - \frac{1}{3}
\end{array} \right. \Rightarrow M\left( {\frac{{11}}{3}; - \frac{1}{3}} \right)\\
\left\{ \begin{array}{l}
x = \frac{1}{3}\\
y = \frac{1}{3}
\end{array} \right. \Rightarrow M\left( {\frac{1}{3};\frac{1}{3}} \right)
\end{array} \right.
\end{array}$
Vì M thuộc BC nên chiều cao của tam giác ABM và ABC bằng nhau
