\begin{enumerate}
\item Placement des points A, B et C :

\begin{center}
\begin{tikzpicture}[>=stealth']
\draw[gray,dotted] (-2,-2) grid (4,2);
\draw[thick,->] (-3,0) -- (5,0);
\draw[thick,->] (0,-3) -- (0,3);
\fill (-2,1) circle (1.5pt) node[above] {A};
\fill (1,-2) circle (1.5pt) node[above] {B};
\fill (4,2) circle (1.5pt) node[above] {C};
\draw (1,0.1) -- (1,-0.1) node[below] {1};
\draw (0.1,1) -- (-0.1,1) node[left] {1};
\end{tikzpicture}
\end{center}

\item $\vv{AB}\dbinom{1-(-2)}{-2-1}$, soit $\vv{AB}\dbinom{3}{-3}$.

$\vv{AC}\dbinom{4-(-2)}{2-1}$, soit $\vv{AC}\dbinom{6}{1}$.

$\vv{BC}\dbinom{4-1}{2-(-2)}$, soit $\vv{BC}\dbinom{3}{4}$.

\item $AB=\|\vv{AB}\|=\sqrt{3^2+(-3)^2}=\sqrt{18}=3\sqrt{2}$.

$AC=\sqrt{6^2+1^2}=\sqrt{37}$.

$BC=\sqrt{3^2+4^2}=\sqrt{25}=5$.

Le triangle ABC n'est manifestement pas rectangle car $37\neq18+25$.

\item $I\coord{\dfrac{1+4}{2}}{\dfrac{-2+2}{2}}$, soit $I\coord{\dfrac{5}{2}}{0}$.
\end{enumerate}