
Dans un repère orthonormé \Rij, on considère les points A\coord{-2}{-3}, B\coord{4}{-1} et\linebreak C\coord{-7}{2}.

\begin{enumerate}
\item 
\begin{enumerate}
\item $\vv{\text{AB}}\dbinom{x_B-x_A}{y_B-y_A}\Leftrightarrow \vv{\text{AB}}\dbinom{4-(-2)}{-1-(-3)} \Leftrightarrow \vv{\text{AB}}\dbinom{6}{2}$.

$\vv{\text{AC}}\dbinom{x_C-x_A}{y_C-y_A}\Leftrightarrow \vv{\text{AC}}\dbinom{-7-(-2)}{2-(-3)} \Leftrightarrow \vv{\text{AC}}\dbinom{-5}{5}$.

Ainsi, $\vv{\text{AB}}+\vv{\text{AC}}\dbinom{6+(-5)}{2+5}\Leftrightarrow\boxed{\vv{\text{AB}}+\vv{\text{AC}}\dbinom{1}{7}}$.

\item $\begin{aligned}[t]
\text{ABDC est un parallélogramme} & \Longleftrightarrow \vv{\text{AB}}+\vv{\text{AC}}=\vv{\text{AD}}\\
& \Longleftrightarrow\dbinom{1}{7}=\dbinom{x_D-(-2)}{y_D-(-3)}\\
& \Longleftrightarrow\dbinom{1}{7}=\dbinom{x_D+2}{y_D+3}\\
& \Longleftrightarrow\begin{cases}
1=x_D+2\\
7=y_D+3
\end{cases}\\
& \Longleftrightarrow\begin{cases}
x_D=1-2=-1\\
y_D=7-3=4
\end{cases}
\end{aligned}$

 
\underline{Ainsi, D\coord{-1}{4}}.
\end{enumerate}

\item $\begin{aligned}[t]
\text{I milieu de [AD]} & \Longleftrightarrow \dfrac{1}{2}\vv{\text{AD}}=\vv{\text{AI}} \Longleftrightarrow \dfrac{1}{2}\dbinom{1}{7}=	\dbinom{x_I-(-2)}{y_I-(-3)}\\
& \Longleftrightarrow\dbinom{\frac{1}{2}}{\frac{7}{2}}=\dbinom{x_I+2}{y_I+3} \Longleftrightarrow \begin{cases}
x_I=\frac{1}{2}-2=-\frac{3}{2}\\
y_I=\frac{7}{2}-3=\frac{1}{2}
\end{cases}
\end{aligned}$

\underline{Ainsi, I\coord{-\dfrac{3}{2}}{\dfrac{1}{2}}}.
\end{enumerate}
