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\begin{center}\section*{\textcolor{red}{Contrôle sur 10 points }}\end{center}




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\subsection{}
Dans un repère $(O~;~I~;~J)$, placer les points $A(2~;~4)$, $B(-1~;~-3)$, $C(0~;~5)$ et $D(3~;~0)$.

\begin{center}
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\subsection{}
On considère les points $A(2~;~7)$ et $B(-3~;~-5)$.

$AB=\sqrt{\left(x_B-x_A\right)^2+\left(y_B-y_A\right)^2}\\
=\sqrt{(-3-2)^2+(-5-7)^2}=\sqrt{(-5)^2+(-12)^2}\\
=\sqrt{25+144}=\sqrt{169}=13$.

$\boxed{\textcolor{red}{AB=13}}$.

\subsection{}
On considère les points $A(2~;~5)$, $B(7~;~1)$, $C(1~;~-3)$.

Déterminons les coordonnées du point $D$ tel que $ABCD$ soit un parallélogramme.

\begin{enumerate}[a)]
\item Soit $M$ le milieu de la diagonale $[AC]$ :

\begin{enumerate}[$\bullet$]
\item $x_M=\dfrac{x_A+x_C}{2}=\dfrac{2+1}{2}=\dfrac{3}{2}$

\item $y_M=\dfrac{y_A+y_C}{2}=\dfrac{5+(-3)}{2}=\dfrac{2}{2}=1$

$M$ a pour coordonnées $M\left(\dfrac{3}{2}~;~1\right)$
\end{enumerate}

\item Pour que $ABCD$ soit un parallélogramme, il faut que $M$ soit aussi le milieu de la diagonale $[BD]$.

\begin{enumerate}[$\bullet$]
\item $x_M=\dfrac{x_B+x_D}{2}$ donne $\dfrac{3}{2}=\dfrac{7+x_D}{2}$ donc $3=7+x_D$ d'où $x_D=3-7=-4$

\item $y_M=\dfrac{y_B+y_D}{2}$ donne $1=\dfrac{1+y_D}{2}$ donc $2=1+y_D$ d'où $y_D=2-1=1$

\end{enumerate}
$D$ a pour coordonnées : $\boxed{\textcolor{red}{D(-4~;~1)}}$
\end{enumerate}

\begin{center}
Figure (non demandée) :


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\subsection{}
Soit $A\left(2~;~\sqrt{3}\right)$.

$OA=\sqrt{\left(x_A-x_O\right)^2+\left(y_Ay_O\right)^2}=\sqrt{(2-0)^2+\left(\sqrt{3}-0\right)^2}\\
=\sqrt{4+3}=\sqrt{7}$ donc A appartient bien au cercle de centre O et de rayon $\sqrt{7}$.


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