2021-2022/2nd/10_Geometrie_reperee/2B_milieu.tex

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\documentclass[a4paper,12pt]{article}
\usepackage{myXsim}
\author{Benjamin Bertrand}
\title{Géométrie repérée - Cours}
\date{2022-01-13}
\pagestyle{empty}
\begin{document}
\maketitle
\bigskip
\section*{Milieu d'un segment}
\begin{propriete}[Coordonnée du milieu d'un segment]
\begin{minipage}{0.5\linewidth}
Soit $M$ et $N$ deux points quelconques et $K$ le milieu du segment $[MN]$. Alors
\begin{itemize}
\item L'abscisse de $K$ est la moyenne des abscisses de $M$ et $N$
\[
x_K = \frac{x_M + x_N}{2}
\]
\item L'ordonnée de $K$ est la moyenne des ordonnées de $M$ et $N$
\[
y_K = \frac{y_M + y_N}{2}
\]
\end{itemize}
\end{minipage}
\hfill
\begin{minipage}{0.4\linewidth}
\begin{tikzpicture}[scale=1.2]
\draw[->, very thick] (-1, 0) -- (4, 0);
\draw[->, very thick] (0, -1) -- (0, 4);
\draw (0, 0) node [below left] {0};
\draw (1, 0) node {+} node [below left] {1};
\draw (0, 1) node {+} node [below left] {1};
\draw (1.3, 1.4) node {+} node [below left] {$M$};
\draw (1.3, 0) node {+} node [below] {$x_M$};
\draw (0, 1.4) node {+} node [left] {$y_M$};
\draw[dashed] (1.3, 1.4) --(1.3, 0);
\draw[dashed] (1.3, 1.4) --(0, 1.4);
\draw (3.3, 3.4) node {+} node [below right] {$N$};
\draw (3.3, 0) node {+} node [below] {$x_N$};
\draw (0, 3.4) node {+} node [left] {$y_N$};
\draw[dashed] (3.3, 3.4) --(3.3, 0);
\draw[dashed] (3.3, 3.4) --(0, 3.4);
\draw (2.3, 2.4) node {+} node [below right] {$K$};
\draw (2.3, 0) node {+} node [below] {$x_K$};
\draw (0, 2.4) node {+} node [left] {$y_K$};
\draw[dashed] (2.3, 2.4) --(2.3, 0);
\draw[dashed] (2.3, 2.4) --(0, 2.4);
\draw (1.3, 1.4) -- node [midway, sloped] {//}
(2.3, 2.4) -- node [midway, sloped] {//}
(3.3, 3.4);
\draw (1.3, 0) -- node [midway, sloped] {$\bullet$}
(2.3, 0) -- node [midway, sloped] {$\bullet$}
(3.3, 0);
\draw (0, 1.4) -- node [midway, sloped] {$\diamond$}
(0, 2.4) -- node [midway, sloped] {$\diamond$}
(0, 3.4);
\end{tikzpicture}
\end{minipage}
\end{propriete}
\paragraph{Exemple}: Coordonnée de $I$ milieu du segment $[AB]$ avec $A(23; 45)$ et $B (-3; 12)$
\[
x_I = \frac{x_A + x_B}{2} = \frac{23 + (-3)}{2} = 10
\qquad \qquad
y_I = \frac{y_A + y_B}{2} = \frac{45 + 12}{2} = 28.5
\]
Les coordonnées de $I$ sont $(10; 28.5)$.
\end{document}