85 lines
2.7 KiB
TeX
85 lines
2.7 KiB
TeX
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\documentclass[a4paper,12pt]{article}
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\usepackage{myXsim}
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\author{Benjamin Bertrand}
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\title{Géométrie repérée - Cours}
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\date{janvier 2023}
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\pagestyle{empty}
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\begin{document}
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\maketitle
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\bigskip
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\section*{Milieu d'un segment}
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\begin{propriete}[Coordonnée du milieu d'un segment]
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\begin{minipage}{0.5\linewidth}
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Soit $M$ et $N$ deux points quelconques et $K$ le milieu du segment $[MN]$. Alors
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\begin{itemize}
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\item L'abscisse de $K$ est la moyenne des abscisses de $M$ et $N$
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\[
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x_K = \frac{x_M + x_N}{2}
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\]
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\item L'ordonnée de $K$ est la moyenne des ordonnées de $M$ et $N$
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\[
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y_K = \frac{y_M + y_N}{2}
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\]
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\end{itemize}
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\end{minipage}
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\hfill
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\begin{minipage}{0.4\linewidth}
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\begin{tikzpicture}[scale=1.2]
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\draw[->, very thick] (-1, 0) -- (4, 0);
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\draw[->, very thick] (0, -1) -- (0, 4);
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\draw (0, 0) node [below left] {0};
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\draw (1, 0) node {+} node [below left] {1};
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\draw (0, 1) node {+} node [below left] {1};
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\draw (1.3, 1.4) node {+} node [below left] {$M$};
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\draw (1.3, 0) node {+} node [below] {$x_M$};
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\draw (0, 1.4) node {+} node [left] {$y_M$};
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\draw[dashed] (1.3, 1.4) --(1.3, 0);
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\draw[dashed] (1.3, 1.4) --(0, 1.4);
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\draw (3.3, 3.4) node {+} node [below right] {$N$};
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\draw (3.3, 0) node {+} node [below] {$x_N$};
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\draw (0, 3.4) node {+} node [left] {$y_N$};
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\draw[dashed] (3.3, 3.4) --(3.3, 0);
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\draw[dashed] (3.3, 3.4) --(0, 3.4);
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\draw (2.3, 2.4) node {+} node [below right] {$K$};
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\draw (2.3, 0) node {+} node [below] {$x_K$};
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\draw (0, 2.4) node {+} node [left] {$y_K$};
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\draw[dashed] (2.3, 2.4) --(2.3, 0);
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\draw[dashed] (2.3, 2.4) --(0, 2.4);
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\draw (1.3, 1.4) -- node [midway, sloped] {//}
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(2.3, 2.4) -- node [midway, sloped] {//}
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(3.3, 3.4);
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\draw (1.3, 0) -- node [midway, sloped] {$\bullet$}
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(2.3, 0) -- node [midway, sloped] {$\bullet$}
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(3.3, 0);
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\draw (0, 1.4) -- node [midway, sloped] {$\diamond$}
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(0, 2.4) -- node [midway, sloped] {$\diamond$}
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(0, 3.4);
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\end{tikzpicture}
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\end{minipage}
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\end{propriete}
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\paragraph{Exemple}: Coordonnée de $I$ milieu du segment $[AB]$ avec $A(23; 45)$ et $B (-3; 12)$
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\[
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x_I = \frac{x_A + x_B}{2} = \frac{23 + (-3)}{2} = 10
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\qquad \qquad
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y_I = \frac{y_A + y_B}{2} = \frac{45 + 12}{2} = 28.5
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\]
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Les coordonnées de $I$ sont $(10; 28.5)$.
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\end{document}
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