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-rw-r--r--macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointwith.tex73
1 files changed, 37 insertions, 36 deletions
diff --git a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointwith.tex b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointwith.tex
index 98c1ef0c6d..8af1881003 100644
--- a/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointwith.tex
+++ b/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointwith.tex
@@ -2,21 +2,21 @@
\subsection{\tkzcname{tkzDefPointWith}}
There are several possibilities to create points that meet certain vector conditions.
-This can be done with \tkzcname{tkzDefPointWith}. The general principle is as follows, two points are passed as arguments, i.e. a vector. The different options allow to obtain a new point forming with the first point ( with some exceptions) a collinear vector or a vector orthogonal to the first vector. Then the length is either proportional to that of the first one, or proportional to the unit. Since this point is only used temporarily, it does not have to be named immediately. The result is in \tkzcname{tkzPointResult}. The macro \tkzNameMacro{tkzGetPoint} allows you to retrieve the point and name it differently.
+This can be done with \tkzcname{tkzDefPointWith}. The general principle is as follows, two points are passed as arguments, i.e. a vector. The different options allow to obtain a new point forming with the first point (with some exceptions) a collinear vector or a vector orthogonal to the first vector. Then the length is either proportional to that of the first one, or proportional to the unit. Since this point is only used temporarily, it does not have to be named immediately. The result is in \tkzname{tkzPointResult}. The macro \tkzNameMacro{tkzGetPoint} allows you to retrieve the point and name it differently.
There are options to define the distance between the given point and the obtained point.
In the general case this distance is the distance between the 2 points given as arguments if the option is of the "normed" type then the distance between the given point and the obtained point is 1 cm. Then the $K$ option allows to obtain multiples.
-\begin{NewMacroBox}{tkzDefPointWith}{\parg{pt1,pt2}}
+\begin{NewMacroBox}{tkzDefPointWith}{\parg{pt1,pt2}}%
It is in fact the definition of a point meeting vectorial conditions.
\medskip
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
arguments & definition & explication \\
\midrule
-\TAline{(pt1,pt2)} {point couple}{the result is a point in \tkzcname{tkzPointResult} } \\
+\TAline{(pt1,pt2)} {point couple}{the result is a point in \tkzname{tkzPointResult} } \\
\bottomrule
\end{tabular}
@@ -24,23 +24,21 @@ arguments & definition & explication \\
\medskip
In what follows, it is assumed that the point is recovered by \tkzNameMacro{tkzGetPoint\{C\}}
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
-options & exemple & explication \\
+options & example & explication \\
\midrule
-\TOline{orthogonal}{[orthogonal](A,B)}{$AC=AB$ et $\overrightarrow{AC} \perp \overrightarrow{AB}$}
-\TOline{orthogonal normed}{[orthogonal normed](A,B)}{$AC=1$ et $\overrightarrow{AC} \perp \overrightarrow{AB}$}
+\TOline{orthogonal}{[orthogonal](A,B)}{$AC=AB$ and $\overrightarrow{AC} \perp \overrightarrow{AB}$}
+\TOline{orthogonal normed}{[orthogonal normed](A,B)}{$AC=1$ and $\overrightarrow{AC} \perp \overrightarrow{AB}$}
\TOline{linear}{[linear](A,B)}{$\overrightarrow{AC}=K \times \overrightarrow{AB}$}
-\TOline{linear normed}{[linear normed](A,B)}{$AC=K$ et $\overrightarrow{AC}=k\times \overrightarrow{AB}$ }
+\TOline{linear normed}{[linear normed](A,B)}{$AC=K$ and $\overrightarrow{AC}=k\times \overrightarrow{AB}$ }
\TOline{colinear= at \#1}{[colinear= at C](A,B)}{$\overrightarrow{CD}= \overrightarrow{AB}$ }
\TOline{colinear normed= at \#1}{[colinear normed= at C](A,B)}{$\overrightarrow{CD}= \overrightarrow{AB}$ }
\TOline{K}{[linear](A,B),K=2}{$\overrightarrow{AC}=2\times \overrightarrow{AB}$}
- \bottomrule
\end{tabular}
-
\end{NewMacroBox}
-\subsubsection{\tkzcname{tkzDefPointWith} et \tkzname{colinear at}}
+\subsubsection{Option \tkzname{colinear at}}
$(\overrightarrow{AB}=\overrightarrow{CD})$
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=1.2,
@@ -56,7 +54,7 @@ options & exemple & explication \\
\end{tkzexample}
-\subsubsection{colinear at}
+\subsubsection{Option \tkzname{colinear at} with $K$}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[vect/.style={->,
@@ -74,8 +72,7 @@ options & exemple & explication \\
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{colinear $K=\frac{\sqrt{2}}{2}$}
-
+\subsubsection{Option \tkzname{colinear at} with $K=\frac{\sqrt{2}}{2}$}
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[vect/.style={->,
shorten >=3pt,>=latex'}]
@@ -89,17 +86,18 @@ options & exemple & explication \\
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{\tkzcname{tkzDefPointWith} et \tkzname{orthogonal}}
-$K=-1$ afin que $(\overrightarrow{AC},\overrightarrow{AB})$ détermine un angle positif. AB=AC puisque $|K|=1$
+\subsubsection{Option \tkzname{orthogonal}}
+AB=AC since $K=1$.
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=1.2,
vect/.style={->,shorten >=3pt,>=latex'}]
\tkzDefPoint(2,3){A}
\tkzDefPoint(4,2){B}
- \tkzDefPointWith[orthogonal,K=-1](A,B)
+ \tkzDefPointWith[orthogonal,K=1](A,B)
\tkzGetPoint{C}
\tkzDrawPoints[color=red](A,B,C)
- \tkzLabelPoints[right=3pt](A,B,C)
+ \tkzLabelPoints[right=3pt](B,C)
+ \tkzLabelPoints[below=3pt](A)
\tkzDrawSegments[vect](A,B A,C)
\tkzMarkRightAngle(B,A,C)
\end{tikzpicture}
@@ -107,7 +105,9 @@ $K=-1$ afin que $(\overrightarrow{AC},\overrightarrow{AB})$ détermine un angle
-\subsubsection{ orthogonal simple}
+\subsubsection{Option \tkzname{orthogonal} with $K=-1$}
+OK=OI since $\lvert K \rvert=1$ then OI=OJ=OK.
+
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.75]
\tkzDefPoint(1,2){O}
@@ -124,7 +124,7 @@ $K=-1$ afin que $(\overrightarrow{AC},\overrightarrow{AB})$ détermine un angle
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{advanced orthogonal}
+\subsubsection{Option \tkzname{orthogonal} more complicated example}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.75]
\tkzDefPoints{0/0/A,6/0/B}
@@ -147,7 +147,7 @@ $K=-1$ afin que $(\overrightarrow{AC},\overrightarrow{AB})$ détermine un angle
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{segment colinear and orthogonal}
+\subsubsection{Options \tkzname{colinear} and \tkzname{orthogonal}}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1.2,
vect/.style={->,shorten >=3pt,>=latex'}]
@@ -163,8 +163,8 @@ $K=-1$ afin que $(\overrightarrow{AC},\overrightarrow{AB})$ détermine un angle
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{\tkzcname{tkzDefPointWith} \tkzname{orthogonal normed}, K=1}
-AC=1
+\subsubsection{Option \tkzname{orthogonal normed}, $K=1$}
+$AC=1$.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1.2,
@@ -178,8 +178,8 @@ AC=1
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{\tkzcname{tkzDefPointWith} et \tkzname{orthogonal normed} K=2}
-$K=2$ donc AC=2.
+\subsubsection{Option \tkzname{orthogonal normed} and $K=2$}
+$K=2$ therefore $AC=2$.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1.2,
@@ -195,8 +195,9 @@ $K=2$ donc AC=2.
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{\tkzcname{tkzDefPointWith} \tkzname{linear} }
- Ici $K=0.5$
+\subsubsection{Option \tkzname{linear}}
+Here $K=0.5$.
+
This amounts to applying a homothety or a multiplication of a vector by a real. Here is the middle of $[AB]$.
\begin{tkzexample}[latex=7cm,small]
@@ -210,8 +211,8 @@ This amounts to applying a homothety or a multiplication of a vector by a real.
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{\tkzcname{tkzDefPointWith} \tkzname{linear normed}}
-In the following example AC=1 and C belongs to $(AB)$.
+\subsubsection{Option \tkzname{linear normed}}
+In the following example $AC=1$ and $C$ belongs to $(AB)$.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1.2]
@@ -233,19 +234,19 @@ In the following example AC=1 and C belongs to $(AB)$.
\subsection{\tkzcname{tkzGetVectxy} }
-Retrieving the coordinates of a vector
+Retrieving the coordinates of a vector.
-\begin{NewMacroBox}{tkzGetVectxy}{\parg{$A,B$}\var{text}}
-Allows to obtain the coordinates of a vector
+\begin{NewMacroBox}{tkzGetVectxy}{\parg{$A,B$}\var{text}}%
+Allows to obtain the coordinates of a vector.
\medskip
-\begin{tabular}{lll}
+\begin{tabular}{lll}%
\toprule
-arguments & example & explication \\
+arguments & example & explication \\
\midrule
-\TAline{(point)\{name of macro\}} {\tkzcname{tkzGetVectxy}(A,B)\{V\}}{\tkzcname{Vx},\tkzcname{Vy} : coordinates of $\overrightarrow{AB}$}
+\TAline{(point)\{name of macro\}} {\tkzcname{tkzGetVectxy}(A,B)\{V\}}{\tkzcname{Vx},\tkzcname{Vy}: coordinates of $\overrightarrow{AB}$}
\end{tabular}
\end{NewMacroBox}