summaryrefslogtreecommitdiff
path: root/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-pointby.tex
blob: 67e357c625f5bd164c50c6fec23a37ec2fe1cf23 (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
\section{Definition of points by transformation}
These transformations are:

\begin{itemize}
   \item translation;
   \item homothety;
   \item orthogonal reflection or symmetry;
   \item central symmetry;
   \item orthogonal projection;
   \item rotation (degrees or radians);
   \item inversion with respect to a circle.
\end{itemize}

\subsection{\tkzcname{tkzDefPointBy}}
The choice of transformations is made through the options. There are two macros, one for the transformation of a single point \tkzcname{tkzDefPointBy} and the other for the transformation of a list of points \tkzcname{tkzDefPointsBy}. By default the image of $A$ is $A'$. For example, we'll write:
\begin{tkzltxexample}[]
\tkzDefPointBy[translation= from A to A'](B) 
\end{tkzltxexample}
The result is in \tkzname{tkzPointResult}

\medskip
\begin{NewMacroBox}{tkzDefPointBy}{\oarg{local options}\parg{pt}}%
The argument is a simple existing point and its image is stored in \tkzname{tkzPointResult}. If you want to keep this point then the macro \tkzcname{tkzGetPoint\{M\}} allows you to assign the name \tkzname{M} to the point.

\begin{tabular}{lll}%
\toprule
arguments &  definition & examples               \\ 
\midrule
\TAline{pt}   {existing point name}   {$(A)$}
\bottomrule
\end{tabular}

\begin{tabular}{lll}%
options     &     & examples                         \\ 
\midrule
\TOline{translation}{= from \#1 to \#2}{[translation=from A to B](E)}
\TOline{homothety}  {= center \#1 ratio \#2}{[homothety=center A ratio .5](E)}
\TOline{reflection} {= over \#1--\#2}{[reflection=over A--B](E)}
\TOline{symmetry }  {= center \#1}{[symmetry=center A](E)}
\TOline{projection }{= onto \#1--\#2}{[projection=onto A--B](E)}
\TOline{rotation }  {= center \#1 angle \#2}{[rotation=center O angle 30](E)}
\TOline{rotation in rad}{= center \#1 angle \#2}{[rotation in rad=center O angle pi/3](E)} 
\TOline{rotation with nodes}{= center \#1 from \#2 to \#3}{[center O from A to B](E)} 
\TOline{inversion}{= center \#1 through \#2}{[inversion =center O through A](E)} 
\TOline{inversion negative}{= center \#1 through \#2}{...} 
\bottomrule
\end{tabular}

\medskip
\emph{The image is only defined and not drawn.}
\end{NewMacroBox} 

\subsubsection{\tkzname{translation}} 

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[>=latex] 
 \tkzDefPoints{0/0/A,3/1/B,3/0/C}
 \tkzDefPointBy[translation= from B to A](C) 
 \tkzGetPoint{D} 
 \tkzDrawPoints[teal](A,B,C,D)
 \tkzLabelPoints[color=teal](A,B,C,D) 
 \tkzDrawSegments[orange,->](A,B D,C) 
\end{tikzpicture} 
\end{tkzexample}

\subsubsection{\tkzname{reflection} (orthogonal symmetry)} 

\begin{tkzexample}[latex=7cm,small] 
\begin{tikzpicture}[scale=.75]
 \tkzDefPoints{-2/-2/A,-1/-1/C,-4/2/D,-4/0/O}    
 \tkzDrawCircle(O,A)
 \tkzDefPointBy[reflection = over C--D](A)
 \tkzGetPoint{A'}
 \tkzDefPointBy[reflection = over C--D](O)
 \tkzGetPoint{O'}
 \tkzDrawCircle(O',A')
 \tkzDrawLine[add= .5 and .5](C,D)
 \tkzDrawPoints(C,D,O,O')
\end{tikzpicture}
\end{tkzexample}


\subsubsection{\tkzname{homothety} and \tkzname{projection}}

\begin{tkzexample}[latex=7cm,small] 
\begin{tikzpicture}
  \tkzDefPoints{0/1/A,5/3/B,3/4/C}
  \tkzDefLine[bisector](B,A,C) \tkzGetPoint{a} 
  \tkzDrawLine[add=0 and 0,color=magenta!50 ](A,a) 
  \tkzDefPointBy[homothety=center A ratio .5](a)   
  \tkzGetPoint{a'} 
  \tkzDefPointBy[projection = onto A--B](a') 
  \tkzGetPoint{k'}  
  \tkzDefPointBy[projection = onto A--B](a) 
  \tkzGetPoint{k} 
  \tkzDrawLines[add= 0 and .3](A,k A,C)   
  \tkzDrawSegments[blue](a',k' a,k) 
  \tkzDrawPoints(a,a',k,k',A)
  \tkzDrawCircles(a',k' a,k)   
  \tkzLabelPoints(a,a',k,A)
\end{tikzpicture}
\end{tkzexample}  


\subsubsection{\tkzname{projection}}
\begin{tkzexample}[latex=7cm,small] 
\begin{tikzpicture}[scale=1.5]  
 \tkzDefPoints{0/0/A,0/4/B}
 \tkzDefTriangle[pythagore](B,A) \tkzGetPoint{C}
 \tkzDefLine[bisector](B,C,A) \tkzGetPoint{c}
 \tkzInterLL(C,c)(A,B)  \tkzGetPoint{D}
 \tkzDefPointBy[projection=onto B--C](D) 
 \tkzGetPoint{G}
 \tkzInterLC(C,D)(D,A) \tkzGetPoints{E}{F}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawSegment(C,D)
 \tkzDrawCircle(D,A)
 \tkzDrawSegment[new](D,G)
 \tkzMarkRightAngle[fill=orange!10,opacity=.4](D,G,B)
 \tkzDrawPoints(A,C,F) \tkzLabelPoints(A,C,F)
 \tkzDrawPoints(B,D,E,G)   
 \tkzLabelPoints[above right](B,D,E)
  \tkzLabelPoints[above](G)
 \end{tikzpicture}
 \end{tkzexample} 

\subsubsection{\tkzname{symmetry} }
\begin{tkzexample}[latex=6cm,small] 
\begin{tikzpicture}[scale=1]
  \tkzDefPoints{2/-1/A,2/2/B,0/0/O}
  \tkzDefPointsBy[symmetry=center O](B,A){}
  \tkzDrawLine(A,A')
  \tkzDrawLine(B,B')
  \tkzMarkAngle[mark=s,arc=lll,
      size=1.5,mkcolor=red](A,O,B) 
  \tkzLabelAngle[pos=2,circle,draw,
    fill=blue!10,font=\scriptsize](A,O,B){$60^{\circ}$}
  \tkzDrawPoints(A,B,O,A',B') 
  \tkzLabelPoints(B,B')
  \tkzLabelPoints[below](A,O,A')  
\end{tikzpicture}   
\end{tkzexample}

\subsubsection{\tkzname{rotation} } 
\begin{tkzexample}[latex=7cm,small] 
\begin{tikzpicture}[scale=0.5] 
 \tkzDefPoints{0/0/A,5/0/B}
 \tkzDrawSegment(A,B)
 \tkzDefPointBy[rotation=center A angle 60](B) 
 \tkzGetPoint{C} 
 \tkzDefPointBy[symmetry=center C](A) 
 \tkzGetPoint{D} 
 \tkzDrawSegment(A,tkzPointResult) 
 \tkzDrawLine(B,D)
 \tkzDrawArc(A,B)(C) \tkzDrawArc(B,C)(A)
 \tkzDrawArc(C,D)(D)  
 \tkzMarkRightAngle(D,B,A) 
 \tkzDrawPoints(A,B) 
 \tkzLabelPoints(A,B)
 \tkzLabelPoints[above](C)
 \tkzLabelPoints[right](D)
\end{tikzpicture}  
\end{tkzexample}  

\subsubsection{\tkzname{rotation in radian}} 
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}
  \tkzDefPoint["$A$" left](1,5){A}
  \tkzDefPoint["$B$" right](4,3){B}
  \tkzDefPointBy[rotation in rad= center A angle pi/3](B)
  \tkzGetPoint{C}  
  \tkzDrawSegment(A,B)
  \tkzDrawPoints(A,B,C) 
  \tkzCompass(A,C)
  \tkzCompass(B,C) 
  \tkzLabelPoints(C)
\end{tikzpicture}
\end{tkzexample} 

\subsubsection{\tkzname{rotation with nodes}} 
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}
 \tkzDefPoint(0,0){O}    
 \tkzDefPoint(0:2){A} 
 \tkzDefPoint(40:2){B}  
 \tkzDefPoint(20:4){C}
 \tkzDrawLine(O,A)
 \tkzDefPointBy[rotation with nodes%
             =center O from A to B](C)  
 \tkzGetPoint{D}
\tkzDrawPoints(A,B,C,D)
\tkzDrawCircle(O,A)
\tkzLabelPoints(A,C,D)
\tkzLabelPoints[above](B)
\end{tikzpicture}
\end{tkzexample} 

\subsubsection{\tkzname{inversion }}

Inversion is the process of transforming points to a corresponding set of points known as their inverse points. Two points $P$ and $P'$ are said to be inverses with respect to an inversion circle having inversion center $O$ and inversion radius $k$ if $P'$ is the perpendicular foot of the altitude of $OQP$, where  $Q$ is a point on the circle such that $OQ$ is perpendicular to $PQ$.\\
 The quantity $k^2$ is known as the circle power (Coxeter 1969, p. 81).
(\url{https://mathworld.wolfram.com/Inversion.html})

Some propositions :
\begin{itemize}
\item The inverse of a circle (not through the center of inversion) is a circle.
\item The inverse of a circle through the center of inversion is a line.
\item The inverse of a line (not through the center of inversion) is a circle through the center of inversion.
\item A circle orthogonal to the circle of inversion is its own inverse.
\item A line through the center of inversion is its own inverse.
\item Angles are preserved in inversion.
\end{itemize}

Explanation:

Directly 
(Center O power=$k^2={OA}^2=OP \times OP'$)

\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.5]
  \tkzDefPoints{4/0/A,6/0/P,0/0/O}
  \tkzDefPointBy[inversion = center O through A](P)
  \tkzGetPoint{P'}
  \tkzDrawSegments(O,P)
  \tkzDrawCircle(O,A)
  \tkzLabelPoints[above right,font=\scriptsize](O,A,P,P')
  \tkzDrawPoints(O,A,P,P')
\end{tikzpicture}
\end{tkzexample} 

\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.5]
  \tkzDefPoints{4/0/A,6/0/P,0/0/O}
  \tkzDefLine[orthogonal=through P](O,P)
  \tkzGetPoint{L}
  \tkzDefLine[tangent from = P](O,A) \tkzGetPoints{R}{Q}
  \tkzDefPointBy[projection=onto O--A](Q) \tkzGetPoint{P'}
  \tkzDrawSegments(O,P O,A)
  \tkzDrawSegments[new](O,P O,Q P,Q Q,P')
  \tkzDrawCircle(O,A)
  \tkzDrawLines[add=1 and 0](P,L)
  \tkzLabelPoints[below,font=\scriptsize](O,P')
  \tkzLabelPoints[above right,font=\scriptsize](P,Q)
  \tkzDrawPoints(O,P) \tkzDrawPoints[new](Q,P')
  \tkzLabelSegment[above](O,Q){$k$}
  \tkzMarkRightAngles(A,P',Q P,Q,O)
  \tkzLabelCircle[above=.5cm,
      font=\scriptsize](O,A)(100){inversion circle}
  \tkzLabelPoint[left,font=\scriptsize](O){inversion center}
  \tkzLabelPoint[left,font=\scriptsize](L){polar}
\end{tikzpicture}
\end{tkzexample} 


\subsubsection{\tkzname{Inversion of lines} ex 1}
\begin{tkzexample}[latex=6cm,small]  
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{0/0/O,3/0/I,4/3/P,6/-3/Q}
\tkzDrawCircle(O,I)
\tkzDefPointBy[projection= onto P--Q](O) \tkzGetPoint{A}
\tkzDefPointBy[inversion = center O through I](A)
\tkzGetPoint{A'}
\tkzDefPointBy[inversion = center O through I](P)
\tkzGetPoint{P'}
\tkzDefCircle[diameter](O,A')\tkzGetPoint{o}
\tkzDrawCircle[new](o,A')
\tkzDrawLines[add=.25 and .25,red](P,Q)
\tkzDrawLines[add=.25 and .25](O,A)
\tkzDrawSegments(O,P)
\tkzDrawPoints(A,P,O) \tkzDrawPoints[new](A',P')
\end{tikzpicture}
\end{tkzexample} 

\subsubsection{\tkzname{inversion of lines} ex 2}
\begin{tkzexample}[latex=6cm,small]  
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{0/0/O,3/0/I,3/2/P,3/-2/Q}
\tkzDrawCircle(O,I)
\tkzDefPointBy[projection= onto P--Q](O) \tkzGetPoint{A}
\tkzDefPointBy[inversion = center O through I](A)
\tkzGetPoint{A'}
\tkzDefPointBy[inversion = center O through I](P)
\tkzGetPoint{P'}
\tkzDefCircle[diameter](O,A')\tkzGetPoint{o}
\tkzDrawCircle[new](o,A')
\tkzDrawLines[add=.25 and .25,red](P,Q)
\tkzDrawLines[add=.25 and .25](O,A)
\tkzDrawSegments(O,P)
\tkzDrawPoints(A,P,O) \tkzDrawPoints[new](A',P')
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\tkzname{inversion of lines} ex 3}
\begin{tkzexample}[latex=6cm,small]  
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{0/0/O,3/0/I,2/1/P,2/-2/Q}
\tkzDrawCircle(O,I)
\tkzDefPointBy[projection= onto P--Q](O) \tkzGetPoint{A}
\tkzDefPointBy[inversion = center O through I](A)
\tkzGetPoint{A'}
\tkzDefPointBy[inversion = center O through I](P)
\tkzGetPoint{P'}
\tkzDefCircle[diameter](O,A')
\tkzDrawCircle[new](I,A')
\tkzDrawLines[add=.25 and .75,red](P,Q)
\tkzDrawLines[add=.25 and .25](O,A')
\tkzDrawSegments(O,P')
\tkzDrawPoints(A,P,O) \tkzDrawPoints[new](A',P')
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\tkzname{inversion} of circle and \tkzname{homothety} }
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.75]
\tkzDefPoints{0/0/O,3/2/A,2/1/P}
\tkzDefLine[tangent from = O](A,P) \tkzGetPoints{T}{X}
\tkzDefPointsBy[homothety = center O%
                ratio 1.25](A,P,T){}
\tkzInterCC(A,P)(A',P') \tkzGetPoints{C}{D}
\tkzCalcLength(A,P)
\tkzGetLength{rAP}
\tkzDefPointOnCircle[R=  center A angle 190 radius \rAP]
\tkzGetPoint{M}
\tkzDefPointBy[inversion = center O through C](M)
\tkzGetPoint{M'}
\tkzDrawCircles[new](A,P A',P')
\tkzDrawCircle(O,C)
\tkzDrawLines[add=0 and .5](O,T' O,A' O,M' O,P')
\tkzDrawPoints(A,A',P,P',O,T,T',M,M')
\tkzLabelPoints(O,T,T',M,M')
\tkzLabelPoints[below](P,P')
\end{tikzpicture}
\end{tkzexample}


\subsubsection{\tkzname{inversion} of Triangle  with respect to the Incircle}
\begin{tkzexample}[latex=6cm,small] 
\begin{tikzpicture}[scale=1]
\tkzDefPoints{0/0/A,5/1/B,3/6/C}
\tkzDefTriangleCenter[in](A,B,C) \tkzGetPoint{O} 
\tkzDefPointBy[projection= onto A--C](O) \tkzGetPoint{b}
\tkzDefPointBy[projection= onto A--C](O) \tkzGetPoint{b}
\tkzDefPointBy[projection= onto B--C](O) \tkzGetPoint{a}
\tkzDefPointBy[projection= onto A--B](O) \tkzGetPoint{c}
\tkzDefPointsBy[inversion = center O through b](a,b,c)%
                                             {Ia,Ib,Ic}
\tkzDefMidPoint(O,Ia) \tkzGetPoint{Ja}
\tkzDefMidPoint(O,Ib) \tkzGetPoint{Jb}
\tkzDefMidPoint(O,Ic) \tkzGetPoint{Jc}
\tkzInterCC(Ja,O)(Jb,O) \tkzGetPoints{O}{x}
\tkzInterCC(Ja,O)(Jc,O) \tkzGetPoints{y}{O}
\tkzInterCC(Jb,O)(Jc,O) \tkzGetPoints{O}{z}
\tkzDrawPolygon(A,B,C)
\tkzDrawCircle(O,b)\tkzDrawPoints(A,B,C,O)
\tkzDrawCircles[dashed,gray](Ja,y Jb,x Jc,z)
\tkzDrawArc[line width=1pt,orange,delta=0](Jb,x)(z)
\tkzDrawArc[line width=1pt,orange,delta=0](Jc,z)(y)
\tkzDrawArc[line width=1pt,orange,delta=0](Ja,y)(x)
\tkzLabelPoint[below](A){$A$}\tkzLabelPoint[above](C){$C$}
\tkzLabelPoint[right](B){$B$}\tkzLabelPoint[below](O){$O$}
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\tkzname{inversion}: orthogonal circle with inversion circle} 
The inversion circle itself, circles orthogonal to it, and lines through the inversion center are invariant under inversion. If the circle meets the reference circle, these invariant points of intersection are also on the inverse circle. See I and J in the next figure.

\begin{tkzexample}[latex=5cm,small]
\begin{tikzpicture}[scale=1]
\tkzDefPoint(0,0){O}\tkzDefPoint(1,0){A}
\tkzDefPoint(-1.5,-1.5){z1} 
\tkzDefPoint(1.5,-1.25){z2} 
\tkzDefCircle[orthogonal through=z1 and z2](O,A)
\tkzGetPoint{c} 
\tkzDrawCircle[new](c,z1) 
\tkzDefPointBy[inversion =  center O through A](z1)
\tkzGetPoint{Z1} 
\tkzInterCC(O,A)(c,z1) \tkzGetPoints{I}{J}
\tkzDefPointBy[inversion =  center O through A](I)
\tkzGetPoint{I'}
\tkzDrawCircle(O,A)
\tkzDrawPoints(O,A,z1,z2) 
\tkzDrawPoints[new](c,Z1,I,J) 
\tkzLabelPoints(O,A,z1,z2,c,Z1,I,J)
\end{tikzpicture}
\end{tkzexample}



For a more complex example see \tkzname{Pappus} \ref{pappus}

\subsubsection{\tkzname{inversion negative}}
It's an inversion followed by a symmetry of center $O$
\begin{tkzexample}[latex=5cm,small]  
\begin{tikzpicture}[scale=1.5]
  \tkzDefPoints{1/0/A,0/0/O}
  \tkzDefPoint(-1.5,-1.5){z1}
  \tkzDefPoint(0.35,-2){z2} 
  \tkzDefPointBy[inversion negative = center O through A](z1)
  \tkzGetPoint{Z1} 
  \tkzDefPointBy[inversion negative = center O through A](z2)
  \tkzGetPoint{Z2}
  \tkzDrawCircle(O,A)  
  \tkzDrawPoints[color=black, fill=red,size=4](Z1,Z2)    
  \tkzDrawSegments(z1,Z1 z2,Z2)
  \tkzDrawPoints[color=black, fill=red,size=4](O,z1,z2)
  \tkzLabelPoints[font=\scriptsize](O,A,z1,z2,Z1,Z2)  
\end{tikzpicture}
\end{tkzexample} 


\newpage
\subsection{Transformation of multiple points; \tkzcname{tkzDefPointsBy} }
Variant of the previous macro for defining multiple images.
You must give the names of the images as arguments, or indicate that the names of the images are formed from the names of the antecedents, leaving the argument empty. 

\begin{tkzltxexample}[]
\tkzDefPointsBy[translation= from A to A'](B,C){}
\end{tkzltxexample}
The images are $B'$ and $C'$.

\begin{tkzltxexample}[]
\tkzDefPointsBy[translation= from A to A'](B,C){D,E} 
\end{tkzltxexample}
The images are $D$ and $E$.

\begin{tkzltxexample}[]
\tkzDefPointsBy[translation= from A to A'](B)
\end{tkzltxexample}
The image is $B'$.
\begin{NewMacroBox}{tkzDefPointsBy}{\oarg{local options}\parg{list of points}\marg{list of points}}%
\begin{tabular}{lll}%
arguments &  examples  &                  \\ 
\midrule
\TAline{\parg{list of points}\marg{list of pts}}{(A,B)\{E,F\}}{$E$,$F$ images of $A$, $B$}   \\
\bottomrule
\end{tabular}

\medskip
If the list of images is empty then the name of the image is the name of the antecedent to which " ' " is added.

\medskip
\begin{tabular}{lll}%
\toprule
options     &     & examples                         \\ 
\midrule
\TOline{translation = from \#1 to \#2}{}{[translation=from A to B](E)\{\}}
\TOline{homothety = center \#1 ratio \#2}{}{[homothety=center A ratio .5](E)\{F\}}
\TOline{reflection = over \#1--\#2}{}{[reflection=over A--B](E)\{F\}}
\TOline{symmetry = center \#1}{}{[symmetry=center A](E)\{F\}}
\TOline{projection = onto \#1--\#2}{}{[projection=onto A--B](E)\{F\}}
\TOline{rotation = center \#1 angle \#2}{}{[rotation=center  angle 30](E)\{F\}}
\TOline{rotation in rad = center \#1 angle \#2}{}{for instance angle pi/3}
\TOline{rotation with nodes = center \#1 from \#2 to \#3}{}{[center O from A to B](E)\{F\}} 
\TOline{inversion = center \#1 through \#2}{}{[inversion = center O through A](E)\{F\}} 
\TOline{inversion negative = center \#1 through \#2}{}{...} 
\bottomrule
\end{tabular}

\medskip
The points are only defined and not drawn.
\end{NewMacroBox}

\subsubsection{\tkzname{translation} of multiple points}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[>=latex] 
 \tkzDefPoints{0/0/A,3/0/B,3/1/A',1/2/C}
 \tkzDefPointsBy[translation= from A to A'](B,C){} 
 \tkzDrawPolygon(A,B,C)
 \tkzDrawPolygon[new](A',B',C')
 \tkzDrawPoints(A,B,C)
 \tkzDrawPoints[new](A',B',C') 
 \tkzLabelPoints(A,B,A',B')  
 \tkzLabelPoints[above](C,C')
 \tkzDrawSegments[color = gray,->,
              style=dashed](A,A' B,B' C,C') 
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\tkzname{symmetry} of multiple points: an oval}

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=0.4]
  \tkzDefPoint(-4,0){I}
  \tkzDefPoint(4,0){J}
  \tkzDefPoint(0,0){O} 
  \tkzInterCC(J,O)(O,J) \tkzGetPoints{L}{H}
  \tkzInterCC(I,O)(O,I) \tkzGetPoints{K}{G} 
  \tkzInterLL(I,K)(J,H) \tkzGetPoint{M}
  \tkzInterLL(I,G)(J,L) \tkzGetPoint{N}
  \tkzDefPointsBy[symmetry=center J](L,H){D,E} 
  \tkzDefPointsBy[symmetry=center I](G,K){C,F}
  \begin{scope}[line style/.style = {very thin,teal}]
    \tkzDrawLines[add=1.5 and 1.5](I,K I,G J,H J,L) 
    \tkzDrawLines[add=.5 and .5](I,J) 
    \tkzDrawCircles(O,I I,O J,O) 
    \tkzDrawArc[delta=0,orange](N,D)(C) 
    \tkzDrawArc[delta=0,orange](M,F)(E) 
    \tkzDrawArc[delta=0,orange](J,E)(D) 
    \tkzDrawArc[delta=0,orange](I,C)(F) 
  \end{scope}   
\end{tikzpicture} 
\end{tkzexample}

\endinput