Reflecting a line with named coordinates
This code does not work with named coordinates (such as the following code). How can I reflect the blue line over the red line by using coordinate names.
documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
tikz-pgf
add a comment |
This code does not work with named coordinates (such as the following code). How can I reflect the blue line over the red line by using coordinate names.
documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
tikz-pgf
add a comment |
This code does not work with named coordinates (such as the following code). How can I reflect the blue line over the red line by using coordinate names.
documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
tikz-pgf
This code does not work with named coordinates (such as the following code). How can I reflect the blue line over the red line by using coordinate names.
documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
tikz-pgf
tikz-pgf
asked 17 hours ago
blackened
1,427713
1,427713
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
One possibility is using the tkz-euclide
package.
To define A1 the mirror image of the point A with respect to the line DE use: tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
documentclass[border=1cm,tikz]{standalone}
usepackage{tkz-euclide}
begin{document}
begin{tikzpicture}
draw[help lines,dashed](0,0)grid(4,4);
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate[label=E] (E) at (2,3);
tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
tkzDefPointBy[reflection=over D--E](B) tkzGetPoint{B1}
tkzDefPointBy[reflection=over D--E](C) tkzGetPoint{C1}
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw [green] (B1)--(A1)--(C1);
end{tikzpicture}
end{document}
add a comment |
A PSTricks solution only for comparison purposes.
documentclass[pstricks,border=12pt]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[PointName=none,PointSymbol=none](8,3)
pstGeonode(1,3){A}(0,0){B}(2,2){C}(4,3){X}(4,0){Y}
pstOrtSym{X}{Y}{A,B,C}[A',B',C']
psline[linecolor=blue](X)(Y)
psline[linecolor=red](A)(B)(C)
psline[linecolor=red](A')(B')(C')
endpspicture
end{document}
add a comment |
AND ONE MORE UPDATE: Doesn't work with rescaling things.
documentclass[tikz]{standalone}
makeatletter
tikzset{get mirror data/.code args={#1--#2}{%pgftransformreset
pgfutil@tempdima=pgf@x
pgfutil@tempdimb=pgf@y
pgfpointanchor{#1}{center}
pgf@xa=pgf@x
pgf@ya=pgf@y
pgfpointanchor{#2}{center}
pgf@xb=pgf@x
pgf@yb=pgf@y
pgfmathsetmacro{tmpt}{2*(-(pgf@ya*(pgf@xb-pgf@xa)) + pgfutil@tempdimb*(pgf@xb-pgf@xa) + (pgf@xa - pgfutil@tempdima)*(pgf@yb-pgf@ya))/((pgf@xb-pgf@xa)^2 + (pgf@yb-pgf@ya)^2)}
advancepgf@xb by-pgf@xa
advancepgf@yb by-pgf@ya
pgfutil@tempdima=tmptpgf@yb
pgfutil@tempdimb=-tmptpgf@xb
},
mirror at/.style args={#1--#2}{get mirror data=#1--#2,xshift=pgfutil@tempdima,
yshift=pgfutil@tempdimb}}
makeatother
begin{document}
begin{tikzpicture}[scale=1]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
path (2,0) coordinate (D) ++ (rnd*120:2) coordinate (E);
draw[blue] (B)--(A)--(C);
draw[blue] ([mirror at=D--E]B)--([mirror at=D--E]A)--([mirror at=D--E]C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
YET ANOTHER UPDATE: (ab)use show path construction
.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.pathreplacing,calc}
makeatletter
tikzset{reflect at/.style args={#1--#2}{decorate,decoration={
show path construction,
lineto code={draw[tikz@textcolor]
($2*($(#1)!(tikzinputsegmentfirst)!(#2)$)-(tikzinputsegmentfirst)$)
-- ($2*($(#1)!(tikzinputsegmentlast)!(#2)$)-(tikzinputsegmentlast)$);}}}}
makeatother
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B)--(A)--(C);
end{tikzpicture}
end{document}
UPDATE: Here is a simple style reflect at
that allows you to reflect straight lines at whatever line you are interested in. No auxiliary coordinates and the like are needed. (Note, however, that you need to use to
instead of --
since this is a transformation that depends on the point, of course.)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect at/.style args={#1--#2}{to path={%
($2*($(#1)!(tikztostart)!(#2)$)-(tikztostart)$)
-- ($2*($(#1)!(tikztotarget)!(#2)$)-(tikztotarget)$)
}}}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B) to (A) (A) to (C);
end{tikzpicture}
end{document}
OLD ANSWER: Paul Gaborit's solution seems to work.
documentclass[tikz]{standalone}
usetikzlibrary{spy,decorations.fractals}
tikzset{
mirror scope/.is family,
mirror scope/angle/.store in=mirrorangle,
mirror scope/center/.store in=mirrorcenter,
mirror setup/.code={tikzset{mirror scope/.cd,#1}},
mirror scope/.style={mirror setup={#1},spy scope={
rectangle,lens={rotate=mirrorangle,yscale=-1,rotate=-1*mirrorangle},size=80cm}},
}
newcommandmirror[1]{spy[overlay,#1] on (mirrorcenter) in node at (mirrorcenter)}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw [help lines] (0,0) grid (4,3);
begin{scope}[mirror scope={center={2,0},angle=90}]
draw[blue] (B) -- (A) -- (C);
draw[red] (D) -- (E);
mirror;
end{scope}
end{tikzpicture}
end{document}
ADDENDUM: A style that computes the reflected coordinates. Unfortunately, the syntax in this version requires to specify the coordinate twice, e.g. there are two B
s in ([reflect=B at D--E]B)
, and it does not work well with global transformations like scale=0.55
. Other than that it uses this answer which shows how to compute the orthogonal projection of a point on a line. (Of course, I could write that I got it from the pgfmanual, but the truth is that I got it from Jake's nice answer...)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect/.style args={#1 at #2--#3}{shift={%
($2*($(#2)!(#1)!(#3)$)-2*(#1)$)
}}}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[orange] ([reflect=B at D--E]B) -- ([reflect=A at D--E]A)
-- ([reflect=C at D--E]C);
end{tikzpicture}
end{document}
Thanks. My main point is that we may have no idea what(D)
and(E)
is; then, I think,[mirror scope={center={2,0},angle=90}]
will have no use. Am I wrong?
– blackened
17 hours ago
@blackenedD
is the mirror center, and sinceE
is aboveD
, the angle is 90 degrees. For general coordinates one could use calc to compute the angle (or write a new style).
– marmot
17 hours ago
Thanks. Among your updated answers, is one superior to another? Or is it just preference?
– blackened
6 hours ago
@blackened I guess the question is what you want to achieve. I think that the second one is rather short. (I do believe that one should be able to simplify it further. I was starting to look attikzoption
for that.)
– marmot
6 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
One possibility is using the tkz-euclide
package.
To define A1 the mirror image of the point A with respect to the line DE use: tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
documentclass[border=1cm,tikz]{standalone}
usepackage{tkz-euclide}
begin{document}
begin{tikzpicture}
draw[help lines,dashed](0,0)grid(4,4);
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate[label=E] (E) at (2,3);
tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
tkzDefPointBy[reflection=over D--E](B) tkzGetPoint{B1}
tkzDefPointBy[reflection=over D--E](C) tkzGetPoint{C1}
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw [green] (B1)--(A1)--(C1);
end{tikzpicture}
end{document}
add a comment |
One possibility is using the tkz-euclide
package.
To define A1 the mirror image of the point A with respect to the line DE use: tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
documentclass[border=1cm,tikz]{standalone}
usepackage{tkz-euclide}
begin{document}
begin{tikzpicture}
draw[help lines,dashed](0,0)grid(4,4);
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate[label=E] (E) at (2,3);
tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
tkzDefPointBy[reflection=over D--E](B) tkzGetPoint{B1}
tkzDefPointBy[reflection=over D--E](C) tkzGetPoint{C1}
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw [green] (B1)--(A1)--(C1);
end{tikzpicture}
end{document}
add a comment |
One possibility is using the tkz-euclide
package.
To define A1 the mirror image of the point A with respect to the line DE use: tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
documentclass[border=1cm,tikz]{standalone}
usepackage{tkz-euclide}
begin{document}
begin{tikzpicture}
draw[help lines,dashed](0,0)grid(4,4);
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate[label=E] (E) at (2,3);
tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
tkzDefPointBy[reflection=over D--E](B) tkzGetPoint{B1}
tkzDefPointBy[reflection=over D--E](C) tkzGetPoint{C1}
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw [green] (B1)--(A1)--(C1);
end{tikzpicture}
end{document}
One possibility is using the tkz-euclide
package.
To define A1 the mirror image of the point A with respect to the line DE use: tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
documentclass[border=1cm,tikz]{standalone}
usepackage{tkz-euclide}
begin{document}
begin{tikzpicture}
draw[help lines,dashed](0,0)grid(4,4);
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate[label=E] (E) at (2,3);
tkzDefPointBy[reflection=over D--E](A) tkzGetPoint{A1}
tkzDefPointBy[reflection=over D--E](B) tkzGetPoint{B1}
tkzDefPointBy[reflection=over D--E](C) tkzGetPoint{C1}
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw [green] (B1)--(A1)--(C1);
end{tikzpicture}
end{document}
edited 17 hours ago
answered 17 hours ago
Hafid Boukhoulda
1,5441516
1,5441516
add a comment |
add a comment |
A PSTricks solution only for comparison purposes.
documentclass[pstricks,border=12pt]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[PointName=none,PointSymbol=none](8,3)
pstGeonode(1,3){A}(0,0){B}(2,2){C}(4,3){X}(4,0){Y}
pstOrtSym{X}{Y}{A,B,C}[A',B',C']
psline[linecolor=blue](X)(Y)
psline[linecolor=red](A)(B)(C)
psline[linecolor=red](A')(B')(C')
endpspicture
end{document}
add a comment |
A PSTricks solution only for comparison purposes.
documentclass[pstricks,border=12pt]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[PointName=none,PointSymbol=none](8,3)
pstGeonode(1,3){A}(0,0){B}(2,2){C}(4,3){X}(4,0){Y}
pstOrtSym{X}{Y}{A,B,C}[A',B',C']
psline[linecolor=blue](X)(Y)
psline[linecolor=red](A)(B)(C)
psline[linecolor=red](A')(B')(C')
endpspicture
end{document}
add a comment |
A PSTricks solution only for comparison purposes.
documentclass[pstricks,border=12pt]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[PointName=none,PointSymbol=none](8,3)
pstGeonode(1,3){A}(0,0){B}(2,2){C}(4,3){X}(4,0){Y}
pstOrtSym{X}{Y}{A,B,C}[A',B',C']
psline[linecolor=blue](X)(Y)
psline[linecolor=red](A)(B)(C)
psline[linecolor=red](A')(B')(C')
endpspicture
end{document}
A PSTricks solution only for comparison purposes.
documentclass[pstricks,border=12pt]{standalone}
usepackage{pst-eucl}
begin{document}
pspicture[PointName=none,PointSymbol=none](8,3)
pstGeonode(1,3){A}(0,0){B}(2,2){C}(4,3){X}(4,0){Y}
pstOrtSym{X}{Y}{A,B,C}[A',B',C']
psline[linecolor=blue](X)(Y)
psline[linecolor=red](A)(B)(C)
psline[linecolor=red](A')(B')(C')
endpspicture
end{document}
answered 15 hours ago
God Must Be Crazy
5,40511039
5,40511039
add a comment |
add a comment |
AND ONE MORE UPDATE: Doesn't work with rescaling things.
documentclass[tikz]{standalone}
makeatletter
tikzset{get mirror data/.code args={#1--#2}{%pgftransformreset
pgfutil@tempdima=pgf@x
pgfutil@tempdimb=pgf@y
pgfpointanchor{#1}{center}
pgf@xa=pgf@x
pgf@ya=pgf@y
pgfpointanchor{#2}{center}
pgf@xb=pgf@x
pgf@yb=pgf@y
pgfmathsetmacro{tmpt}{2*(-(pgf@ya*(pgf@xb-pgf@xa)) + pgfutil@tempdimb*(pgf@xb-pgf@xa) + (pgf@xa - pgfutil@tempdima)*(pgf@yb-pgf@ya))/((pgf@xb-pgf@xa)^2 + (pgf@yb-pgf@ya)^2)}
advancepgf@xb by-pgf@xa
advancepgf@yb by-pgf@ya
pgfutil@tempdima=tmptpgf@yb
pgfutil@tempdimb=-tmptpgf@xb
},
mirror at/.style args={#1--#2}{get mirror data=#1--#2,xshift=pgfutil@tempdima,
yshift=pgfutil@tempdimb}}
makeatother
begin{document}
begin{tikzpicture}[scale=1]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
path (2,0) coordinate (D) ++ (rnd*120:2) coordinate (E);
draw[blue] (B)--(A)--(C);
draw[blue] ([mirror at=D--E]B)--([mirror at=D--E]A)--([mirror at=D--E]C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
YET ANOTHER UPDATE: (ab)use show path construction
.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.pathreplacing,calc}
makeatletter
tikzset{reflect at/.style args={#1--#2}{decorate,decoration={
show path construction,
lineto code={draw[tikz@textcolor]
($2*($(#1)!(tikzinputsegmentfirst)!(#2)$)-(tikzinputsegmentfirst)$)
-- ($2*($(#1)!(tikzinputsegmentlast)!(#2)$)-(tikzinputsegmentlast)$);}}}}
makeatother
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B)--(A)--(C);
end{tikzpicture}
end{document}
UPDATE: Here is a simple style reflect at
that allows you to reflect straight lines at whatever line you are interested in. No auxiliary coordinates and the like are needed. (Note, however, that you need to use to
instead of --
since this is a transformation that depends on the point, of course.)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect at/.style args={#1--#2}{to path={%
($2*($(#1)!(tikztostart)!(#2)$)-(tikztostart)$)
-- ($2*($(#1)!(tikztotarget)!(#2)$)-(tikztotarget)$)
}}}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B) to (A) (A) to (C);
end{tikzpicture}
end{document}
OLD ANSWER: Paul Gaborit's solution seems to work.
documentclass[tikz]{standalone}
usetikzlibrary{spy,decorations.fractals}
tikzset{
mirror scope/.is family,
mirror scope/angle/.store in=mirrorangle,
mirror scope/center/.store in=mirrorcenter,
mirror setup/.code={tikzset{mirror scope/.cd,#1}},
mirror scope/.style={mirror setup={#1},spy scope={
rectangle,lens={rotate=mirrorangle,yscale=-1,rotate=-1*mirrorangle},size=80cm}},
}
newcommandmirror[1]{spy[overlay,#1] on (mirrorcenter) in node at (mirrorcenter)}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw [help lines] (0,0) grid (4,3);
begin{scope}[mirror scope={center={2,0},angle=90}]
draw[blue] (B) -- (A) -- (C);
draw[red] (D) -- (E);
mirror;
end{scope}
end{tikzpicture}
end{document}
ADDENDUM: A style that computes the reflected coordinates. Unfortunately, the syntax in this version requires to specify the coordinate twice, e.g. there are two B
s in ([reflect=B at D--E]B)
, and it does not work well with global transformations like scale=0.55
. Other than that it uses this answer which shows how to compute the orthogonal projection of a point on a line. (Of course, I could write that I got it from the pgfmanual, but the truth is that I got it from Jake's nice answer...)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect/.style args={#1 at #2--#3}{shift={%
($2*($(#2)!(#1)!(#3)$)-2*(#1)$)
}}}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[orange] ([reflect=B at D--E]B) -- ([reflect=A at D--E]A)
-- ([reflect=C at D--E]C);
end{tikzpicture}
end{document}
Thanks. My main point is that we may have no idea what(D)
and(E)
is; then, I think,[mirror scope={center={2,0},angle=90}]
will have no use. Am I wrong?
– blackened
17 hours ago
@blackenedD
is the mirror center, and sinceE
is aboveD
, the angle is 90 degrees. For general coordinates one could use calc to compute the angle (or write a new style).
– marmot
17 hours ago
Thanks. Among your updated answers, is one superior to another? Or is it just preference?
– blackened
6 hours ago
@blackened I guess the question is what you want to achieve. I think that the second one is rather short. (I do believe that one should be able to simplify it further. I was starting to look attikzoption
for that.)
– marmot
6 hours ago
add a comment |
AND ONE MORE UPDATE: Doesn't work with rescaling things.
documentclass[tikz]{standalone}
makeatletter
tikzset{get mirror data/.code args={#1--#2}{%pgftransformreset
pgfutil@tempdima=pgf@x
pgfutil@tempdimb=pgf@y
pgfpointanchor{#1}{center}
pgf@xa=pgf@x
pgf@ya=pgf@y
pgfpointanchor{#2}{center}
pgf@xb=pgf@x
pgf@yb=pgf@y
pgfmathsetmacro{tmpt}{2*(-(pgf@ya*(pgf@xb-pgf@xa)) + pgfutil@tempdimb*(pgf@xb-pgf@xa) + (pgf@xa - pgfutil@tempdima)*(pgf@yb-pgf@ya))/((pgf@xb-pgf@xa)^2 + (pgf@yb-pgf@ya)^2)}
advancepgf@xb by-pgf@xa
advancepgf@yb by-pgf@ya
pgfutil@tempdima=tmptpgf@yb
pgfutil@tempdimb=-tmptpgf@xb
},
mirror at/.style args={#1--#2}{get mirror data=#1--#2,xshift=pgfutil@tempdima,
yshift=pgfutil@tempdimb}}
makeatother
begin{document}
begin{tikzpicture}[scale=1]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
path (2,0) coordinate (D) ++ (rnd*120:2) coordinate (E);
draw[blue] (B)--(A)--(C);
draw[blue] ([mirror at=D--E]B)--([mirror at=D--E]A)--([mirror at=D--E]C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
YET ANOTHER UPDATE: (ab)use show path construction
.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.pathreplacing,calc}
makeatletter
tikzset{reflect at/.style args={#1--#2}{decorate,decoration={
show path construction,
lineto code={draw[tikz@textcolor]
($2*($(#1)!(tikzinputsegmentfirst)!(#2)$)-(tikzinputsegmentfirst)$)
-- ($2*($(#1)!(tikzinputsegmentlast)!(#2)$)-(tikzinputsegmentlast)$);}}}}
makeatother
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B)--(A)--(C);
end{tikzpicture}
end{document}
UPDATE: Here is a simple style reflect at
that allows you to reflect straight lines at whatever line you are interested in. No auxiliary coordinates and the like are needed. (Note, however, that you need to use to
instead of --
since this is a transformation that depends on the point, of course.)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect at/.style args={#1--#2}{to path={%
($2*($(#1)!(tikztostart)!(#2)$)-(tikztostart)$)
-- ($2*($(#1)!(tikztotarget)!(#2)$)-(tikztotarget)$)
}}}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B) to (A) (A) to (C);
end{tikzpicture}
end{document}
OLD ANSWER: Paul Gaborit's solution seems to work.
documentclass[tikz]{standalone}
usetikzlibrary{spy,decorations.fractals}
tikzset{
mirror scope/.is family,
mirror scope/angle/.store in=mirrorangle,
mirror scope/center/.store in=mirrorcenter,
mirror setup/.code={tikzset{mirror scope/.cd,#1}},
mirror scope/.style={mirror setup={#1},spy scope={
rectangle,lens={rotate=mirrorangle,yscale=-1,rotate=-1*mirrorangle},size=80cm}},
}
newcommandmirror[1]{spy[overlay,#1] on (mirrorcenter) in node at (mirrorcenter)}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw [help lines] (0,0) grid (4,3);
begin{scope}[mirror scope={center={2,0},angle=90}]
draw[blue] (B) -- (A) -- (C);
draw[red] (D) -- (E);
mirror;
end{scope}
end{tikzpicture}
end{document}
ADDENDUM: A style that computes the reflected coordinates. Unfortunately, the syntax in this version requires to specify the coordinate twice, e.g. there are two B
s in ([reflect=B at D--E]B)
, and it does not work well with global transformations like scale=0.55
. Other than that it uses this answer which shows how to compute the orthogonal projection of a point on a line. (Of course, I could write that I got it from the pgfmanual, but the truth is that I got it from Jake's nice answer...)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect/.style args={#1 at #2--#3}{shift={%
($2*($(#2)!(#1)!(#3)$)-2*(#1)$)
}}}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[orange] ([reflect=B at D--E]B) -- ([reflect=A at D--E]A)
-- ([reflect=C at D--E]C);
end{tikzpicture}
end{document}
Thanks. My main point is that we may have no idea what(D)
and(E)
is; then, I think,[mirror scope={center={2,0},angle=90}]
will have no use. Am I wrong?
– blackened
17 hours ago
@blackenedD
is the mirror center, and sinceE
is aboveD
, the angle is 90 degrees. For general coordinates one could use calc to compute the angle (or write a new style).
– marmot
17 hours ago
Thanks. Among your updated answers, is one superior to another? Or is it just preference?
– blackened
6 hours ago
@blackened I guess the question is what you want to achieve. I think that the second one is rather short. (I do believe that one should be able to simplify it further. I was starting to look attikzoption
for that.)
– marmot
6 hours ago
add a comment |
AND ONE MORE UPDATE: Doesn't work with rescaling things.
documentclass[tikz]{standalone}
makeatletter
tikzset{get mirror data/.code args={#1--#2}{%pgftransformreset
pgfutil@tempdima=pgf@x
pgfutil@tempdimb=pgf@y
pgfpointanchor{#1}{center}
pgf@xa=pgf@x
pgf@ya=pgf@y
pgfpointanchor{#2}{center}
pgf@xb=pgf@x
pgf@yb=pgf@y
pgfmathsetmacro{tmpt}{2*(-(pgf@ya*(pgf@xb-pgf@xa)) + pgfutil@tempdimb*(pgf@xb-pgf@xa) + (pgf@xa - pgfutil@tempdima)*(pgf@yb-pgf@ya))/((pgf@xb-pgf@xa)^2 + (pgf@yb-pgf@ya)^2)}
advancepgf@xb by-pgf@xa
advancepgf@yb by-pgf@ya
pgfutil@tempdima=tmptpgf@yb
pgfutil@tempdimb=-tmptpgf@xb
},
mirror at/.style args={#1--#2}{get mirror data=#1--#2,xshift=pgfutil@tempdima,
yshift=pgfutil@tempdimb}}
makeatother
begin{document}
begin{tikzpicture}[scale=1]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
path (2,0) coordinate (D) ++ (rnd*120:2) coordinate (E);
draw[blue] (B)--(A)--(C);
draw[blue] ([mirror at=D--E]B)--([mirror at=D--E]A)--([mirror at=D--E]C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
YET ANOTHER UPDATE: (ab)use show path construction
.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.pathreplacing,calc}
makeatletter
tikzset{reflect at/.style args={#1--#2}{decorate,decoration={
show path construction,
lineto code={draw[tikz@textcolor]
($2*($(#1)!(tikzinputsegmentfirst)!(#2)$)-(tikzinputsegmentfirst)$)
-- ($2*($(#1)!(tikzinputsegmentlast)!(#2)$)-(tikzinputsegmentlast)$);}}}}
makeatother
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B)--(A)--(C);
end{tikzpicture}
end{document}
UPDATE: Here is a simple style reflect at
that allows you to reflect straight lines at whatever line you are interested in. No auxiliary coordinates and the like are needed. (Note, however, that you need to use to
instead of --
since this is a transformation that depends on the point, of course.)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect at/.style args={#1--#2}{to path={%
($2*($(#1)!(tikztostart)!(#2)$)-(tikztostart)$)
-- ($2*($(#1)!(tikztotarget)!(#2)$)-(tikztotarget)$)
}}}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B) to (A) (A) to (C);
end{tikzpicture}
end{document}
OLD ANSWER: Paul Gaborit's solution seems to work.
documentclass[tikz]{standalone}
usetikzlibrary{spy,decorations.fractals}
tikzset{
mirror scope/.is family,
mirror scope/angle/.store in=mirrorangle,
mirror scope/center/.store in=mirrorcenter,
mirror setup/.code={tikzset{mirror scope/.cd,#1}},
mirror scope/.style={mirror setup={#1},spy scope={
rectangle,lens={rotate=mirrorangle,yscale=-1,rotate=-1*mirrorangle},size=80cm}},
}
newcommandmirror[1]{spy[overlay,#1] on (mirrorcenter) in node at (mirrorcenter)}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw [help lines] (0,0) grid (4,3);
begin{scope}[mirror scope={center={2,0},angle=90}]
draw[blue] (B) -- (A) -- (C);
draw[red] (D) -- (E);
mirror;
end{scope}
end{tikzpicture}
end{document}
ADDENDUM: A style that computes the reflected coordinates. Unfortunately, the syntax in this version requires to specify the coordinate twice, e.g. there are two B
s in ([reflect=B at D--E]B)
, and it does not work well with global transformations like scale=0.55
. Other than that it uses this answer which shows how to compute the orthogonal projection of a point on a line. (Of course, I could write that I got it from the pgfmanual, but the truth is that I got it from Jake's nice answer...)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect/.style args={#1 at #2--#3}{shift={%
($2*($(#2)!(#1)!(#3)$)-2*(#1)$)
}}}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[orange] ([reflect=B at D--E]B) -- ([reflect=A at D--E]A)
-- ([reflect=C at D--E]C);
end{tikzpicture}
end{document}
AND ONE MORE UPDATE: Doesn't work with rescaling things.
documentclass[tikz]{standalone}
makeatletter
tikzset{get mirror data/.code args={#1--#2}{%pgftransformreset
pgfutil@tempdima=pgf@x
pgfutil@tempdimb=pgf@y
pgfpointanchor{#1}{center}
pgf@xa=pgf@x
pgf@ya=pgf@y
pgfpointanchor{#2}{center}
pgf@xb=pgf@x
pgf@yb=pgf@y
pgfmathsetmacro{tmpt}{2*(-(pgf@ya*(pgf@xb-pgf@xa)) + pgfutil@tempdimb*(pgf@xb-pgf@xa) + (pgf@xa - pgfutil@tempdima)*(pgf@yb-pgf@ya))/((pgf@xb-pgf@xa)^2 + (pgf@yb-pgf@ya)^2)}
advancepgf@xb by-pgf@xa
advancepgf@yb by-pgf@ya
pgfutil@tempdima=tmptpgf@yb
pgfutil@tempdimb=-tmptpgf@xb
},
mirror at/.style args={#1--#2}{get mirror data=#1--#2,xshift=pgfutil@tempdima,
yshift=pgfutil@tempdimb}}
makeatother
begin{document}
begin{tikzpicture}[scale=1]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
path (2,0) coordinate (D) ++ (rnd*120:2) coordinate (E);
draw[blue] (B)--(A)--(C);
draw[blue] ([mirror at=D--E]B)--([mirror at=D--E]A)--([mirror at=D--E]C);
draw[red] (D)--(E);
end{tikzpicture}
end{document}
YET ANOTHER UPDATE: (ab)use show path construction
.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{decorations.pathreplacing,calc}
makeatletter
tikzset{reflect at/.style args={#1--#2}{decorate,decoration={
show path construction,
lineto code={draw[tikz@textcolor]
($2*($(#1)!(tikzinputsegmentfirst)!(#2)$)-(tikzinputsegmentfirst)$)
-- ($2*($(#1)!(tikzinputsegmentlast)!(#2)$)-(tikzinputsegmentlast)$);}}}}
makeatother
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B)--(A)--(C);
end{tikzpicture}
end{document}
UPDATE: Here is a simple style reflect at
that allows you to reflect straight lines at whatever line you are interested in. No auxiliary coordinates and the like are needed. (Note, however, that you need to use to
instead of --
since this is a transformation that depends on the point, of course.)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect at/.style args={#1--#2}{to path={%
($2*($(#1)!(tikztostart)!(#2)$)-(tikztostart)$)
-- ($2*($(#1)!(tikztotarget)!(#2)$)-(tikztotarget)$)
}}}
begin{document}
begin{tikzpicture}[scale=0.55]
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[blue,reflect at=D--E] (B) to (A) (A) to (C);
end{tikzpicture}
end{document}
OLD ANSWER: Paul Gaborit's solution seems to work.
documentclass[tikz]{standalone}
usetikzlibrary{spy,decorations.fractals}
tikzset{
mirror scope/.is family,
mirror scope/angle/.store in=mirrorangle,
mirror scope/center/.store in=mirrorcenter,
mirror setup/.code={tikzset{mirror scope/.cd,#1}},
mirror scope/.style={mirror setup={#1},spy scope={
rectangle,lens={rotate=mirrorangle,yscale=-1,rotate=-1*mirrorangle},size=80cm}},
}
newcommandmirror[1]{spy[overlay,#1] on (mirrorcenter) in node at (mirrorcenter)}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw [help lines] (0,0) grid (4,3);
begin{scope}[mirror scope={center={2,0},angle=90}]
draw[blue] (B) -- (A) -- (C);
draw[red] (D) -- (E);
mirror;
end{scope}
end{tikzpicture}
end{document}
ADDENDUM: A style that computes the reflected coordinates. Unfortunately, the syntax in this version requires to specify the coordinate twice, e.g. there are two B
s in ([reflect=B at D--E]B)
, and it does not work well with global transformations like scale=0.55
. Other than that it uses this answer which shows how to compute the orthogonal projection of a point on a line. (Of course, I could write that I got it from the pgfmanual, but the truth is that I got it from Jake's nice answer...)
documentclass[tikz]{standalone}
usetikzlibrary{calc}
tikzset{reflect/.style args={#1 at #2--#3}{shift={%
($2*($(#2)!(#1)!(#3)$)-2*(#1)$)
}}}
begin{document}
begin{tikzpicture}
coordinate (A) at (0,0);
coordinate (B) at (1,1);
coordinate (C) at (1,2);
coordinate (D) at (2,0);
coordinate (E) at (2,3);
draw[blue] (B)--(A)--(C);
draw[red] (D)--(E);
draw[orange] ([reflect=B at D--E]B) -- ([reflect=A at D--E]A)
-- ([reflect=C at D--E]C);
end{tikzpicture}
end{document}
edited 4 hours ago
answered 17 hours ago
marmot
85.9k499183
85.9k499183
Thanks. My main point is that we may have no idea what(D)
and(E)
is; then, I think,[mirror scope={center={2,0},angle=90}]
will have no use. Am I wrong?
– blackened
17 hours ago
@blackenedD
is the mirror center, and sinceE
is aboveD
, the angle is 90 degrees. For general coordinates one could use calc to compute the angle (or write a new style).
– marmot
17 hours ago
Thanks. Among your updated answers, is one superior to another? Or is it just preference?
– blackened
6 hours ago
@blackened I guess the question is what you want to achieve. I think that the second one is rather short. (I do believe that one should be able to simplify it further. I was starting to look attikzoption
for that.)
– marmot
6 hours ago
add a comment |
Thanks. My main point is that we may have no idea what(D)
and(E)
is; then, I think,[mirror scope={center={2,0},angle=90}]
will have no use. Am I wrong?
– blackened
17 hours ago
@blackenedD
is the mirror center, and sinceE
is aboveD
, the angle is 90 degrees. For general coordinates one could use calc to compute the angle (or write a new style).
– marmot
17 hours ago
Thanks. Among your updated answers, is one superior to another? Or is it just preference?
– blackened
6 hours ago
@blackened I guess the question is what you want to achieve. I think that the second one is rather short. (I do believe that one should be able to simplify it further. I was starting to look attikzoption
for that.)
– marmot
6 hours ago
Thanks. My main point is that we may have no idea what
(D)
and (E)
is; then, I think, [mirror scope={center={2,0},angle=90}]
will have no use. Am I wrong?– blackened
17 hours ago
Thanks. My main point is that we may have no idea what
(D)
and (E)
is; then, I think, [mirror scope={center={2,0},angle=90}]
will have no use. Am I wrong?– blackened
17 hours ago
@blackened
D
is the mirror center, and since E
is above D
, the angle is 90 degrees. For general coordinates one could use calc to compute the angle (or write a new style).– marmot
17 hours ago
@blackened
D
is the mirror center, and since E
is above D
, the angle is 90 degrees. For general coordinates one could use calc to compute the angle (or write a new style).– marmot
17 hours ago
Thanks. Among your updated answers, is one superior to another? Or is it just preference?
– blackened
6 hours ago
Thanks. Among your updated answers, is one superior to another? Or is it just preference?
– blackened
6 hours ago
@blackened I guess the question is what you want to achieve. I think that the second one is rather short. (I do believe that one should be able to simplify it further. I was starting to look at
tikzoption
for that.)– marmot
6 hours ago
@blackened I guess the question is what you want to achieve. I think that the second one is rather short. (I do believe that one should be able to simplify it further. I was starting to look at
tikzoption
for that.)– marmot
6 hours ago
add a comment |
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