Reflecting a line with named coordinates












5














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}









share|improve this question



























    5














    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}









    share|improve this question

























      5












      5








      5


      0





      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}









      share|improve this question













      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






      share|improve this question













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      share|improve this question




      share|improve this question










      asked 2 days ago









      blackened

      1,427714




      1,427714






















          3 Answers
          3






          active

          oldest

          votes


















          3














          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}


          enter image description here



          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}


          enter image description here



          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}


          enter image description here



          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}


          enter image description here



          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 Bs 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}


          enter image description here






          share|improve this answer























          • 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
            2 days 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
            2 days ago












          • Thanks. Among your updated answers, is one superior to another? Or is it just preference?
            – blackened
            2 days 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
            2 days ago










          • (1) I also think the second version is best (works with scale and has clean syntax. (2) Do you think that it is best to remove "the old answer." (It does not seem to serve any purpose, especially for a new-comer.) (3) Is it possible to incorporate "point reflection" in your answer? (Reflect point (C) over line, etc...)
            – blackened
            4 hours ago



















          4














          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}


          enter image description here






          share|improve this answer































            4














            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}


            enter image description here






            share|improve this answer





















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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3














              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}


              enter image description here



              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}


              enter image description here



              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}


              enter image description here



              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}


              enter image description here



              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 Bs 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}


              enter image description here






              share|improve this answer























              • 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
                2 days 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
                2 days ago












              • Thanks. Among your updated answers, is one superior to another? Or is it just preference?
                – blackened
                2 days 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
                2 days ago










              • (1) I also think the second version is best (works with scale and has clean syntax. (2) Do you think that it is best to remove "the old answer." (It does not seem to serve any purpose, especially for a new-comer.) (3) Is it possible to incorporate "point reflection" in your answer? (Reflect point (C) over line, etc...)
                – blackened
                4 hours ago
















              3














              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}


              enter image description here



              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}


              enter image description here



              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}


              enter image description here



              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}


              enter image description here



              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 Bs 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}


              enter image description here






              share|improve this answer























              • 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
                2 days 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
                2 days ago












              • Thanks. Among your updated answers, is one superior to another? Or is it just preference?
                – blackened
                2 days 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
                2 days ago










              • (1) I also think the second version is best (works with scale and has clean syntax. (2) Do you think that it is best to remove "the old answer." (It does not seem to serve any purpose, especially for a new-comer.) (3) Is it possible to incorporate "point reflection" in your answer? (Reflect point (C) over line, etc...)
                – blackened
                4 hours ago














              3












              3








              3






              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}


              enter image description here



              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}


              enter image description here



              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}


              enter image description here



              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}


              enter image description here



              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 Bs 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}


              enter image description here






              share|improve this answer














              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}


              enter image description here



              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}


              enter image description here



              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}


              enter image description here



              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}


              enter image description here



              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 Bs 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}


              enter image description here







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 2 days ago

























              answered 2 days ago









              marmot

              86.5k499184




              86.5k499184












              • 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
                2 days 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
                2 days ago












              • Thanks. Among your updated answers, is one superior to another? Or is it just preference?
                – blackened
                2 days 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
                2 days ago










              • (1) I also think the second version is best (works with scale and has clean syntax. (2) Do you think that it is best to remove "the old answer." (It does not seem to serve any purpose, especially for a new-comer.) (3) Is it possible to incorporate "point reflection" in your answer? (Reflect point (C) over line, etc...)
                – blackened
                4 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
                2 days 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
                2 days ago












              • Thanks. Among your updated answers, is one superior to another? Or is it just preference?
                – blackened
                2 days 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
                2 days ago










              • (1) I also think the second version is best (works with scale and has clean syntax. (2) Do you think that it is best to remove "the old answer." (It does not seem to serve any purpose, especially for a new-comer.) (3) Is it possible to incorporate "point reflection" in your answer? (Reflect point (C) over line, etc...)
                – blackened
                4 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
              2 days 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
              2 days 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
              2 days 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
              2 days ago














              Thanks. Among your updated answers, is one superior to another? Or is it just preference?
              – blackened
              2 days ago




              Thanks. Among your updated answers, is one superior to another? Or is it just preference?
              – blackened
              2 days 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
              2 days 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
              2 days ago












              (1) I also think the second version is best (works with scale and has clean syntax. (2) Do you think that it is best to remove "the old answer." (It does not seem to serve any purpose, especially for a new-comer.) (3) Is it possible to incorporate "point reflection" in your answer? (Reflect point (C) over line, etc...)
              – blackened
              4 hours ago




              (1) I also think the second version is best (works with scale and has clean syntax. (2) Do you think that it is best to remove "the old answer." (It does not seem to serve any purpose, especially for a new-comer.) (3) Is it possible to incorporate "point reflection" in your answer? (Reflect point (C) over line, etc...)
              – blackened
              4 hours ago











              4














              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}


              enter image description here






              share|improve this answer




























                4














                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}


                enter image description here






                share|improve this answer


























                  4












                  4








                  4






                  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}


                  enter image description here






                  share|improve this answer














                  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}


                  enter image description here







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  Hafid Boukhoulda

                  1,5891516




                  1,5891516























                      4














                      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}


                      enter image description here






                      share|improve this answer


























                        4














                        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}


                        enter image description here






                        share|improve this answer
























                          4












                          4








                          4






                          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}


                          enter image description here






                          share|improve this answer












                          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}


                          enter image description here







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 2 days ago









                          God Must Be Crazy

                          5,54011039




                          5,54011039






























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