Determine homogeneous transformation matrix for reflection about the line $y = mx + b$, or specifically $y =...











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Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$, or specifically $ y = 2x – 6$.




I use $mx - y +b =0$: $text{slope} = m$, $tan(theta)= m$



intersection with the axes:



$x =0$ is $y = -b$ and $y =0$ is $x = dfrac{b}{m}$



My question is, what can I do next?






linear-algebra







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  • 1




    What are your thoughts on the problem so far? What have you tried?
    – Cameron Buie
    Nov 4 '13 at 22:21






  • 1




    zebragraph.com/Geometers_Corner_files/Reflection.pdf
    – Mickey Tin
    Apr 2 '14 at 15:14










  • Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
    – Rory Daulton
    Dec 31 '14 at 4:29















up vote
3
down vote

favorite
1













Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$, or specifically $ y = 2x – 6$.




I use $mx - y +b =0$: $text{slope} = m$, $tan(theta)= m$



intersection with the axes:



$x =0$ is $y = -b$ and $y =0$ is $x = dfrac{b}{m}$



My question is, what can I do next?






linear-algebra







share|cite|improve this question




















  • 1




    What are your thoughts on the problem so far? What have you tried?
    – Cameron Buie
    Nov 4 '13 at 22:21






  • 1




    zebragraph.com/Geometers_Corner_files/Reflection.pdf
    – Mickey Tin
    Apr 2 '14 at 15:14










  • Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
    – Rory Daulton
    Dec 31 '14 at 4:29













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1






Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$, or specifically $ y = 2x – 6$.




I use $mx - y +b =0$: $text{slope} = m$, $tan(theta)= m$



intersection with the axes:



$x =0$ is $y = -b$ and $y =0$ is $x = dfrac{b}{m}$



My question is, what can I do next?






linear-algebra







share|cite|improve this question
















Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$, or specifically $ y = 2x – 6$.




I use $mx - y +b =0$: $text{slope} = m$, $tan(theta)= m$



intersection with the axes:



$x =0$ is $y = -b$ and $y =0$ is $x = dfrac{b}{m}$



My question is, what can I do next?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 7 '15 at 6:27









coldnumber

3,2841820




3,2841820










asked Nov 4 '13 at 22:18









Adel Hassan

9317




9317








  • 1




    What are your thoughts on the problem so far? What have you tried?
    – Cameron Buie
    Nov 4 '13 at 22:21






  • 1




    zebragraph.com/Geometers_Corner_files/Reflection.pdf
    – Mickey Tin
    Apr 2 '14 at 15:14










  • Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
    – Rory Daulton
    Dec 31 '14 at 4:29














  • 1




    What are your thoughts on the problem so far? What have you tried?
    – Cameron Buie
    Nov 4 '13 at 22:21






  • 1




    zebragraph.com/Geometers_Corner_files/Reflection.pdf
    – Mickey Tin
    Apr 2 '14 at 15:14










  • Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
    – Rory Daulton
    Dec 31 '14 at 4:29








1




1




What are your thoughts on the problem so far? What have you tried?
– Cameron Buie
Nov 4 '13 at 22:21




What are your thoughts on the problem so far? What have you tried?
– Cameron Buie
Nov 4 '13 at 22:21




1




1




zebragraph.com/Geometers_Corner_files/Reflection.pdf
– Mickey Tin
Apr 2 '14 at 15:14




zebragraph.com/Geometers_Corner_files/Reflection.pdf
– Mickey Tin
Apr 2 '14 at 15:14












Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
– Rory Daulton
Dec 31 '14 at 4:29




Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
– Rory Daulton
Dec 31 '14 at 4:29










2 Answers
2






active

oldest

votes

















up vote
0
down vote













Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:



$$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.



This can be simplified to:
$$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).



Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.



Now, you can simply write the matrix out:



$$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$



In our specific case:
$$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$




You can see it's way easier, if you use the canonical implicit form for the line. If you have
$$ax+by=c$$
where $a^2+b^2=1$, then the matrix simplifies to
$$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
or, if you want, $I-2(a,b,0)^T(a,b,c)$.






share|cite|improve this answer






























    up vote
    -1
    down vote













    Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.






    share|cite|improve this answer





















    • We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
      – orion
      Feb 19 '16 at 8:03











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    Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:



    $$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
    I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.



    This can be simplified to:
    $$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
    The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).



    Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.



    Now, you can simply write the matrix out:



    $$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$



    In our specific case:
    $$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$




    You can see it's way easier, if you use the canonical implicit form for the line. If you have
    $$ax+by=c$$
    where $a^2+b^2=1$, then the matrix simplifies to
    $$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
    or, if you want, $I-2(a,b,0)^T(a,b,c)$.






    share|cite|improve this answer



























      up vote
      0
      down vote













      Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:



      $$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
      I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.



      This can be simplified to:
      $$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
      The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).



      Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.



      Now, you can simply write the matrix out:



      $$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$



      In our specific case:
      $$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$




      You can see it's way easier, if you use the canonical implicit form for the line. If you have
      $$ax+by=c$$
      where $a^2+b^2=1$, then the matrix simplifies to
      $$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
      or, if you want, $I-2(a,b,0)^T(a,b,c)$.






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:



        $$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
        I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.



        This can be simplified to:
        $$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
        The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).



        Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.



        Now, you can simply write the matrix out:



        $$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$



        In our specific case:
        $$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$




        You can see it's way easier, if you use the canonical implicit form for the line. If you have
        $$ax+by=c$$
        where $a^2+b^2=1$, then the matrix simplifies to
        $$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
        or, if you want, $I-2(a,b,0)^T(a,b,c)$.






        share|cite|improve this answer














        Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:



        $$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
        I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.



        This can be simplified to:
        $$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
        The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).



        Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.



        Now, you can simply write the matrix out:



        $$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$



        In our specific case:
        $$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$




        You can see it's way easier, if you use the canonical implicit form for the line. If you have
        $$ax+by=c$$
        where $a^2+b^2=1$, then the matrix simplifies to
        $$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
        or, if you want, $I-2(a,b,0)^T(a,b,c)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 19 '16 at 8:32

























        answered Feb 19 '16 at 8:19









        orion

        12.9k11836




        12.9k11836






















            up vote
            -1
            down vote













            Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.






            share|cite|improve this answer





















            • We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
              – orion
              Feb 19 '16 at 8:03















            up vote
            -1
            down vote













            Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.






            share|cite|improve this answer





















            • We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
              – orion
              Feb 19 '16 at 8:03













            up vote
            -1
            down vote










            up vote
            -1
            down vote









            Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.






            share|cite|improve this answer












            Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 4 '13 at 22:50









            Tim Ratigan

            6,1151028




            6,1151028












            • We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
              – orion
              Feb 19 '16 at 8:03


















            • We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
              – orion
              Feb 19 '16 at 8:03
















            We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
            – orion
            Feb 19 '16 at 8:03




            We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
            – orion
            Feb 19 '16 at 8:03


















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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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