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












3















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
















3















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














3












3








3


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

3,2941820




3,2941820










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


















0














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































    -1














    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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    2 Answers
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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




























      0














      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


























        0












        0








        0






        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

        13k11836




        13k11836























            -1














            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
















            -1














            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














            -1












            -1








            -1






            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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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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