How to convert a straight line into polar coordinates?












0












$begingroup$


I'm trying to go through a simple exercise for the Hough transform where I have a simple straight line in the form of $;y=-x+5;$ and I want to obtain polar coordinates $;(rho,theta);$, I know polar coordinates can be represented by $;rho = x⋅cos(theta) + y⋅sin(theta),;$ what are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format



edit: maybe I am not wording it clearly, this is the question word for word:
enter image description here






geometry polar-coordinates







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$endgroup$












  • $begingroup$
    Are you sure that’s what $p$ is supposed to be?
    $endgroup$
    – amd
    Nov 30 '18 at 20:15










  • $begingroup$
    thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
    $endgroup$
    – dshawn
    Nov 30 '18 at 23:02








  • 2




    $begingroup$
    "I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
    $endgroup$
    – Yves Daoust
    Nov 30 '18 at 23:16










  • $begingroup$
    Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
    $endgroup$
    – dshawn
    Dec 1 '18 at 1:57


















0












$begingroup$


I'm trying to go through a simple exercise for the Hough transform where I have a simple straight line in the form of $;y=-x+5;$ and I want to obtain polar coordinates $;(rho,theta);$, I know polar coordinates can be represented by $;rho = x⋅cos(theta) + y⋅sin(theta),;$ what are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format



edit: maybe I am not wording it clearly, this is the question word for word:
enter image description here






geometry polar-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure that’s what $p$ is supposed to be?
    $endgroup$
    – amd
    Nov 30 '18 at 20:15










  • $begingroup$
    thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
    $endgroup$
    – dshawn
    Nov 30 '18 at 23:02








  • 2




    $begingroup$
    "I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
    $endgroup$
    – Yves Daoust
    Nov 30 '18 at 23:16










  • $begingroup$
    Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
    $endgroup$
    – dshawn
    Dec 1 '18 at 1:57
















0












0








0





$begingroup$


I'm trying to go through a simple exercise for the Hough transform where I have a simple straight line in the form of $;y=-x+5;$ and I want to obtain polar coordinates $;(rho,theta);$, I know polar coordinates can be represented by $;rho = x⋅cos(theta) + y⋅sin(theta),;$ what are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format



edit: maybe I am not wording it clearly, this is the question word for word:
enter image description here






geometry polar-coordinates







share|cite|improve this question











$endgroup$




I'm trying to go through a simple exercise for the Hough transform where I have a simple straight line in the form of $;y=-x+5;$ and I want to obtain polar coordinates $;(rho,theta);$, I know polar coordinates can be represented by $;rho = x⋅cos(theta) + y⋅sin(theta),;$ what are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format



edit: maybe I am not wording it clearly, this is the question word for word:
enter image description here






geometry polar-coordinates




geometry polar-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 20:56









user376343

3,7883827




3,7883827










asked Nov 30 '18 at 17:25









dshawndshawn

324




324












  • $begingroup$
    Are you sure that’s what $p$ is supposed to be?
    $endgroup$
    – amd
    Nov 30 '18 at 20:15










  • $begingroup$
    thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
    $endgroup$
    – dshawn
    Nov 30 '18 at 23:02








  • 2




    $begingroup$
    "I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
    $endgroup$
    – Yves Daoust
    Nov 30 '18 at 23:16










  • $begingroup$
    Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
    $endgroup$
    – dshawn
    Dec 1 '18 at 1:57




















  • $begingroup$
    Are you sure that’s what $p$ is supposed to be?
    $endgroup$
    – amd
    Nov 30 '18 at 20:15










  • $begingroup$
    thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
    $endgroup$
    – dshawn
    Nov 30 '18 at 23:02








  • 2




    $begingroup$
    "I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
    $endgroup$
    – Yves Daoust
    Nov 30 '18 at 23:16










  • $begingroup$
    Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
    $endgroup$
    – dshawn
    Dec 1 '18 at 1:57


















$begingroup$
Are you sure that’s what $p$ is supposed to be?
$endgroup$
– amd
Nov 30 '18 at 20:15




$begingroup$
Are you sure that’s what $p$ is supposed to be?
$endgroup$
– amd
Nov 30 '18 at 20:15












$begingroup$
thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
$endgroup$
– dshawn
Nov 30 '18 at 23:02






$begingroup$
thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
$endgroup$
– dshawn
Nov 30 '18 at 23:02






2




2




$begingroup$
"I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
$endgroup$
– Yves Daoust
Nov 30 '18 at 23:16




$begingroup$
"I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
$endgroup$
– Yves Daoust
Nov 30 '18 at 23:16












$begingroup$
Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
$endgroup$
– dshawn
Dec 1 '18 at 1:57






$begingroup$
Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
$endgroup$
– dshawn
Dec 1 '18 at 1:57












3 Answers
3






active

oldest

votes


















0












$begingroup$

After the definition of the Hough transform is




  • $rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$


  • $theta=frac pi4;$ is the angle between $x-$ axis and $OA.$







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Using polar coordinates, a line is represented as



    $$ax+by+c=arhocostheta+brhosintheta+c=0$$



    or



    $$rho=-frac c{acostheta+bsintheta}.$$




    With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten



    $$rho=frac p{cos(theta-theta_0)},$$



    where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.



      For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        is it $tan ^{-1}?$
        $endgroup$
        – user376343
        Dec 1 '18 at 20:58










      • $begingroup$
        @user376343 Yes, fixed it. Thanks,
        $endgroup$
        – Aditya Dua
        Dec 2 '18 at 3:41











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      After the definition of the Hough transform is




      • $rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$


      • $theta=frac pi4;$ is the angle between $x-$ axis and $OA.$







      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        After the definition of the Hough transform is




        • $rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$


        • $theta=frac pi4;$ is the angle between $x-$ axis and $OA.$







        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          After the definition of the Hough transform is




          • $rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$


          • $theta=frac pi4;$ is the angle between $x-$ axis and $OA.$







          share|cite|improve this answer









          $endgroup$



          After the definition of the Hough transform is




          • $rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$


          • $theta=frac pi4;$ is the angle between $x-$ axis and $OA.$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 21:17









          user376343user376343

          3,7883827




          3,7883827























              0












              $begingroup$

              Using polar coordinates, a line is represented as



              $$ax+by+c=arhocostheta+brhosintheta+c=0$$



              or



              $$rho=-frac c{acostheta+bsintheta}.$$




              With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten



              $$rho=frac p{cos(theta-theta_0)},$$



              where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Using polar coordinates, a line is represented as



                $$ax+by+c=arhocostheta+brhosintheta+c=0$$



                or



                $$rho=-frac c{acostheta+bsintheta}.$$




                With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten



                $$rho=frac p{cos(theta-theta_0)},$$



                where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Using polar coordinates, a line is represented as



                  $$ax+by+c=arhocostheta+brhosintheta+c=0$$



                  or



                  $$rho=-frac c{acostheta+bsintheta}.$$




                  With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten



                  $$rho=frac p{cos(theta-theta_0)},$$



                  where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.






                  share|cite|improve this answer











                  $endgroup$



                  Using polar coordinates, a line is represented as



                  $$ax+by+c=arhocostheta+brhosintheta+c=0$$



                  or



                  $$rho=-frac c{acostheta+bsintheta}.$$




                  With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten



                  $$rho=frac p{cos(theta-theta_0)},$$



                  where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 23:20

























                  answered Nov 30 '18 at 23:13









                  Yves DaoustYves Daoust

                  127k673226




                  127k673226























                      0












                      $begingroup$

                      You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.



                      For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        is it $tan ^{-1}?$
                        $endgroup$
                        – user376343
                        Dec 1 '18 at 20:58










                      • $begingroup$
                        @user376343 Yes, fixed it. Thanks,
                        $endgroup$
                        – Aditya Dua
                        Dec 2 '18 at 3:41
















                      0












                      $begingroup$

                      You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.



                      For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        is it $tan ^{-1}?$
                        $endgroup$
                        – user376343
                        Dec 1 '18 at 20:58










                      • $begingroup$
                        @user376343 Yes, fixed it. Thanks,
                        $endgroup$
                        – Aditya Dua
                        Dec 2 '18 at 3:41














                      0












                      0








                      0





                      $begingroup$

                      You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.



                      For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.






                      share|cite|improve this answer











                      $endgroup$



                      You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.



                      For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 2 '18 at 3:41

























                      answered Nov 30 '18 at 17:39









                      Aditya DuaAditya Dua

                      1,16918




                      1,16918












                      • $begingroup$
                        is it $tan ^{-1}?$
                        $endgroup$
                        – user376343
                        Dec 1 '18 at 20:58










                      • $begingroup$
                        @user376343 Yes, fixed it. Thanks,
                        $endgroup$
                        – Aditya Dua
                        Dec 2 '18 at 3:41


















                      • $begingroup$
                        is it $tan ^{-1}?$
                        $endgroup$
                        – user376343
                        Dec 1 '18 at 20:58










                      • $begingroup$
                        @user376343 Yes, fixed it. Thanks,
                        $endgroup$
                        – Aditya Dua
                        Dec 2 '18 at 3:41
















                      $begingroup$
                      is it $tan ^{-1}?$
                      $endgroup$
                      – user376343
                      Dec 1 '18 at 20:58




                      $begingroup$
                      is it $tan ^{-1}?$
                      $endgroup$
                      – user376343
                      Dec 1 '18 at 20:58












                      $begingroup$
                      @user376343 Yes, fixed it. Thanks,
                      $endgroup$
                      – Aditya Dua
                      Dec 2 '18 at 3:41




                      $begingroup$
                      @user376343 Yes, fixed it. Thanks,
                      $endgroup$
                      – Aditya Dua
                      Dec 2 '18 at 3:41


















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