How do I choose the signs and angles when converting the equation of a straight line from general form to...












0














The general form of a straight line is $ Ax + By + C = 0 $. One way to convert this into the normal form ($ x cos omega + y sin omega = p $) is to find out $ cos omega $ , $ sin omega$ and p individually using the formulae:
$$ cos omega = pm frac{A}{sqrt{A^2 + B^2}} $$
$$ sin omega = pm frac{B}{sqrt{A^2 + B^2}} $$ and
$$ p = pm frac{C}{sqrt{A^2 + B^2}} $$



My question is, once you have these expressions, how do you choose which signs to use in the equation? For example, $ x cos 30^0 + y cos 30^0 = 5$ is not the same as $ x cos 30^0 - y cos 30^0 = 5$. So, how do I choose the signs while converting this way?



I understand that there are simpler alternatives to transform the equation from general form to normal form, but I'd like to know how to do it by this method as well.



Further, how do I choose which the angle $ omega $ if two angles satisfy the expressions obtained. For example, if we consider the equation $ y - 2 = 0$, then $ p = pm 2 $, $ cos omega = 0 $ and $ sin omega = pm 1 $. In this case, we can see that both the angles $ 90^0 $ and $ 270^0 $ satisfy the expressions. (At 90 degrees, cos is 0 and sin is 1, while at 270, cos is 0 and sin is -1).






linear-algebra







share|cite|improve this question



























    0














    The general form of a straight line is $ Ax + By + C = 0 $. One way to convert this into the normal form ($ x cos omega + y sin omega = p $) is to find out $ cos omega $ , $ sin omega$ and p individually using the formulae:
    $$ cos omega = pm frac{A}{sqrt{A^2 + B^2}} $$
    $$ sin omega = pm frac{B}{sqrt{A^2 + B^2}} $$ and
    $$ p = pm frac{C}{sqrt{A^2 + B^2}} $$



    My question is, once you have these expressions, how do you choose which signs to use in the equation? For example, $ x cos 30^0 + y cos 30^0 = 5$ is not the same as $ x cos 30^0 - y cos 30^0 = 5$. So, how do I choose the signs while converting this way?



    I understand that there are simpler alternatives to transform the equation from general form to normal form, but I'd like to know how to do it by this method as well.



    Further, how do I choose which the angle $ omega $ if two angles satisfy the expressions obtained. For example, if we consider the equation $ y - 2 = 0$, then $ p = pm 2 $, $ cos omega = 0 $ and $ sin omega = pm 1 $. In this case, we can see that both the angles $ 90^0 $ and $ 270^0 $ satisfy the expressions. (At 90 degrees, cos is 0 and sin is 1, while at 270, cos is 0 and sin is -1).






    linear-algebra







    share|cite|improve this question

























      0












      0








      0







      The general form of a straight line is $ Ax + By + C = 0 $. One way to convert this into the normal form ($ x cos omega + y sin omega = p $) is to find out $ cos omega $ , $ sin omega$ and p individually using the formulae:
      $$ cos omega = pm frac{A}{sqrt{A^2 + B^2}} $$
      $$ sin omega = pm frac{B}{sqrt{A^2 + B^2}} $$ and
      $$ p = pm frac{C}{sqrt{A^2 + B^2}} $$



      My question is, once you have these expressions, how do you choose which signs to use in the equation? For example, $ x cos 30^0 + y cos 30^0 = 5$ is not the same as $ x cos 30^0 - y cos 30^0 = 5$. So, how do I choose the signs while converting this way?



      I understand that there are simpler alternatives to transform the equation from general form to normal form, but I'd like to know how to do it by this method as well.



      Further, how do I choose which the angle $ omega $ if two angles satisfy the expressions obtained. For example, if we consider the equation $ y - 2 = 0$, then $ p = pm 2 $, $ cos omega = 0 $ and $ sin omega = pm 1 $. In this case, we can see that both the angles $ 90^0 $ and $ 270^0 $ satisfy the expressions. (At 90 degrees, cos is 0 and sin is 1, while at 270, cos is 0 and sin is -1).






      linear-algebra







      share|cite|improve this question













      The general form of a straight line is $ Ax + By + C = 0 $. One way to convert this into the normal form ($ x cos omega + y sin omega = p $) is to find out $ cos omega $ , $ sin omega$ and p individually using the formulae:
      $$ cos omega = pm frac{A}{sqrt{A^2 + B^2}} $$
      $$ sin omega = pm frac{B}{sqrt{A^2 + B^2}} $$ and
      $$ p = pm frac{C}{sqrt{A^2 + B^2}} $$



      My question is, once you have these expressions, how do you choose which signs to use in the equation? For example, $ x cos 30^0 + y cos 30^0 = 5$ is not the same as $ x cos 30^0 - y cos 30^0 = 5$. So, how do I choose the signs while converting this way?



      I understand that there are simpler alternatives to transform the equation from general form to normal form, but I'd like to know how to do it by this method as well.



      Further, how do I choose which the angle $ omega $ if two angles satisfy the expressions obtained. For example, if we consider the equation $ y - 2 = 0$, then $ p = pm 2 $, $ cos omega = 0 $ and $ sin omega = pm 1 $. In this case, we can see that both the angles $ 90^0 $ and $ 270^0 $ satisfy the expressions. (At 90 degrees, cos is 0 and sin is 1, while at 270, cos is 0 and sin is -1).






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 24 at 22:56









      WorldGov

      2489




      2489






















          1 Answer
          1






          active

          oldest

          votes


















          2














          Divide the entire equation by $sqrt{A^2+B^2}$. You get (after moving the free term to the other side): $$xfrac A{sqrt{A^2+B^2}}+yfrac B{sqrt{A^2+B^2}}=-frac C{sqrt{A^2+B^2}}$$
          So if $costheta=frac A{sqrt{A^2+B^2}}$, you need to keep the same sign for $sintheta$, and choose the opposite sign for $p$






          share|cite|improve this answer





















          • Thanks for the answer. I still don't completely understand. Would you mind explaining this for an example? Let's say, we have $ x - sqrt{3} y+ 8 = 0 $. After dividing by $ sqrt{A^2 + B^2} $ and moving the free term, we have, $ frac x2 - y frac { sqrt{3}}{2} = - frac 82 $. The cos term is positive; so, should I make the sin term positive as well? And we should make the p term positive?
            – WorldGov
            Nov 25 at 13:34












          • Sorry, I could have been a little clearer. No. Keep $cosomega=1/2$, $sinomega=-sqrt 3/2$, and $p=-4$. Alternatively, you can switch sign for all of them: $cosomega =-1/2$, $sinomega=sqrt 3/2$, and $p=4$. The meaning of this change can be understood as following: In the first case, you have an angle in the second quadrant, and then a negative displacement. That is equivalent to going in exactly opposite direction, with a positive displacement.
            – Andrei
            Nov 25 at 14:23











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012203%2fhow-do-i-choose-the-signs-and-angles-when-converting-the-equation-of-a-straight%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Divide the entire equation by $sqrt{A^2+B^2}$. You get (after moving the free term to the other side): $$xfrac A{sqrt{A^2+B^2}}+yfrac B{sqrt{A^2+B^2}}=-frac C{sqrt{A^2+B^2}}$$
          So if $costheta=frac A{sqrt{A^2+B^2}}$, you need to keep the same sign for $sintheta$, and choose the opposite sign for $p$






          share|cite|improve this answer





















          • Thanks for the answer. I still don't completely understand. Would you mind explaining this for an example? Let's say, we have $ x - sqrt{3} y+ 8 = 0 $. After dividing by $ sqrt{A^2 + B^2} $ and moving the free term, we have, $ frac x2 - y frac { sqrt{3}}{2} = - frac 82 $. The cos term is positive; so, should I make the sin term positive as well? And we should make the p term positive?
            – WorldGov
            Nov 25 at 13:34












          • Sorry, I could have been a little clearer. No. Keep $cosomega=1/2$, $sinomega=-sqrt 3/2$, and $p=-4$. Alternatively, you can switch sign for all of them: $cosomega =-1/2$, $sinomega=sqrt 3/2$, and $p=4$. The meaning of this change can be understood as following: In the first case, you have an angle in the second quadrant, and then a negative displacement. That is equivalent to going in exactly opposite direction, with a positive displacement.
            – Andrei
            Nov 25 at 14:23
















          2














          Divide the entire equation by $sqrt{A^2+B^2}$. You get (after moving the free term to the other side): $$xfrac A{sqrt{A^2+B^2}}+yfrac B{sqrt{A^2+B^2}}=-frac C{sqrt{A^2+B^2}}$$
          So if $costheta=frac A{sqrt{A^2+B^2}}$, you need to keep the same sign for $sintheta$, and choose the opposite sign for $p$






          share|cite|improve this answer





















          • Thanks for the answer. I still don't completely understand. Would you mind explaining this for an example? Let's say, we have $ x - sqrt{3} y+ 8 = 0 $. After dividing by $ sqrt{A^2 + B^2} $ and moving the free term, we have, $ frac x2 - y frac { sqrt{3}}{2} = - frac 82 $. The cos term is positive; so, should I make the sin term positive as well? And we should make the p term positive?
            – WorldGov
            Nov 25 at 13:34












          • Sorry, I could have been a little clearer. No. Keep $cosomega=1/2$, $sinomega=-sqrt 3/2$, and $p=-4$. Alternatively, you can switch sign for all of them: $cosomega =-1/2$, $sinomega=sqrt 3/2$, and $p=4$. The meaning of this change can be understood as following: In the first case, you have an angle in the second quadrant, and then a negative displacement. That is equivalent to going in exactly opposite direction, with a positive displacement.
            – Andrei
            Nov 25 at 14:23














          2












          2








          2






          Divide the entire equation by $sqrt{A^2+B^2}$. You get (after moving the free term to the other side): $$xfrac A{sqrt{A^2+B^2}}+yfrac B{sqrt{A^2+B^2}}=-frac C{sqrt{A^2+B^2}}$$
          So if $costheta=frac A{sqrt{A^2+B^2}}$, you need to keep the same sign for $sintheta$, and choose the opposite sign for $p$






          share|cite|improve this answer












          Divide the entire equation by $sqrt{A^2+B^2}$. You get (after moving the free term to the other side): $$xfrac A{sqrt{A^2+B^2}}+yfrac B{sqrt{A^2+B^2}}=-frac C{sqrt{A^2+B^2}}$$
          So if $costheta=frac A{sqrt{A^2+B^2}}$, you need to keep the same sign for $sintheta$, and choose the opposite sign for $p$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 23:08









          Andrei

          11.1k21026




          11.1k21026












          • Thanks for the answer. I still don't completely understand. Would you mind explaining this for an example? Let's say, we have $ x - sqrt{3} y+ 8 = 0 $. After dividing by $ sqrt{A^2 + B^2} $ and moving the free term, we have, $ frac x2 - y frac { sqrt{3}}{2} = - frac 82 $. The cos term is positive; so, should I make the sin term positive as well? And we should make the p term positive?
            – WorldGov
            Nov 25 at 13:34












          • Sorry, I could have been a little clearer. No. Keep $cosomega=1/2$, $sinomega=-sqrt 3/2$, and $p=-4$. Alternatively, you can switch sign for all of them: $cosomega =-1/2$, $sinomega=sqrt 3/2$, and $p=4$. The meaning of this change can be understood as following: In the first case, you have an angle in the second quadrant, and then a negative displacement. That is equivalent to going in exactly opposite direction, with a positive displacement.
            – Andrei
            Nov 25 at 14:23


















          • Thanks for the answer. I still don't completely understand. Would you mind explaining this for an example? Let's say, we have $ x - sqrt{3} y+ 8 = 0 $. After dividing by $ sqrt{A^2 + B^2} $ and moving the free term, we have, $ frac x2 - y frac { sqrt{3}}{2} = - frac 82 $. The cos term is positive; so, should I make the sin term positive as well? And we should make the p term positive?
            – WorldGov
            Nov 25 at 13:34












          • Sorry, I could have been a little clearer. No. Keep $cosomega=1/2$, $sinomega=-sqrt 3/2$, and $p=-4$. Alternatively, you can switch sign for all of them: $cosomega =-1/2$, $sinomega=sqrt 3/2$, and $p=4$. The meaning of this change can be understood as following: In the first case, you have an angle in the second quadrant, and then a negative displacement. That is equivalent to going in exactly opposite direction, with a positive displacement.
            – Andrei
            Nov 25 at 14:23
















          Thanks for the answer. I still don't completely understand. Would you mind explaining this for an example? Let's say, we have $ x - sqrt{3} y+ 8 = 0 $. After dividing by $ sqrt{A^2 + B^2} $ and moving the free term, we have, $ frac x2 - y frac { sqrt{3}}{2} = - frac 82 $. The cos term is positive; so, should I make the sin term positive as well? And we should make the p term positive?
          – WorldGov
          Nov 25 at 13:34






          Thanks for the answer. I still don't completely understand. Would you mind explaining this for an example? Let's say, we have $ x - sqrt{3} y+ 8 = 0 $. After dividing by $ sqrt{A^2 + B^2} $ and moving the free term, we have, $ frac x2 - y frac { sqrt{3}}{2} = - frac 82 $. The cos term is positive; so, should I make the sin term positive as well? And we should make the p term positive?
          – WorldGov
          Nov 25 at 13:34














          Sorry, I could have been a little clearer. No. Keep $cosomega=1/2$, $sinomega=-sqrt 3/2$, and $p=-4$. Alternatively, you can switch sign for all of them: $cosomega =-1/2$, $sinomega=sqrt 3/2$, and $p=4$. The meaning of this change can be understood as following: In the first case, you have an angle in the second quadrant, and then a negative displacement. That is equivalent to going in exactly opposite direction, with a positive displacement.
          – Andrei
          Nov 25 at 14:23




          Sorry, I could have been a little clearer. No. Keep $cosomega=1/2$, $sinomega=-sqrt 3/2$, and $p=-4$. Alternatively, you can switch sign for all of them: $cosomega =-1/2$, $sinomega=sqrt 3/2$, and $p=4$. The meaning of this change can be understood as following: In the first case, you have an angle in the second quadrant, and then a negative displacement. That is equivalent to going in exactly opposite direction, with a positive displacement.
          – Andrei
          Nov 25 at 14:23


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012203%2fhow-do-i-choose-the-signs-and-angles-when-converting-the-equation-of-a-straight%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Plaza Victoria

          Image
          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
          Read more

          In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
          Read more

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

          Image
          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
          Read more