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












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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).






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    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







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      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).






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      asked Nov 24 at 22:56









      WorldGov

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          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











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          1 Answer
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          active

          oldest

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          active

          oldest

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          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


















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