2D Coordinate - Plot a line parallel to another line












0












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enter image description here



I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, $h$, $m$ and $L$.



I want to find out the coordinate of the Unknown Point of Interest, which is the end point of line2 which is parallel to line1.



How can I do this?






geometry trigonometry coordinate-systems







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    0












    $begingroup$


    enter image description here



    I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, $h$, $m$ and $L$.



    I want to find out the coordinate of the Unknown Point of Interest, which is the end point of line2 which is parallel to line1.



    How can I do this?






    geometry trigonometry coordinate-systems







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      enter image description here



      I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, $h$, $m$ and $L$.



      I want to find out the coordinate of the Unknown Point of Interest, which is the end point of line2 which is parallel to line1.



      How can I do this?






      geometry trigonometry coordinate-systems







      share|cite|improve this question









      $endgroup$




      enter image description here



      I am given values of $x_1, y_1, x_2, y_2$, $theta_{1}$, $theta_{2}$, $h$, $m$ and $L$.



      I want to find out the coordinate of the Unknown Point of Interest, which is the end point of line2 which is parallel to line1.



      How can I do this?






      geometry trigonometry coordinate-systems




      geometry trigonometry coordinate-systems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 '18 at 8:44









      Eric KimEric Kim

      1073




      1073






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          Consider a unitary vector $hat{u}_1$ along the direction of line 1,



          $$
          hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
          $$



          The location you are looking for is



          $$
          mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
            $endgroup$
            – Eric Kim
            Nov 26 '18 at 8:55






          • 1




            $begingroup$
            @EricKim Happy to help
            $endgroup$
            – caverac
            Nov 26 '18 at 8:56



















          1












          $begingroup$

          First you know x2 and y2 (the coordinates of the beginnning of line 2)

          Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.

          If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$

          All you need to do is move point using the vector (adding them together):
          $$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
          Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed



          Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Consider a unitary vector $hat{u}_1$ along the direction of line 1,



            $$
            hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
            $$



            The location you are looking for is



            $$
            mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
              $endgroup$
              – Eric Kim
              Nov 26 '18 at 8:55






            • 1




              $begingroup$
              @EricKim Happy to help
              $endgroup$
              – caverac
              Nov 26 '18 at 8:56
















            1












            $begingroup$

            Consider a unitary vector $hat{u}_1$ along the direction of line 1,



            $$
            hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
            $$



            The location you are looking for is



            $$
            mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
              $endgroup$
              – Eric Kim
              Nov 26 '18 at 8:55






            • 1




              $begingroup$
              @EricKim Happy to help
              $endgroup$
              – caverac
              Nov 26 '18 at 8:56














            1












            1








            1





            $begingroup$

            Consider a unitary vector $hat{u}_1$ along the direction of line 1,



            $$
            hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
            $$



            The location you are looking for is



            $$
            mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
            $$






            share|cite|improve this answer









            $endgroup$



            Consider a unitary vector $hat{u}_1$ along the direction of line 1,



            $$
            hat{u}_1 = pmatrix{cos theta_2 \ sintheta_2}
            $$



            The location you are looking for is



            $$
            mbox{unkowon point of interest} = pmatrix{x_2 \ y_2} + L hat{u}_1 = pmatrix{x_2 + Lcostheta_2 \ y_2 + L sintheta_2}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 '18 at 8:50









            caveraccaverac

            14.4k31130




            14.4k31130












            • $begingroup$
              thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
              $endgroup$
              – Eric Kim
              Nov 26 '18 at 8:55






            • 1




              $begingroup$
              @EricKim Happy to help
              $endgroup$
              – caverac
              Nov 26 '18 at 8:56


















            • $begingroup$
              thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
              $endgroup$
              – Eric Kim
              Nov 26 '18 at 8:55






            • 1




              $begingroup$
              @EricKim Happy to help
              $endgroup$
              – caverac
              Nov 26 '18 at 8:56
















            $begingroup$
            thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
            $endgroup$
            – Eric Kim
            Nov 26 '18 at 8:55




            $begingroup$
            thanks for answering two of my questions in a row. your answers really helped me. I'm a python programmer who sucks at trigonometry... haha. I will accept your answers as soon as i can
            $endgroup$
            – Eric Kim
            Nov 26 '18 at 8:55




            1




            1




            $begingroup$
            @EricKim Happy to help
            $endgroup$
            – caverac
            Nov 26 '18 at 8:56




            $begingroup$
            @EricKim Happy to help
            $endgroup$
            – caverac
            Nov 26 '18 at 8:56











            1












            $begingroup$

            First you know x2 and y2 (the coordinates of the beginnning of line 2)

            Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.

            If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$

            All you need to do is move point using the vector (adding them together):
            $$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
            Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed



            Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              First you know x2 and y2 (the coordinates of the beginnning of line 2)

              Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.

              If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$

              All you need to do is move point using the vector (adding them together):
              $$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
              Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed



              Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                First you know x2 and y2 (the coordinates of the beginnning of line 2)

                Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.

                If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$

                All you need to do is move point using the vector (adding them together):
                $$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
                Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed



                Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$






                share|cite|improve this answer









                $endgroup$



                First you know x2 and y2 (the coordinates of the beginnning of line 2)

                Because line 1 and line 2 are parallel, all you need to do is slide the point $(begin{matrix}x2\y2end{matrix})$ along line 2.

                If you consider line 2 as a vector: $vec{v} = (begin{matrix} L*cos(theta_2) \ L*sin(theta_2) end{matrix})$

                All you need to do is move point using the vector (adding them together):
                $$Point = (begin{matrix} x2+L*cos(theta_2)\ y2+L*sin(theta_2)end{matrix})$$
                Note: i think $theta_1$ and $theta_2$ are mixed up on the picture, here i used the picture as my reference. Feel free to swap $theta_1$ and $theta_2$ if needed



                Note2: in theory the slope $m=frac{sin(theta_2)}{cos(theta_2)} = tan(theta_2)$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 '18 at 9:06









                TheD0ubleTTheD0ubleT

                39218




                39218






























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