polyfit line coordinates - Matlab










1















I calculated a line based on two points, how can I store all the line coordinates (x and y) in two separated arrays?



Script:



x1=50;
x2=130;
y1=30;
y2=200;
coefficients = polyfit([x1, x2], [y1, y2], 1);
a = coefficients (1);
b = coefficients (2);
plot([x1, x2], [y1, y2], 'b','LineWidth',2)









share|improve this question


























    1















    I calculated a line based on two points, how can I store all the line coordinates (x and y) in two separated arrays?



    Script:



    x1=50;
    x2=130;
    y1=30;
    y2=200;
    coefficients = polyfit([x1, x2], [y1, y2], 1);
    a = coefficients (1);
    b = coefficients (2);
    plot([x1, x2], [y1, y2], 'b','LineWidth',2)









    share|improve this question
























      1












      1








      1








      I calculated a line based on two points, how can I store all the line coordinates (x and y) in two separated arrays?



      Script:



      x1=50;
      x2=130;
      y1=30;
      y2=200;
      coefficients = polyfit([x1, x2], [y1, y2], 1);
      a = coefficients (1);
      b = coefficients (2);
      plot([x1, x2], [y1, y2], 'b','LineWidth',2)









      share|improve this question














      I calculated a line based on two points, how can I store all the line coordinates (x and y) in two separated arrays?



      Script:



      x1=50;
      x2=130;
      y1=30;
      y2=200;
      coefficients = polyfit([x1, x2], [y1, y2], 1);
      a = coefficients (1);
      b = coefficients (2);
      plot([x1, x2], [y1, y2], 'b','LineWidth',2)






      matlab






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 14 '18 at 22:12









      janejane

      697




      697






















          2 Answers
          2






          active

          oldest

          votes


















          3














          Read about polyval. You ca generate the points as shown in the below code:



          x1=50;
          x2=130;
          y1=30;
          y2=200;
          p = polyfit([x1, x2], [y1, y2], 1);

          N = 100 ; % can be changed
          xi = linspace(x1,x2,N) ;
          yi = polyval(p,xi) ;

          plot(xi,yi,'.-r')


          Alternatively you can also use slope/ intercept obtained from polyfit to get the coordinates.



          x1=50;
          x2=130;
          y1=30;
          y2=200;
          p = polyfit([x1, x2], [y1, y2], 1);

          N = 100 ; % can be changed
          xi = linspace(x1,x2,N) ;
          yi = p(2)+p(1)*xi ;
          plot(xi,yi,'.-r')





          share|improve this answer






























            1














            If you have two points, (x1, y1) and (x2, y2), you can directly obtain the line coordinates by slicing:



            n = 100;
            xx=x1:(x2-x1)/n:x2
            yy=y1:(y2-y1)/n:y2


            Where n specifies how many points of coordinate. xx and yy are two arrays storing and coordinates on the line.



            You can also plot the line by



            plot(xx,yy)





            share|improve this answer






















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3














              Read about polyval. You ca generate the points as shown in the below code:



              x1=50;
              x2=130;
              y1=30;
              y2=200;
              p = polyfit([x1, x2], [y1, y2], 1);

              N = 100 ; % can be changed
              xi = linspace(x1,x2,N) ;
              yi = polyval(p,xi) ;

              plot(xi,yi,'.-r')


              Alternatively you can also use slope/ intercept obtained from polyfit to get the coordinates.



              x1=50;
              x2=130;
              y1=30;
              y2=200;
              p = polyfit([x1, x2], [y1, y2], 1);

              N = 100 ; % can be changed
              xi = linspace(x1,x2,N) ;
              yi = p(2)+p(1)*xi ;
              plot(xi,yi,'.-r')





              share|improve this answer



























                3














                Read about polyval. You ca generate the points as shown in the below code:



                x1=50;
                x2=130;
                y1=30;
                y2=200;
                p = polyfit([x1, x2], [y1, y2], 1);

                N = 100 ; % can be changed
                xi = linspace(x1,x2,N) ;
                yi = polyval(p,xi) ;

                plot(xi,yi,'.-r')


                Alternatively you can also use slope/ intercept obtained from polyfit to get the coordinates.



                x1=50;
                x2=130;
                y1=30;
                y2=200;
                p = polyfit([x1, x2], [y1, y2], 1);

                N = 100 ; % can be changed
                xi = linspace(x1,x2,N) ;
                yi = p(2)+p(1)*xi ;
                plot(xi,yi,'.-r')





                share|improve this answer

























                  3












                  3








                  3







                  Read about polyval. You ca generate the points as shown in the below code:



                  x1=50;
                  x2=130;
                  y1=30;
                  y2=200;
                  p = polyfit([x1, x2], [y1, y2], 1);

                  N = 100 ; % can be changed
                  xi = linspace(x1,x2,N) ;
                  yi = polyval(p,xi) ;

                  plot(xi,yi,'.-r')


                  Alternatively you can also use slope/ intercept obtained from polyfit to get the coordinates.



                  x1=50;
                  x2=130;
                  y1=30;
                  y2=200;
                  p = polyfit([x1, x2], [y1, y2], 1);

                  N = 100 ; % can be changed
                  xi = linspace(x1,x2,N) ;
                  yi = p(2)+p(1)*xi ;
                  plot(xi,yi,'.-r')





                  share|improve this answer













                  Read about polyval. You ca generate the points as shown in the below code:



                  x1=50;
                  x2=130;
                  y1=30;
                  y2=200;
                  p = polyfit([x1, x2], [y1, y2], 1);

                  N = 100 ; % can be changed
                  xi = linspace(x1,x2,N) ;
                  yi = polyval(p,xi) ;

                  plot(xi,yi,'.-r')


                  Alternatively you can also use slope/ intercept obtained from polyfit to get the coordinates.



                  x1=50;
                  x2=130;
                  y1=30;
                  y2=200;
                  p = polyfit([x1, x2], [y1, y2], 1);

                  N = 100 ; % can be changed
                  xi = linspace(x1,x2,N) ;
                  yi = p(2)+p(1)*xi ;
                  plot(xi,yi,'.-r')






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 15 '18 at 11:48









                  Siva Srinivas KolukulaSiva Srinivas Kolukula

                  9851512




                  9851512























                      1














                      If you have two points, (x1, y1) and (x2, y2), you can directly obtain the line coordinates by slicing:



                      n = 100;
                      xx=x1:(x2-x1)/n:x2
                      yy=y1:(y2-y1)/n:y2


                      Where n specifies how many points of coordinate. xx and yy are two arrays storing and coordinates on the line.



                      You can also plot the line by



                      plot(xx,yy)





                      share|improve this answer



























                        1














                        If you have two points, (x1, y1) and (x2, y2), you can directly obtain the line coordinates by slicing:



                        n = 100;
                        xx=x1:(x2-x1)/n:x2
                        yy=y1:(y2-y1)/n:y2


                        Where n specifies how many points of coordinate. xx and yy are two arrays storing and coordinates on the line.



                        You can also plot the line by



                        plot(xx,yy)





                        share|improve this answer

























                          1












                          1








                          1







                          If you have two points, (x1, y1) and (x2, y2), you can directly obtain the line coordinates by slicing:



                          n = 100;
                          xx=x1:(x2-x1)/n:x2
                          yy=y1:(y2-y1)/n:y2


                          Where n specifies how many points of coordinate. xx and yy are two arrays storing and coordinates on the line.



                          You can also plot the line by



                          plot(xx,yy)





                          share|improve this answer













                          If you have two points, (x1, y1) and (x2, y2), you can directly obtain the line coordinates by slicing:



                          n = 100;
                          xx=x1:(x2-x1)/n:x2
                          yy=y1:(y2-y1)/n:y2


                          Where n specifies how many points of coordinate. xx and yy are two arrays storing and coordinates on the line.



                          You can also plot the line by



                          plot(xx,yy)






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 15 '18 at 3:00









                          Banghua ZhaoBanghua Zhao

                          1,2851720




                          1,2851720



























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