Intersection point of 2 lines defined by 2 points each












3












$begingroup$


I'm implementing this in code, but I'll rewrite it so that it is easier understood (like pseudocode):



# a = pt 1 on line 1
# b = pt 2 on line 1
# c = pt 1 on line 2
# d = pt 2 on line 2
def intersect(a,b,c,d):

# stuff for line 1
a1 = b.y-a.y
b1 = a.x-b.x
c1 = a1*a.x + b1*a.y

# stuff for line 2
a2 = d.y-c.y
b2 = c.x-d.x
c2 = a2*c.x + b2*c.y

determinant = a1*b2 - a2*b1

if (determinant == 0):
# Return (infinity, infinity) if they never intersect
# By "never intersect", I mean that the lines are parallel to each other
return math.inf, math,inf
else:
x = (b2*c1 - b1*c2)/determinant
y = (a1*c2 - a2*c1)/determinant
return x,y


All the above works, ... but only does by assuming that the lines extend infinitely in each direction, like a linear equation. I'll show what I mean here.



There are the 2 lines, red and green, and the gold dot is what is returned when I test this code ... but the lines don't actually intersect. What can be used to test whether the lines truly intersect?



Heres the actual Python code if needed.










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I'm implementing this in code, but I'll rewrite it so that it is easier understood (like pseudocode):



    # a = pt 1 on line 1
    # b = pt 2 on line 1
    # c = pt 1 on line 2
    # d = pt 2 on line 2
    def intersect(a,b,c,d):

    # stuff for line 1
    a1 = b.y-a.y
    b1 = a.x-b.x
    c1 = a1*a.x + b1*a.y

    # stuff for line 2
    a2 = d.y-c.y
    b2 = c.x-d.x
    c2 = a2*c.x + b2*c.y

    determinant = a1*b2 - a2*b1

    if (determinant == 0):
    # Return (infinity, infinity) if they never intersect
    # By "never intersect", I mean that the lines are parallel to each other
    return math.inf, math,inf
    else:
    x = (b2*c1 - b1*c2)/determinant
    y = (a1*c2 - a2*c1)/determinant
    return x,y


    All the above works, ... but only does by assuming that the lines extend infinitely in each direction, like a linear equation. I'll show what I mean here.



    There are the 2 lines, red and green, and the gold dot is what is returned when I test this code ... but the lines don't actually intersect. What can be used to test whether the lines truly intersect?



    Heres the actual Python code if needed.










    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I'm implementing this in code, but I'll rewrite it so that it is easier understood (like pseudocode):



      # a = pt 1 on line 1
      # b = pt 2 on line 1
      # c = pt 1 on line 2
      # d = pt 2 on line 2
      def intersect(a,b,c,d):

      # stuff for line 1
      a1 = b.y-a.y
      b1 = a.x-b.x
      c1 = a1*a.x + b1*a.y

      # stuff for line 2
      a2 = d.y-c.y
      b2 = c.x-d.x
      c2 = a2*c.x + b2*c.y

      determinant = a1*b2 - a2*b1

      if (determinant == 0):
      # Return (infinity, infinity) if they never intersect
      # By "never intersect", I mean that the lines are parallel to each other
      return math.inf, math,inf
      else:
      x = (b2*c1 - b1*c2)/determinant
      y = (a1*c2 - a2*c1)/determinant
      return x,y


      All the above works, ... but only does by assuming that the lines extend infinitely in each direction, like a linear equation. I'll show what I mean here.



      There are the 2 lines, red and green, and the gold dot is what is returned when I test this code ... but the lines don't actually intersect. What can be used to test whether the lines truly intersect?



      Heres the actual Python code if needed.










      share|cite|improve this question











      $endgroup$




      I'm implementing this in code, but I'll rewrite it so that it is easier understood (like pseudocode):



      # a = pt 1 on line 1
      # b = pt 2 on line 1
      # c = pt 1 on line 2
      # d = pt 2 on line 2
      def intersect(a,b,c,d):

      # stuff for line 1
      a1 = b.y-a.y
      b1 = a.x-b.x
      c1 = a1*a.x + b1*a.y

      # stuff for line 2
      a2 = d.y-c.y
      b2 = c.x-d.x
      c2 = a2*c.x + b2*c.y

      determinant = a1*b2 - a2*b1

      if (determinant == 0):
      # Return (infinity, infinity) if they never intersect
      # By "never intersect", I mean that the lines are parallel to each other
      return math.inf, math,inf
      else:
      x = (b2*c1 - b1*c2)/determinant
      y = (a1*c2 - a2*c1)/determinant
      return x,y


      All the above works, ... but only does by assuming that the lines extend infinitely in each direction, like a linear equation. I'll show what I mean here.



      There are the 2 lines, red and green, and the gold dot is what is returned when I test this code ... but the lines don't actually intersect. What can be used to test whether the lines truly intersect?



      Heres the actual Python code if needed.







      linear-algebra matrices python






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 5 hours ago









      Ethan Bolker

      45.6k553120




      45.6k553120










      asked 6 hours ago









      crazicrafter1crazicrafter1

      247




      247






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          I think you are asking for the intersection point (if any) of two line segments, not two lines.



          Once you find the intersection point $P$ as you have, you can check that it is between the endpoints $A$ and $B$ of a segment by solving the equation
          $$
          tA + (1-t)B = P
          $$

          for $t$ and checking that $t$ is between $0$ and $1$. That equation will have a solution because you know $P$ is on the line. Do that for each of the two segments.



          Warning: you may have numerical instability if the determinant is close to $0$. That will happen when the lines are nearly parallel.



          (There may be a shorter way to do this from scratch, but this will work.)






          share|cite|improve this answer











          $endgroup$





















            3












            $begingroup$

            You have the point $x$ where the infinite lines intersect. You need to check whether that point is on both finite line segments.



            Line segment 1 has endpoints $a$ and $b$. Use these to make a vector $vec{ab}=b-a$. If the dot product $vec{ab}cdotvec{ax}$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vec{ab}cdotvec{bx}$ is positive, then $x$ is forward of $b$. The point $x$ is on the segment if it's between $a$ and $b$.



            Do the same test for the other line segment.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              This is an elaboration on Ethan Bolker's post.



              Every point on the line segment from $a$ to $b$ can be expressed as $$a + t(b-a)$$ for some $t$ between $0$ and $1$. Likewise, every point on the line segment from $c$ to $d$ can be expressed as $$c + s(d-c)$$ for some $s$ between $0$ and $1$. Any intersection point can be expressed in both forms, so we will have



              $$a + t(b-a) = c + s(d - c)$$
              re-arranging,
              $$t(b - a) + s(c - d) = c - a$$



              Letting $n = b - a, m = c-d$, and $p = c - a$, and switching to coordinates, we have
              $$n_xt + m_xs = p_x\n_yt + m_ys = p_y$$



              Instead of solving directly for the coordinates of the point of intersection, you should solve for $t$ and $s$. Letting
              $$D = n_xm_y - n_ym_x\Q_x = m_yp_x - m_xp_y\Q_y =n_xp_y - n_yp_x$$



              We have $$t=frac {Q_x}D, quad s = frac {Q_y}D$$



              In particular, we can figure out the following before doing the divisions:




              • If $D$ is $0$, then the segments are parallel - the only way they can intersect in a unique point is if the two segments are adjacent and have a common endpoint (for instance $b = c$, and $a$ and $d$ are on opposite sides of the common point).

              • If $D$ is the opposite sign of either $Q_x$ or $Q_y$, then one of $t$ or $s < 0$, so the line segments do not intersect.

              • If $|D|$ is larger than either $|Q_x|$ or $|Q_y|$, then one of $t$ or $s > 1$, and again the line segments do not intersect.

              • Otherwise, the segments do intersect, and you either calculate $t$ and $a + t(b -a)$ or calculate $s$ and $c + s(d - c)$ to get the point of intersection.


              This is a more numerically stable approach to the problem.






              share|cite|improve this answer









              $endgroup$














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                3 Answers
                3






                active

                oldest

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                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                I think you are asking for the intersection point (if any) of two line segments, not two lines.



                Once you find the intersection point $P$ as you have, you can check that it is between the endpoints $A$ and $B$ of a segment by solving the equation
                $$
                tA + (1-t)B = P
                $$

                for $t$ and checking that $t$ is between $0$ and $1$. That equation will have a solution because you know $P$ is on the line. Do that for each of the two segments.



                Warning: you may have numerical instability if the determinant is close to $0$. That will happen when the lines are nearly parallel.



                (There may be a shorter way to do this from scratch, but this will work.)






                share|cite|improve this answer











                $endgroup$


















                  3












                  $begingroup$

                  I think you are asking for the intersection point (if any) of two line segments, not two lines.



                  Once you find the intersection point $P$ as you have, you can check that it is between the endpoints $A$ and $B$ of a segment by solving the equation
                  $$
                  tA + (1-t)B = P
                  $$

                  for $t$ and checking that $t$ is between $0$ and $1$. That equation will have a solution because you know $P$ is on the line. Do that for each of the two segments.



                  Warning: you may have numerical instability if the determinant is close to $0$. That will happen when the lines are nearly parallel.



                  (There may be a shorter way to do this from scratch, but this will work.)






                  share|cite|improve this answer











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    I think you are asking for the intersection point (if any) of two line segments, not two lines.



                    Once you find the intersection point $P$ as you have, you can check that it is between the endpoints $A$ and $B$ of a segment by solving the equation
                    $$
                    tA + (1-t)B = P
                    $$

                    for $t$ and checking that $t$ is between $0$ and $1$. That equation will have a solution because you know $P$ is on the line. Do that for each of the two segments.



                    Warning: you may have numerical instability if the determinant is close to $0$. That will happen when the lines are nearly parallel.



                    (There may be a shorter way to do this from scratch, but this will work.)






                    share|cite|improve this answer











                    $endgroup$



                    I think you are asking for the intersection point (if any) of two line segments, not two lines.



                    Once you find the intersection point $P$ as you have, you can check that it is between the endpoints $A$ and $B$ of a segment by solving the equation
                    $$
                    tA + (1-t)B = P
                    $$

                    for $t$ and checking that $t$ is between $0$ and $1$. That equation will have a solution because you know $P$ is on the line. Do that for each of the two segments.



                    Warning: you may have numerical instability if the determinant is close to $0$. That will happen when the lines are nearly parallel.



                    (There may be a shorter way to do this from scratch, but this will work.)







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 5 hours ago

























                    answered 5 hours ago









                    Ethan BolkerEthan Bolker

                    45.6k553120




                    45.6k553120























                        3












                        $begingroup$

                        You have the point $x$ where the infinite lines intersect. You need to check whether that point is on both finite line segments.



                        Line segment 1 has endpoints $a$ and $b$. Use these to make a vector $vec{ab}=b-a$. If the dot product $vec{ab}cdotvec{ax}$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vec{ab}cdotvec{bx}$ is positive, then $x$ is forward of $b$. The point $x$ is on the segment if it's between $a$ and $b$.



                        Do the same test for the other line segment.






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          You have the point $x$ where the infinite lines intersect. You need to check whether that point is on both finite line segments.



                          Line segment 1 has endpoints $a$ and $b$. Use these to make a vector $vec{ab}=b-a$. If the dot product $vec{ab}cdotvec{ax}$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vec{ab}cdotvec{bx}$ is positive, then $x$ is forward of $b$. The point $x$ is on the segment if it's between $a$ and $b$.



                          Do the same test for the other line segment.






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            You have the point $x$ where the infinite lines intersect. You need to check whether that point is on both finite line segments.



                            Line segment 1 has endpoints $a$ and $b$. Use these to make a vector $vec{ab}=b-a$. If the dot product $vec{ab}cdotvec{ax}$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vec{ab}cdotvec{bx}$ is positive, then $x$ is forward of $b$. The point $x$ is on the segment if it's between $a$ and $b$.



                            Do the same test for the other line segment.






                            share|cite|improve this answer









                            $endgroup$



                            You have the point $x$ where the infinite lines intersect. You need to check whether that point is on both finite line segments.



                            Line segment 1 has endpoints $a$ and $b$. Use these to make a vector $vec{ab}=b-a$. If the dot product $vec{ab}cdotvec{ax}$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vec{ab}cdotvec{bx}$ is positive, then $x$ is forward of $b$. The point $x$ is on the segment if it's between $a$ and $b$.



                            Do the same test for the other line segment.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 5 hours ago









                            mr_e_manmr_e_man

                            1,1501424




                            1,1501424























                                0












                                $begingroup$

                                This is an elaboration on Ethan Bolker's post.



                                Every point on the line segment from $a$ to $b$ can be expressed as $$a + t(b-a)$$ for some $t$ between $0$ and $1$. Likewise, every point on the line segment from $c$ to $d$ can be expressed as $$c + s(d-c)$$ for some $s$ between $0$ and $1$. Any intersection point can be expressed in both forms, so we will have



                                $$a + t(b-a) = c + s(d - c)$$
                                re-arranging,
                                $$t(b - a) + s(c - d) = c - a$$



                                Letting $n = b - a, m = c-d$, and $p = c - a$, and switching to coordinates, we have
                                $$n_xt + m_xs = p_x\n_yt + m_ys = p_y$$



                                Instead of solving directly for the coordinates of the point of intersection, you should solve for $t$ and $s$. Letting
                                $$D = n_xm_y - n_ym_x\Q_x = m_yp_x - m_xp_y\Q_y =n_xp_y - n_yp_x$$



                                We have $$t=frac {Q_x}D, quad s = frac {Q_y}D$$



                                In particular, we can figure out the following before doing the divisions:




                                • If $D$ is $0$, then the segments are parallel - the only way they can intersect in a unique point is if the two segments are adjacent and have a common endpoint (for instance $b = c$, and $a$ and $d$ are on opposite sides of the common point).

                                • If $D$ is the opposite sign of either $Q_x$ or $Q_y$, then one of $t$ or $s < 0$, so the line segments do not intersect.

                                • If $|D|$ is larger than either $|Q_x|$ or $|Q_y|$, then one of $t$ or $s > 1$, and again the line segments do not intersect.

                                • Otherwise, the segments do intersect, and you either calculate $t$ and $a + t(b -a)$ or calculate $s$ and $c + s(d - c)$ to get the point of intersection.


                                This is a more numerically stable approach to the problem.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  This is an elaboration on Ethan Bolker's post.



                                  Every point on the line segment from $a$ to $b$ can be expressed as $$a + t(b-a)$$ for some $t$ between $0$ and $1$. Likewise, every point on the line segment from $c$ to $d$ can be expressed as $$c + s(d-c)$$ for some $s$ between $0$ and $1$. Any intersection point can be expressed in both forms, so we will have



                                  $$a + t(b-a) = c + s(d - c)$$
                                  re-arranging,
                                  $$t(b - a) + s(c - d) = c - a$$



                                  Letting $n = b - a, m = c-d$, and $p = c - a$, and switching to coordinates, we have
                                  $$n_xt + m_xs = p_x\n_yt + m_ys = p_y$$



                                  Instead of solving directly for the coordinates of the point of intersection, you should solve for $t$ and $s$. Letting
                                  $$D = n_xm_y - n_ym_x\Q_x = m_yp_x - m_xp_y\Q_y =n_xp_y - n_yp_x$$



                                  We have $$t=frac {Q_x}D, quad s = frac {Q_y}D$$



                                  In particular, we can figure out the following before doing the divisions:




                                  • If $D$ is $0$, then the segments are parallel - the only way they can intersect in a unique point is if the two segments are adjacent and have a common endpoint (for instance $b = c$, and $a$ and $d$ are on opposite sides of the common point).

                                  • If $D$ is the opposite sign of either $Q_x$ or $Q_y$, then one of $t$ or $s < 0$, so the line segments do not intersect.

                                  • If $|D|$ is larger than either $|Q_x|$ or $|Q_y|$, then one of $t$ or $s > 1$, and again the line segments do not intersect.

                                  • Otherwise, the segments do intersect, and you either calculate $t$ and $a + t(b -a)$ or calculate $s$ and $c + s(d - c)$ to get the point of intersection.


                                  This is a more numerically stable approach to the problem.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    This is an elaboration on Ethan Bolker's post.



                                    Every point on the line segment from $a$ to $b$ can be expressed as $$a + t(b-a)$$ for some $t$ between $0$ and $1$. Likewise, every point on the line segment from $c$ to $d$ can be expressed as $$c + s(d-c)$$ for some $s$ between $0$ and $1$. Any intersection point can be expressed in both forms, so we will have



                                    $$a + t(b-a) = c + s(d - c)$$
                                    re-arranging,
                                    $$t(b - a) + s(c - d) = c - a$$



                                    Letting $n = b - a, m = c-d$, and $p = c - a$, and switching to coordinates, we have
                                    $$n_xt + m_xs = p_x\n_yt + m_ys = p_y$$



                                    Instead of solving directly for the coordinates of the point of intersection, you should solve for $t$ and $s$. Letting
                                    $$D = n_xm_y - n_ym_x\Q_x = m_yp_x - m_xp_y\Q_y =n_xp_y - n_yp_x$$



                                    We have $$t=frac {Q_x}D, quad s = frac {Q_y}D$$



                                    In particular, we can figure out the following before doing the divisions:




                                    • If $D$ is $0$, then the segments are parallel - the only way they can intersect in a unique point is if the two segments are adjacent and have a common endpoint (for instance $b = c$, and $a$ and $d$ are on opposite sides of the common point).

                                    • If $D$ is the opposite sign of either $Q_x$ or $Q_y$, then one of $t$ or $s < 0$, so the line segments do not intersect.

                                    • If $|D|$ is larger than either $|Q_x|$ or $|Q_y|$, then one of $t$ or $s > 1$, and again the line segments do not intersect.

                                    • Otherwise, the segments do intersect, and you either calculate $t$ and $a + t(b -a)$ or calculate $s$ and $c + s(d - c)$ to get the point of intersection.


                                    This is a more numerically stable approach to the problem.






                                    share|cite|improve this answer









                                    $endgroup$



                                    This is an elaboration on Ethan Bolker's post.



                                    Every point on the line segment from $a$ to $b$ can be expressed as $$a + t(b-a)$$ for some $t$ between $0$ and $1$. Likewise, every point on the line segment from $c$ to $d$ can be expressed as $$c + s(d-c)$$ for some $s$ between $0$ and $1$. Any intersection point can be expressed in both forms, so we will have



                                    $$a + t(b-a) = c + s(d - c)$$
                                    re-arranging,
                                    $$t(b - a) + s(c - d) = c - a$$



                                    Letting $n = b - a, m = c-d$, and $p = c - a$, and switching to coordinates, we have
                                    $$n_xt + m_xs = p_x\n_yt + m_ys = p_y$$



                                    Instead of solving directly for the coordinates of the point of intersection, you should solve for $t$ and $s$. Letting
                                    $$D = n_xm_y - n_ym_x\Q_x = m_yp_x - m_xp_y\Q_y =n_xp_y - n_yp_x$$



                                    We have $$t=frac {Q_x}D, quad s = frac {Q_y}D$$



                                    In particular, we can figure out the following before doing the divisions:




                                    • If $D$ is $0$, then the segments are parallel - the only way they can intersect in a unique point is if the two segments are adjacent and have a common endpoint (for instance $b = c$, and $a$ and $d$ are on opposite sides of the common point).

                                    • If $D$ is the opposite sign of either $Q_x$ or $Q_y$, then one of $t$ or $s < 0$, so the line segments do not intersect.

                                    • If $|D|$ is larger than either $|Q_x|$ or $|Q_y|$, then one of $t$ or $s > 1$, and again the line segments do not intersect.

                                    • Otherwise, the segments do intersect, and you either calculate $t$ and $a + t(b -a)$ or calculate $s$ and $c + s(d - c)$ to get the point of intersection.


                                    This is a more numerically stable approach to the problem.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 2 hours ago









                                    Paul SinclairPaul Sinclair

                                    20.7k21543




                                    20.7k21543






























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