Intersection point of 2 lines defined by 2 points each The 2019 Stack Overflow Developer Survey Results Are InIntersection between two linesParallel Lines, One point on each.Intersection between 2 linesStraight lines - point of intersectionFinding the intersection point between two lines using a matrixCalculate intersection point between two linescollision point of circle and lineFind intersection point of two straight linesIntersection point of multiple 3D linesFour Dimensional intersection point
Manuscript was "unsubmitted" because the manuscript was deposited in Arxiv Preprints
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Intersection point of 2 lines defined by 2 points each
The 2019 Stack Overflow Developer Survey Results Are InIntersection between two linesParallel Lines, One point on each.Intersection between 2 linesStraight lines - point of intersectionFinding the intersection point between two lines using a matrixCalculate intersection point between two linescollision point of circle and lineFind intersection point of two straight linesIntersection point of multiple 3D linesFour Dimensional intersection point
$begingroup$
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.
linear-algebra matrices python
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra matrices python
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra matrices python
$endgroup$
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
linear-algebra matrices python
edited Apr 6 at 1:29
Ethan Bolker
45.9k553120
45.9k553120
asked Apr 6 at 0:13
crazicrafter1crazicrafter1
247
247
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.)
$endgroup$
add a comment |
$begingroup$
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 $vecab=b-a$. If the dot product $vecabcdotvecax$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vecabcdotvecbx$ 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.
$endgroup$
add a comment |
$begingroup$
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 $$t=frac Q_xD, quad s = frac Q_yD$$
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 smaller than $|Q_x|$ or smaller than $|Q_y|$, 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.
$endgroup$
$begingroup$
@crazicrafter1 This is a much better algorithm than mine built on yours and you should implement it.
$endgroup$
– Ethan Bolker
Apr 6 at 10:48
$begingroup$
@EthanBolker - the problem is one I already had personal experience with.
$endgroup$
– Paul Sinclair
Apr 6 at 16:10
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.)
$endgroup$
add a comment |
$begingroup$
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.)
$endgroup$
add a comment |
$begingroup$
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.)
$endgroup$
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.)
edited Apr 6 at 1:46
answered Apr 6 at 1:34
Ethan BolkerEthan Bolker
45.9k553120
45.9k553120
add a comment |
add a comment |
$begingroup$
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 $vecab=b-a$. If the dot product $vecabcdotvecax$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vecabcdotvecbx$ 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.
$endgroup$
add a comment |
$begingroup$
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 $vecab=b-a$. If the dot product $vecabcdotvecax$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vecabcdotvecbx$ 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.
$endgroup$
add a comment |
$begingroup$
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 $vecab=b-a$. If the dot product $vecabcdotvecax$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vecabcdotvecbx$ 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.
$endgroup$
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 $vecab=b-a$. If the dot product $vecabcdotvecax$ is positive, then $x$ is forward of $a$; if it's negative, then $x$ is behind $a$. Likewise, if $vecabcdotvecbx$ 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.
answered Apr 6 at 1:42
mr_e_manmr_e_man
1,2001424
1,2001424
add a comment |
add a comment |
$begingroup$
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 $$t=frac Q_xD, quad s = frac Q_yD$$
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 smaller than $|Q_x|$ or smaller than $|Q_y|$, 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.
$endgroup$
$begingroup$
@crazicrafter1 This is a much better algorithm than mine built on yours and you should implement it.
$endgroup$
– Ethan Bolker
Apr 6 at 10:48
$begingroup$
@EthanBolker - the problem is one I already had personal experience with.
$endgroup$
– Paul Sinclair
Apr 6 at 16:10
add a comment |
$begingroup$
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 $$t=frac Q_xD, quad s = frac Q_yD$$
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 smaller than $|Q_x|$ or smaller than $|Q_y|$, 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.
$endgroup$
$begingroup$
@crazicrafter1 This is a much better algorithm than mine built on yours and you should implement it.
$endgroup$
– Ethan Bolker
Apr 6 at 10:48
$begingroup$
@EthanBolker - the problem is one I already had personal experience with.
$endgroup$
– Paul Sinclair
Apr 6 at 16:10
add a comment |
$begingroup$
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 $$t=frac Q_xD, quad s = frac Q_yD$$
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 smaller than $|Q_x|$ or smaller than $|Q_y|$, 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.
$endgroup$
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 $$t=frac Q_xD, quad s = frac Q_yD$$
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 smaller than $|Q_x|$ or smaller than $|Q_y|$, 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.
edited 6 hours ago
answered Apr 6 at 4:00
Paul SinclairPaul Sinclair
20.8k21543
20.8k21543
$begingroup$
@crazicrafter1 This is a much better algorithm than mine built on yours and you should implement it.
$endgroup$
– Ethan Bolker
Apr 6 at 10:48
$begingroup$
@EthanBolker - the problem is one I already had personal experience with.
$endgroup$
– Paul Sinclair
Apr 6 at 16:10
add a comment |
$begingroup$
@crazicrafter1 This is a much better algorithm than mine built on yours and you should implement it.
$endgroup$
– Ethan Bolker
Apr 6 at 10:48
$begingroup$
@EthanBolker - the problem is one I already had personal experience with.
$endgroup$
– Paul Sinclair
Apr 6 at 16:10
$begingroup$
@crazicrafter1 This is a much better algorithm than mine built on yours and you should implement it.
$endgroup$
– Ethan Bolker
Apr 6 at 10:48
$begingroup$
@crazicrafter1 This is a much better algorithm than mine built on yours and you should implement it.
$endgroup$
– Ethan Bolker
Apr 6 at 10:48
$begingroup$
@EthanBolker - the problem is one I already had personal experience with.
$endgroup$
– Paul Sinclair
Apr 6 at 16:10
$begingroup$
@EthanBolker - the problem is one I already had personal experience with.
$endgroup$
– Paul Sinclair
Apr 6 at 16:10
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