Intersection point of 2 lines defined by 2 points each
$begingroup$
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
edited 1 hour ago
Ethan Bolker
45.6k553120
asked 3 hours ago
crazicrafter1crazicrafter1
197
$endgroup$
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
edited 1 hour ago
Ethan Bolker
45.6k553120
asked 3 hours ago
crazicrafter1crazicrafter1
197
$endgroup$
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
edited 1 hour ago
Ethan Bolker
45.6k553120
asked 3 hours ago
crazicrafter1crazicrafter1
197
$endgroup$
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 1 hour ago
Ethan Bolker
45.6k553120
asked 3 hours ago
crazicrafter1crazicrafter1
197
edited 1 hour ago
Ethan Bolker
45.6k553120
asked 3 hours ago
crazicrafter1crazicrafter1
197
edited 1 hour ago
Ethan Bolker
45.6k553120
edited 1 hour ago
Ethan Bolker
45.6k553120
edited 1 hour ago
Ethan Bolker
45.6k553120
45.6k553120
asked 3 hours ago
crazicrafter1crazicrafter1
197
asked 3 hours ago
crazicrafter1crazicrafter1
197
asked 3 hours ago
crazicrafter1crazicrafter1
197
197
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
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.)
answered 1 hour ago
Ethan BolkerEthan Bolker
45.6k553120
$endgroup$
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 $vec{ab}=b-a$. 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.
answered 1 hour ago
mr_e_manmr_e_man
1,1401424
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
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.)
answered 1 hour ago
Ethan BolkerEthan Bolker
45.6k553120
$endgroup$
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.)
answered 1 hour ago
Ethan BolkerEthan Bolker
45.6k553120
$endgroup$
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.)
answered 1 hour ago
Ethan BolkerEthan Bolker
45.6k553120
$endgroup$
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.)
answered 1 hour ago
Ethan BolkerEthan Bolker
45.6k553120
edited 1 hour ago
answered 1 hour ago
Ethan BolkerEthan Bolker
45.6k553120
answered 1 hour ago
Ethan BolkerEthan Bolker
45.6k553120
answered 1 hour ago
Ethan BolkerEthan Bolker
45.6k553120
45.6k553120
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 $vec{ab}=b-a$. 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.
answered 1 hour ago
mr_e_manmr_e_man
1,1401424
$endgroup$
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 $vec{ab}=b-a$. 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.
answered 1 hour ago
mr_e_manmr_e_man
1,1401424
$endgroup$
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 $vec{ab}=b-a$. 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.
answered 1 hour ago
mr_e_manmr_e_man
1,1401424
$endgroup$
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.
answered 1 hour ago
mr_e_manmr_e_man
1,1401424
answered 1 hour ago
mr_e_manmr_e_man
1,1401424
answered 1 hour ago
mr_e_manmr_e_man
1,1401424
answered 1 hour ago
mr_e_manmr_e_man
1,1401424
1,1401424
add a comment |
add a comment |
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