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
$endgroup$
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 1 hour ago
Ethan Bolker
45.6k553120
45.6k553120
asked 2 hours ago
crazicrafter1crazicrafter1
197
197
add a comment |
add a comment |
2 Answers
2
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 $vec{ab}=b-a$. 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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3176543%2fintersection-point-of-2-lines-defined-by-2-points-each%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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.)
$endgroup$
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 1 hour ago
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
1,1401424
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3176543%2fintersection-point-of-2-lines-defined-by-2-points-each%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown