Talk:Line-line intersection Source: en.wikipedia.org/wiki/Talk:Line-line_intersection

This  level-5 vital article is rated C-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics Low‑priority This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics Low This article has been rated as Low-priority on the project's priority scale.

This article was nominated for deletion on 9 March 2013 (UTC). The result of the discussion was keep.

I am being bold in updating, and hopefully not making a total donkeys butt of myself at the same time. --Chase-san (talk) 06:49, 10 January 2008 (UTC)[reply]

So what happens when the denominator is zero? Article's still lacking some information. -- Mike Kamermans, Nov. 2010 —Preceding unsigned comment added by 77.167.120.32 (talk) 21:25, 15 November 2010 (UTC)[reply]

I think that will only happen in truly degenerate cases. Up to rotational and reflective symmetry, I think it only happens when ${\displaystyle x_{1}=x_{2}}$ and ${\displaystyle x_{3}=x_{4}}$. This corresponds to parallel lines which either don't intersect (or intersect at infinity, if you like that point of view) or they are coincident, in which case all points on one line are on the other. You are right, this needs to be added. In projective space, this all works out naturally, btw., since it allows us to represent the point at infinity and coincident lines. See Projective_plane#Lines_joining_points_and_intersection_of_lines_.28using_duality.29. —Ben FrantzDale (talk) 22:11, 15 November 2010 (UTC)[reply]

When the denominator is zero it always means that the lines are parallel. Here's my proof: Let the denominator equal zero: (x1 - x2)(y3 - y4) - (y1 - y2)(x3 - x4) = 0. Now rearrange the terms: (y2 - y1)/(x2 - x1) = (y4 - y3)/(x4 - x3). Now notice that the term on the left is the slope of line 1 and the term on the right is the slope of line 2. This equation says that the slopes of the lines are equal, and therefore they must be parallel. If you guys concur I'll add this to the page. --Chrisanderman (talk) 01:48, 13 December 2011 (UTC)[reply]

We really need the match for a case where the end points of the lines are specified in terms of x, y and z coordinates, not just x and y. But I'd keep the 2D case too, as that's no doubt simpler and will be sufficient for most peoples needs. Drkirkby (talk) 12:28, 13 November 2011 (UTC)[reply]

It is covered in the n-line intersection section. —Ben FrantzDale (talk) 12:44, 14 November 2011 (UTC)[reply]

This was added to the article and I removed it for discussion here.

Note: The equation above do not return the intersection of two lines. There may be some unknown additional step needed to get the correct intersection point, or the determinant may be written incorrectly. Here is a version that works correctly. This corresponds to the simplified version in the citation. This code is for MatLab.

x = [0 0; 6 6];  %# Starting points in first row, ending points in second row

y = [0 6; 6 0];

The equations from the above source can then be coded up as follows:

dx = diff(x);  %# Take the differences down each column

dy = diff(y);

den = dx(1)*dy(2)-dy(1)*dx(2);  %# Precompute the denominator

ua = (dx(2)*(y(1)-y(3))-dy(2)*(x(1)-x(3)))/den;

ub = (dx(1)*(y(1)-y(3))-dy(1)*(x(1)-x(3)))/den;

And you can now compute the intersection point of the two lines:

xi = x(1)+ua*dx(1);

yi = y(1)+ua*dy(1);

For the example in the question, the above code gives xi = 3 and yi = 3, as expected. If you want to check that the intersection point lies between the endpoints of the lines (i.e. they are finite line segments), you just have to check that the values ua and ub both lie between 0 and 1:

isInSegment = all(([ua ub] >= 0) & ([ua ub] <= 1));

A couple more points from the tutorial I linked to above:

If the denominator den is 0 then the two lines are parallel.

If the denominator and numerator for the equations for ua and ub are 0 then the two lines are coincident.

-- 06:46, 26 January 2014‎ 69.230.99.150

I am leaving it to others to straighten out any problems. Bill Cherowitzo (talk) 05:34, 27 January 2014 (UTC)[reply]

By the way, I had no trouble in calculating (3,3) as the intersection point in your example using the determinental form in the article - what do you think may be wrong with it? Bill Cherowitzo (talk) 23:25, 28 January 2014 (UTC)[reply]

The presented formula for computing the line–line intersection is wrong. The comment below also notes this. How is this article still around? Please try the proposed computations yourself and then remove this article. — Preceding unsigned comment added by 94.217.39.180 (talk) 23:18, 9 July 2014 (UTC)[reply]

It is common practice to put new sections at the end of a talk page (I realize that this is not the norm outside of WP). As to your comment ... there is nothing wrong with the formula presented. It can be derived from the standard determinental form for the equations of the two lines by applying Cramer's rule and then manipulating the terms to get rid of several minuses. If you think there is a problem you should provide a counterexample. Bill Cherowitzo (talk) 04:23, 10 July 2014 (UTC)[reply]

Look at the following example. Line1: x1,y1 = 15,15 to x2,y2 = 15, 100. Line2: x3,y3 = 10,20 to x4,y4 = 20,20.

Obviously our intersection point should habe an x-coordinate of 15. This not the case with the given Formular for Px. — Preceding unsigned comment added by 94.217.46.136 (talk) 13:22, 10 July 2014 (UTC)[reply]

And the formula for P_x gives 15. Please check your arithmetic. Bill Cherowitzo (talk) 18:04, 10 July 2014 (UTC)[reply]

${\displaystyle P_{x}={\frac {\begin{vmatrix}{\begin{vmatrix}15&15\\15&100\end{vmatrix}}&{\begin{vmatrix}15&1\\15&1\end{vmatrix}}\\\\{\begin{vmatrix}10&20\\20&20\end{vmatrix}}&{\begin{vmatrix}10&1\\20&1\end{vmatrix}}\end{vmatrix}}{\begin{vmatrix}{\begin{vmatrix}15&1\\15&1\end{vmatrix}}&{\begin{vmatrix}15&1\\100&1\end{vmatrix}}\\\\{\begin{vmatrix}10&1\\20&1\end{vmatrix}}&{\begin{vmatrix}20&1\\20&1\end{vmatrix}}\end{vmatrix}}}={\frac {\begin{vmatrix}15(85)&0\\-200&-10\end{vmatrix}}{\begin{vmatrix}0&-85\\-10&0\end{vmatrix}}}={\frac {-150(85)}{-(-10(-85))}}=15.}$

And the other part of the formula correctly gives the y coordinate as 20. Bill Cherowitzo (talk) 18:28, 10 July 2014 (UTC)[reply]

Converting one of the point-vector pairs to two planes intersection as the line representation, gives two line-plane intersections to test. — Preceding unsigned comment added by Goofyseeker311 (talk • contribs) 11:36, 11 October 2023 (UTC)[reply]