Line–line intersection

{{see also|Skew lines#Formulas}}

A necessary condition for two lines to intersect is that they are in the same plane—that is, are not skew lines. Satisfaction of this condition is equivalent to the tetrahedron with vertices at two of the points on one line and two of the points on the other line being degenerate in the sense of having zero volume. For the algebraic form of this condition, see {{slink|Skew lines|Testing for skewness}}.

First we consider the intersection of two lines {{math|L₁}} and {{math|L₂}} in two-dimensional space, with line {{math|L₁}} being defined by two distinct points {{math|(x₁, y₁)}} and {{math|(x₂, y₂)}}, and line {{math|L₂}} being defined by two distinct points {{math|(x₃, y₃)}} and {{math|(x₄, y₄)}}.{{MathWorld|title=Line-Line Intersection|id=Line-LineIntersection|access-date=2008-01-10}}

The intersection {{mvar|P}} of line {{math|L₁}} and {{math|L₂}} can be defined using determinants.

:$$

P_x = \frac{\begin{vmatrix} \begin{vmatrix} x_1 & y_1\\x_2 & y_2\end{vmatrix} & \begin{vmatrix} x_1 & 1\\x_2 & 1\end{vmatrix} \\\\ \begin{vmatrix} x_3 & y_3\\x_4 & y_4\end{vmatrix} & \begin{vmatrix} x_3 & 1\\x_4 & 1\end{vmatrix} \end{vmatrix} }

{\begin{vmatrix} \begin{vmatrix} x_1 & 1\\x_2 & 1\end{vmatrix} & \begin{vmatrix} y_1 & 1\\y_2 & 1\end{vmatrix} \\\\ \begin{vmatrix} x_3 & 1\\x_4 & 1\end{vmatrix} & \begin{vmatrix} y_3 & 1\\y_4 & 1\end{vmatrix} \end{vmatrix}}\,\!

\qquad

P_y = \frac{\begin{vmatrix} \begin{vmatrix} x_1 & y_1\\x_2 & y_2\end{vmatrix} & \begin{vmatrix} y_1 & 1\\y_2 & 1\end{vmatrix} \\\\ \begin{vmatrix} x_3 & y_3\\x_4 & y_4\end{vmatrix} & \begin{vmatrix} y_3 & 1\\y_4 & 1\end{vmatrix} \end{vmatrix} }

{\begin{vmatrix} \begin{vmatrix} x_1 & 1\\x_2 & 1\end{vmatrix} & \begin{vmatrix} y_1 & 1\\y_2 & 1\end{vmatrix} \\\\ \begin{vmatrix} x_3 & 1\\x_4 & 1\end{vmatrix} & \begin{vmatrix} y_3 & 1\\y_4 & 1\end{vmatrix} \end{vmatrix}}\,\!

The determinants can be written out as:

: $$

\begin{align}

P_x&= \frac{(x_1 y_2-y_1 x_2)(x_3-x_4)-(x_1-x_2)(x_3 y_4-y_3 x_4)}{(x_1-x_2) (y_3-y_4) - (y_1-y_2) (x_3-x_4)}

\\[4px]

P_y&= \frac{(x_1 y_2-y_1 x_2)(y_3-y_4)-(y_1-y_2)(x_3 y_4-y_3 x_4)}{(x_1-x_2) (y_3-y_4) - (y_1-y_2) (x_3-x_4)}

\end{align}

When the two lines are parallel or coincident, the denominator is zero.

{{see also|Intersection_(geometry)#Two_line_segments}}

The intersection point above is for the infinitely long lines defined by the points, rather than the line segments between the points, and can produce an intersection point not contained in either of the two line segments. In order to find the position of the intersection in respect to the line segments, we can define lines {{math|L₁}} and {{math|L₂}} in terms of first degree Bézier parameters:

: $$

L_1 = \begin{bmatrix}x_1 \\ y_1\end{bmatrix}

+ t \begin{bmatrix}x_2-x_1 \\ y_2-y_1\end{bmatrix},

\qquad

L_2 = \begin{bmatrix}x_3 \\ y_3\end{bmatrix}

+ u \begin{bmatrix}x_4-x_3 \\ y_4-y_3\end{bmatrix}

(where {{mvar|t}} and {{mvar|u}} are real numbers). The intersection point of the lines is found with one of the following values of {{mvar|t}} or {{mvar|u}}, where

: $$

t = \frac{\begin{vmatrix} x_1-x_3 & x_3-x_4\\y_1-y_3 & y_3-y_4\end{vmatrix}}{\begin{vmatrix} x_1-x_2 & x_3-x_4\\y_1-y_2 & y_3-y_4\end{vmatrix}} = \frac{(x_1 - x_3)(y_3-y_4)-(y_1-y_3)(x_3-x_4)}{(x_1-x_2)(y_3-y_4)-(y_1-y_2)(x_3-x_4)}

and

: $$

u = - \frac{\begin{vmatrix} x_1-x_2 & x_1-x_3\\y_1-y_2 & y_1-y_3\end{vmatrix}}{\begin{vmatrix} x_1-x_2 & x_3-x_4\\y_1-y_2 & y_3-y_4\end{vmatrix}} = -\frac{(x_1 - x_2)(y_1-y_3)-(y_1-y_2)(x_1-x_3)}{(x_1-x_2)(y_3-y_4)-(y_1-y_2)(x_3-x_4)},

with

: $$

(P_x, P_y)= \bigl(x_1 + t (x_2-x_1),\; y_1 + t (y_2-y_1)\bigr) \quad \text{or} \quad (P_x, P_y) = \bigl(x_3 + u (x_4-x_3),\; y_3 + u (y_4-y_3)\bigr)

There will be an intersection if {{math|0 ≤ t ≤ 1}} and {{math|0 ≤ u ≤ 1}}. The intersection point falls within the first line segment if {{math|0 ≤ t ≤ 1}}, and it falls within the second line segment if {{math|0 ≤ u ≤ 1}}. These inequalities can be tested without the need for division, allowing rapid determination of the existence of any line segment intersection before calculating its exact point.{{cite book |last1=Antonio |first1=Franklin |editor-last=Kirk |editor-first=David |title=Graphics Gems III |publisher=Academic Press, Inc. |date=1992 |pages=199–202 |chapter=Chapter IV.6: Faster Line Segment Intersection |isbn=0-12-059756-X}}

In the case where the two line segments share an x axis and $x\_2\; =\; x\_1\; +\; 1$, $t$ and $u$ simplify to

t = u = \frac{y_1 - y_3}{y_1 - y_2 - y_3 + y_4},

with

(P_x, P_y)= \bigl(x_1 + t,\; y_1 + t (y_2-y_1)\bigr) \quad \text{or} \quad (P_x, P_y) = \bigl(x_1 + t,\; y_3 + t (y_4-y_3)\bigr).

{{see also|Intersection_(geometry)#Two_lines}}

The {{mvar|x}} and {{mvar|y}} coordinates of the point of intersection of two non-vertical lines can easily be found using the following substitutions and rearrangements.

Suppose that two lines have the equations {{math|y {{=}} ax + c}} and {{math|y {{=}} bx + d}} where {{mvar|a}} and {{mvar|b}} are the slopes (gradients) of the lines and where {{mvar|c}} and {{mvar|d}} are the {{mvar|y}}-intercepts of the lines. At the point where the two lines intersect (if they do), both {{mvar|y}} coordinates will be the same, hence the following equality:

:$ax+c\; =\; bx+d.$

We can rearrange this expression in order to extract the value of {{mvar|x}},

:$ax-bx\; =\; d-c,$

and so,

:$x\; =\; \backslash frac\{d-c\}\{a-b\}.$

To find the {{mvar|y}} coordinate, all we need to do is substitute the value of {{mvar|x}} into either one of the two line equations, for example, into the first:

:$y\; =\; a\backslash frac\{d-c\}\{a-b\}+c.$

Hence, the point of intersection is

:$P\; =\; \backslash left(\; \backslash frac\{d-c\}\{a-b\},\; a\backslash frac\{d-c\}\{a-b\}+c\; \backslash right)\; .$

Note that if {{math|a {{=}} b}} then the two lines are parallel and they do not intersect, unless {{math|c {{=}} d}} as well, in which case the lines are coincident and they intersect at every point.

By using homogeneous coordinates, the intersection point of two implicitly defined lines can be determined quite easily. In 2D, every point can be defined as a projection of a 3D point, given as the ordered triple {{math|(x, y, w)}}. The mapping from 3D to 2D coordinates is {{math|(x′, y′) {{=}} ({{sfrac|x|w}}, {{sfrac|y|w}})}}. We can convert 2D points to homogeneous coordinates by defining them as {{math|(x, y, 1)}}.

Assume that we want to find intersection of two infinite lines in 2-dimensional space, defined as {{math|a₁x + b₁y + c₁ {{=}} 0}} and {{math|a₂x + b₂y + c₂ {{=}} 0}}. We can represent these two lines in line coordinates as {{math|U₁ {{=}} (a₁, b₁, c₁)}} and {{math|U₂ {{=}} (a₂, b₂, c₂)}}. The intersection {{math|P′}} of two lines is then simply given by{{Cite web |last=Birchfield |first=Stanley |date=1998-04-23 |title=Homogeneous coordinates |url=http://robotics.stanford.edu/~birch/projective/node4.html |url-status=live |archive-url=https://web.archive.org/web/20000929025956/http://robotics.stanford.edu/~birch/projective/node4.html |archive-date=2000-09-29 |website=robotics.stanford.edu |accessdate=2015-08-18}}

:$P\text{'}\; =\; (a\_p,\; b\_p,\; c\_p)\; =\; U\_1\; \backslash times\; U\_2\; =\; (b\_1\; c\_2\; -\; b\_2\; c\_1,\; a\_2\; c\_1-a\_1\; c\_2,\; a\_1\; b\_2\; -\; a\_2\; b\_1)$

If {{math|c_p {{=}} 0}}, the lines do not intersect.

{{see also|Skew lines#More than two lines}}

The intersection of two lines can be generalized to involve additional lines. The existence of and expression for the {{mvar|n}}-line intersection problem are as follows.

In two dimensions, more than two lines almost certainly do not intersect at a single point. To determine if they do and, if so, to find the intersection point, write the {{mvar|i}}th equation ({{math|i {{=}} 1, …, n}}) as

:

and stack these equations into matrix form as

:$\backslash mathbf\{A\}\backslash mathbf\{w\}=\backslash mathbf\{b\},$

where the {{mvar|i}}th row of the {{math|n × 2}} matrix {{math|A}} is {{math|[a_i1, a_i2]}}, {{math|w}} is the 2 × 1 vector {{math|[{{su|p=x|b=y}}]}}, and the {{mvar|i}}th element of the column vector {{math|b}} is {{math|b_i}}. If {{math|A}} has independent columns, its rank is 2. Then if and only if the rank of the augmented matrix {{math|[A {{!}} b]}} is also 2, there exists a solution of the matrix equation and thus an intersection point of the {{mvar|n}} lines. The intersection point, if it exists, is given by

:$\backslash mathbf\{w\}\; =\; \backslash mathbf\{A\}^\backslash mathrm\{g\}\; \backslash mathbf\{b\}\; =\; \backslash left(\backslash mathbf\{A\}^\backslash mathsf\{T\}\; \backslash mathbf\{A\}\backslash right)^\{-1\}\; \backslash mathbf\{A\}^\backslash mathsf\{T\}\; \backslash mathbf\{b\},$

where {{math|A^g}} is the Moore–Penrose generalized inverse of {{math|A}} (which has the form shown because {{math|A}} has full column rank). Alternatively, the solution can be found by jointly solving any two independent equations. But if the rank of {{math|A}} is only 1, then if the rank of the augmented matrix is 2 there is no solution but if its rank is 1 then all of the lines coincide with each other.

The above approach can be readily extended to three dimensions. In three or more dimensions, even two lines almost certainly do not intersect; pairs of non-parallel lines that do not intersect are called skew lines. But if an intersection does exist it can be found, as follows.

In three dimensions a line is represented by the intersection of two planes, each of which has an equation of the form

:

Thus a set of {{mvar|n}} lines can be represented by {{math|2n}} equations in the 3-dimensional coordinate vector {{math|w}}:

:$\backslash mathbf\{A\}\backslash mathbf\{w\}=\backslash mathbf\{b\}$

where now {{math|A}} is {{math|2n × 3}} and {{math|b}} is {{math|2n × 1}}. As before there is a unique intersection point if and only if {{math|A}} has full column rank and the augmented matrix {{math|[A {{!}} b]}} does not, and the unique intersection if it exists is given by

:$\backslash mathbf\{w\}\; =\; \backslash left(\backslash mathbf\{A\}^\backslash mathsf\{T\}\; \backslash mathbf\{A\}\; \backslash right)^\{-1\}\; \backslash mathbf\{A\}^\backslash mathsf\{T\}\; \backslash mathbf\{b\}.$

File:skew_lines_shortest_distance.svg

{{main|Skew lines#Nearest points}}

In two or more dimensions, we can usually find a point that is mutually closest to two or more lines in a least-squares sense.

In the two-dimensional case, first, represent line {{mvar|i}} as a point {{math|p_i}} on the line and a unit normal vector {{math|n̂_i}}, perpendicular to that line. That is, if {{math|x₁}} and {{math|x₂}} are points on line 1, then let {{math|p₁ {{=}} x₁}} and let

:

which is the unit vector along the line, rotated by a right angle.

The distance from a point {{math|x}} to the line {{math|(p, n̂)}} is given by

:$d\backslash bigl(\backslash mathbf\{x\},(\backslash mathbf\{p\},\backslash mathbf\{\backslash hat\; n\})\backslash bigr)\; =\; \backslash bigl|(\backslash mathbf\{x\}-\backslash mathbf\{p\})\backslash cdot\; \backslash mathbf\{\backslash hat\; n\}\backslash bigr|\; =\; \backslash left|(\backslash mathbf\{x\}-\backslash mathbf\{p\})^\backslash mathsf\{T\}\; \backslash mathbf\{\backslash hat\; n\}\backslash right|\; =\; \backslash left|\backslash mathbf\{\backslash hat\; n\}\; ^\backslash mathsf\{T\}\; (\backslash mathbf\{x\}-\backslash mathbf\{p\})\backslash right|\; =\; \backslash sqrt\{(\backslash mathbf\{x\}-\backslash mathbf\{p\})^\backslash mathsf\{T\}\; \backslash mathbf\{\backslash hat\; n\}\; \backslash mathbf\{\backslash hat\; n\}^\backslash mathsf\{T\}\; (\backslash mathbf\{x\}-\backslash mathbf\{p\})\}.$

And so the squared distance from a point {{math|x}} to a line is

:$d\backslash bigl(\backslash mathbf\{x\},(\backslash mathbf\{p\},\backslash mathbf\{\backslash hat\; n\})\backslash bigr)^2\; =\; (\backslash mathbf\{x\}-\backslash mathbf\{p\})^\backslash mathsf\{T\}\; \backslash left(\backslash mathbf\{\backslash hat\; n\}\; \backslash mathbf\{\backslash hat\; n\}^\backslash mathsf\{T\}\; \backslash right)\; (\backslash mathbf\{x\}-\backslash mathbf\{p\}).$

The sum of squared distances to many lines is the cost function:

:$E(\backslash mathbf\{x\})\; =\; \backslash sum\_i\; (\backslash mathbf\{x\}-\backslash mathbf\{p\}\_i)^\backslash mathsf\{T\}\; \backslash left(\backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\backslash right)\; (\backslash mathbf\{x\}-\backslash mathbf\{p\}\_i).$

This can be rearranged:

:$$

\begin{align}

E(\mathbf{x}) & = \sum_i \mathbf{x}^\mathsf{T} \mathbf{\hat n}_i \mathbf{\hat n}_i^\mathsf{T} \mathbf{x} - \mathbf{x}^\mathsf{T} \mathbf{\hat n}_i \mathbf{\hat n}_i^\mathsf{T} \mathbf{p}_i - \mathbf{p}_i^\mathsf{T} \mathbf{\hat n}_i \mathbf{\hat n}_i^\mathsf{T} \mathbf{x} + \mathbf{p}_i^\mathsf{T} \mathbf{\hat n}_i \mathbf{\hat n}_i^\mathsf{T} \mathbf{p}_i \\

& = \mathbf{x}^\mathsf{T} \left(\sum_i \mathbf{\hat n}_i \mathbf{\hat n}_i^\mathsf{T}\right) \mathbf{x} - 2 \mathbf{x}^\mathsf{T} \left(\sum_i \mathbf{\hat n}_i \mathbf{\hat n}_i^\mathsf{T} \mathbf{p}_i\right) + \sum_i \mathbf{p}_i^\mathsf{T} \mathbf{\hat n}_i \mathbf{\hat n}_i^\mathsf{T} \mathbf{p}_i.

\end{align}

To find the minimum, we differentiate with respect to {{math|x}} and set the result equal to the zero vector:

:$\backslash frac\{\backslash partial\; E(\backslash mathbf\{x\})\}\{\backslash partial\; \backslash mathbf\{x\}\}\; =\; \backslash boldsymbol\{0\}\; =\; 2\; \backslash left(\backslash sum\_i\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\backslash right)\; \backslash mathbf\{x\}\; -\; 2\; \backslash left(\backslash sum\_i\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\; \backslash mathbf\{p\}\_i\backslash right)$

so

:$\backslash left(\backslash sum\_i\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\backslash right)\; \backslash mathbf\{x\}\; =\; \backslash sum\_i\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\; \backslash mathbf\{p\}\_i$

and so

:$\backslash mathbf\{x\}\; =\; \backslash left(\backslash sum\_i\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\backslash right)^\{-1\}\; \backslash left(\backslash sum\_i\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\; \backslash mathbf\{p\}\_i\backslash right).$

While {{math|n̂_i}} is not well-defined in more than two dimensions, this can be generalized to any number of dimensions by noting that {{math|n̂_i n̂_i^T}} is simply the symmetric matrix with all eigenvalues unity except for a zero eigenvalue in the direction along the line providing a seminorm on the distance between {{math|p_i}} and another point giving the distance to the line. In any number of dimensions, if {{math|v̂_i}} is a unit vector along the {{mvar|i}}th line, then

: $\backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}$ becomes $\backslash mathbf\{I\}\; -\; \backslash mathbf\{\backslash hat\; v\}\_i\; \backslash mathbf\{\backslash hat\; v\}\_i^\backslash mathsf\{T\}$

where {{math|I}} is the identity matrix, and so{{cite web |last1=Traa |first1=Johannes |date=2013 |title=Least-Squares Intersection of Lines |url=http://cal.cs.illinois.edu/~johannes/research/LS_line_intersect.pdf |url-status=dead |archive-url=https://web.archive.org/web/20170912055605/http://cal.cs.illinois.edu/~johannes/research/LS_line_intersect.pdf |archive-date=2017-09-12 |website=cal.cs.illinois.edu |accessdate=2018-08-30}}

: $x=\; \backslash left(\backslash sum\_i\; \backslash mathbf\{I\}-\backslash mathbf\{\backslash hat\; v\}\_i\; \backslash mathbf\{\backslash hat\; v\}\_i^\backslash mathsf\{T\}\backslash right)^\{-1\}\; \backslash left(\backslash sum\_i\; \backslash left(\backslash mathbf\{I\}-\backslash mathbf\{\backslash hat\; v\}\_i\; \backslash mathbf\{\backslash hat\; v\}\_i^\backslash mathsf\{T\}\; \backslash right)\; \backslash mathbf\{p\}\_i\backslash right).$

In order to find the intersection point of a set of lines, we calculate the point with minimum distance to them. Each line is defined by an origin {{math|a_i}} and a unit direction vector {{math|n̂_i}}. The square of the distance from a point {{math|p}} to one of the lines is given from Pythagoras:

: $d\_i^2\; =\; \backslash left\backslash |\; \backslash mathbf\{p\}\; -\; \backslash mathbf\{a\}\_i\; \backslash right\backslash |^2\; -\; \backslash left(\; \backslash left(\; \backslash mathbf\{p\}\; -\; \backslash mathbf\{a\}\_i\; \backslash right)^\backslash mathsf\{T\}\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash right)^2$

= \left( \mathbf{p} - \mathbf{a}_i \right)^\mathsf{T} \left( \mathbf{p} - \mathbf{a}_i \right) - \left( \left( \mathbf{p} - \mathbf{a}_i \right)^\mathsf{T} \mathbf{\hat n}_i \right)^2

where {{math|(p − a_i)^T n̂_i}} is the projection of {{math|p − a_i}} on line {{mvar|i}}. The sum of distances to the square to all lines is

: $\backslash sum\_i\; d\_i^2\; =\; \backslash sum\_i\; \backslash left(\; \{\backslash left(\; \backslash mathbf\{p\}-\; \backslash mathbf\{a\}\_i\; \backslash right)^\backslash mathsf\{T\}\}\; \backslash left(\; \backslash mathbf\{p\}-\; \backslash mathbf\{a\}\_i\; \backslash right)-\; \{\backslash left(\; \backslash left(\; \backslash mathbf\{p\}-\; \backslash mathbf\{a\}\_i\; \backslash right)^\backslash mathsf\{T\}\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash right)^2\}\; \backslash right)$

To minimize this expression, we differentiate it with respect to {{math|p}}.

: $\backslash sum\_i\; \backslash left(\; 2\backslash left(\; \backslash mathbf\{p\}\; -\; \backslash mathbf\{a\}\_i\; \backslash right)-\; 2\; \backslash left(\backslash left(\; \backslash mathbf\{p\}\; -\; \backslash mathbf\{a\}\_i\; \backslash right)^\backslash mathsf\{T\}\; \backslash mathbf\{\backslash hat\; n\}\_i\backslash right)\; \backslash mathbf\{\backslash hat\; n\}\_i\backslash right)=\backslash boldsymbol\{0\}$

: $\backslash sum\_i\; \backslash left(\; \backslash mathbf\{p\}\; -\; \backslash mathbf\{a\}\_i\; \backslash right)\; =\; \backslash sum\_i\; \backslash left(\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\; \backslash right)\; \backslash left(\; \backslash mathbf\{p\}\; -\; \backslash mathbf\{a\}\_i\; \backslash right)$

which results in

: $\backslash left(\backslash sum\_i\backslash left(\backslash mathbf\{I\}\; -\; \backslash mathbf\{\backslash hat\; n\}\_i\; \backslash mathbf\{\backslash hat\; n\}\_i^\backslash mathsf\{T\}\; \backslash right)\backslash right)\; \backslash mathbf\{p\}$

= \sum_i \left(\mathbf{I} - \mathbf{\hat n}_i \mathbf{\hat n}_i^\mathsf{T} \right) \mathbf{a}_i

where {{math|I}} is the identity matrix. This is a matrix {{math|Sp {{=}} C}}, with solution {{math|p {{=}} S⁺C}}, where {{math|S⁺}} is the pseudo-inverse of {{math|S}}.

{{See also|Parallel postulate}}{{Expand section|date=June 2022}}

File:Euclidian and non euclidian geometry.png

In spherical geometry, any two great circles intersect.{{Cite web |title=Exploring Hyperbolic Space |url=https://math.berkeley.edu/~kpmann/Hyperbolic%20and%20Spherical%20geometries.pdf |access-date=2022-06-03 |website=math.berkeley.edu}}

In hyperbolic geometry, given any line and any point, there are infinitely many lines through that point that do not intersect the given line.

• Line segment intersection
• Line intersection in projective space
• Distance between two parallel lines
• Distance from a point to a line
• Line–plane intersection
• Parallel postulate
• Triangulation (computer vision)
• {{slink|Intersection (Euclidean geometry)|Two line segments}}

• [http://softsurfer.com/Archive/algorithm_0106/algorithm_0106.htm Distance between Lines and Segments with their Closest Point of Approach], applicable to two, three, or more dimensions.

{{DEFAULTSORT:Line-line intersection}}

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