How To Find Intersection Of 2 Lines

The intersection of lines.

In Euclidean geometry, the intersection of a line and a line can exist the empty gear up, a bespeak, or a line. Distinguishing these cases and finding the intersection betoken take use, for example, in computer graphics, motion planning, and collision detection.

In three-dimensional Euclidean geometry, if ii lines are non in the same aeroplane they are called skew lines and have no point of intersection. If they are in the same plane there are 3 possibilities: if they coincide (are non singled-out lines) they take an infinitude of points in common (namely all of the points on either of them); if they are distinct but take the same slope they are said to exist parallel and have no points in common; otherwise they have a single betoken of intersection.

The distinguishing features of non-Euclidean geometry are the number and locations of possible intersections between 2 lines and the number of possible lines with no intersections (parallel lines) with a given line.

Formulas [edit]

A necessary condition for 2 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 ane line and 2 of the points on the other line beingness degenerate in the sense of having aught volume. For the algebraic form of this condition, see Skew lines § Testing for skewness.

Given two points on each line [edit]

First nosotros consider the intersection of ii lines ${\displaystyle L_{i}}$ and ${\displaystyle L_{2}}$ in two-dimensional space, with line ${\displaystyle L_{1}}$ beingness defined by two distinct points ${\displaystyle (x_{i},y_{1})}$ and ${\displaystyle (x_{2},y_{two})}$ , and line ${\displaystyle L_{2}}$ existence defined past two distinct points ${\displaystyle (x_{3},y_{three})}$ and ${\displaystyle (x_{four},y_{4})}$ .^[1]

The intersection ${\displaystyle P}$ of line ${\displaystyle L_{1}}$ and ${\displaystyle L_{two}}$ tin be defined using determinants.

${\displaystyle P_{x}={\frac {\begin{vmatrix}{\begin{vmatrix}x_{1}&y_{i}\\x_{2}&y_{2}\finish{vmatrix}}&{\begin{vmatrix}x_{1}&ane\\x_{2}&1\end{vmatrix}}\\\\{\brainstorm{vmatrix}x_{three}&y_{3}\\x_{4}&y_{4}\cease{vmatrix}}&{\begin{vmatrix}x_{3}&1\\x_{4}&i\end{vmatrix}}\stop{vmatrix}}{\begin{vmatrix}{\begin{vmatrix}x_{one}&1\\x_{2}&i\finish{vmatrix}}&{\begin{vmatrix}y_{ane}&1\\y_{2}&1\end{vmatrix}}\\\\{\begin{vmatrix}x_{three}&1\\x_{iv}&i\terminate{vmatrix}}&{\begin{vmatrix}y_{three}&1\\y_{4}&1\end{vmatrix}}\cease{vmatrix}}}\,\!\qquad P_{y}={\frac {\brainstorm{vmatrix}{\begin{vmatrix}x_{1}&y_{i}\\x_{2}&y_{2}\end{vmatrix}}&{\begin{vmatrix}y_{one}&1\\y_{two}&1\end{vmatrix}}\\\\{\begin{vmatrix}x_{3}&y_{3}\\x_{4}&y_{4}\end{vmatrix}}&{\begin{vmatrix}y_{three}&1\\y_{4}&1\stop{vmatrix}}\terminate{vmatrix}}{\begin{vmatrix}{\begin{vmatrix}x_{1}&one\\x_{two}&one\end{vmatrix}}&{\begin{vmatrix}y_{1}&i\\y_{2}&1\end{vmatrix}}\\\\{\begin{vmatrix}x_{3}&1\\x_{4}&1\stop{vmatrix}}&{\begin{vmatrix}y_{3}&1\\y_{4}&1\end{vmatrix}}\end{vmatrix}}}\,\!}$

The determinants can be written out as:

${\displaystyle {\begin{aligned}(P_{10},P_{y})={\Biggl (}&{\frac {(x_{1}y_{2}-y_{1}x_{two})(x_{3}-x_{four})-(x_{one}-x_{2})(x_{3}y_{iv}-y_{3}x_{4})}{D}},\\&{\frac {(x_{1}y_{ii}-y_{1}x_{ii})(y_{3}-y_{four})-(y_{i}-y_{2})(x_{iii}y_{4}-y_{three}x_{4})}{D}}{\Biggr )}\end{aligned}}}$

where the denominator is:

${\displaystyle D=(x_{ane}-x_{ii})(y_{3}-y_{iv})-(y_{1}-y_{two})(x_{iii}-x_{4})}$

When the two lines are parallel or coincident, the denominator is zip. If the lines are almost parallel, then a computer solution might encounter numeric problems implementing the solution described higher up: the recognition of this condition might crave an approximate test in a practical application. An alternate approach might be to rotate the line segments so that one of them is horizontal, whence the solution of the rotated parametric form of the 2d line is easily obtained. Conscientious discussion of the special cases is required (parallel lines/coincident lines, overlapping/non-overlapping intervals).

Given two points on each line segment [edit]

Note that the intersection bespeak in a higher place is for the infinitely long lines defined by the points, rather than the line segments between the points, and can produce an intersection signal not contained in either of the 2 line segments. In lodge to observe the position of the intersection in respect to the line segments, we tin define lines ${\displaystyle L_{i}}$ and ${\displaystyle L_{ii}}$ in terms of showtime caste Bézier parameters:

${\displaystyle L_{1}={\brainstorm{bmatrix}x_{1}\\y_{1}\end{bmatrix}}+t{\begin{bmatrix}x_{two}-x_{i}\\y_{ii}-y_{one}\finish{bmatrix}},\qquad L_{ii}={\begin{bmatrix}x_{3}\\y_{3}\end{bmatrix}}+u{\begin{bmatrix}x_{4}-x_{3}\\y_{4}-y_{3}\stop{bmatrix}}}$

(where t and u are real numbers). The intersection point of the lines is constitute with ane of the following values of t or u, where

${\displaystyle t={\frac {\begin{vmatrix}x_{1}-x_{iii}&x_{3}-x_{four}\\y_{one}-y_{3}&y_{three}-y_{four}\end{vmatrix}}{\brainstorm{vmatrix}x_{i}-x_{2}&x_{iii}-x_{iv}\\y_{1}-y_{2}&y_{3}-y_{4}\stop{vmatrix}}}={\frac {(x_{1}-x_{3})(y_{3}-y_{iv})-(y_{i}-y_{three})(x_{3}-x_{4})}{(x_{one}-x_{2})(y_{3}-y_{4})-(y_{1}-y_{2})(x_{3}-x_{4})}}}$

and

${\displaystyle u={\frac {\begin{vmatrix}x_{1}-x_{iii}&x_{i}-x_{2}\\y_{one}-y_{3}&y_{ane}-y_{2}\end{vmatrix}}{\begin{vmatrix}x_{1}-x_{2}&x_{3}-x_{four}\\y_{one}-y_{2}&y_{3}-y_{4}\end{vmatrix}}}={\frac {(x_{1}-x_{3})(y_{ane}-y_{2})-(y_{one}-y_{three})(x_{i}-x_{2})}{(x_{1}-x_{2})(y_{three}-y_{4})-(y_{1}-y_{ii})(x_{3}-x_{four})}},}$

with:

${\displaystyle (P_{x},P_{y})=(x_{1}+t(x_{2}-x_{1}),\;y_{i}+t(y_{2}-y_{1}))\quad {\text{or}}\quad (P_{x},P_{y})=(x_{iii}+u(x_{4}-x_{three}),\;y_{3}+u(y_{4}-y_{3}))}$

There volition be an intersection if 0.0 ≤t ≤ 1.0 and 0.0 ≤u ≤ 1.0. The intersection bespeak falls inside the commencement line segment if 0.0 ≤t ≤ 1.0, and it falls within the 2d line segment if 0.0 ≤u ≤ 1.0. These inequalities can be tested without the demand for division, allowing rapid determination of the existence of any line segment intersection earlier calculating its exact bespeak.^[2]

Given two line equations [edit]

The ${\displaystyle 10}$ and ${\displaystyle y}$ coordinates of the point of intersection of two not-vertical lines tin can easily be found using the post-obit substitutions and rearrangements.

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

${\displaystyle ax+c=bx+d.}$

Nosotros can rearrange this expression in order to extract the value of ${\displaystyle x}$ ,

${\displaystyle ax-bx=d-c,}$

and and then,

${\displaystyle x={\frac {d-c}{a-b}}.}$

To observe the y coordinate, all we need to exercise is substitute the value of x into either one of the ii line equations, for example, into the first:

${\displaystyle y=a{\frac {d-c}{a-b}}+c.}$

Hence, the point of intersection is

${\displaystyle P\left({\frac {d-c}{a-b}},a{\frac {d-c}{a-b}}+c\correct).}$

Notation if a = b so the two lines are parallel. If c ≠ d as well, the lines are different and there is no intersection, otherwise the two lines are identical.

Using homogeneous coordinates [edit]

By using homogeneous coordinates, the intersection point of two implicitly defined lines tin can be adamant quite easily. In 2d, every point can be defined as a project of a 3D point, given every bit the ordered triple ${\displaystyle (ten,y,w)}$ . The mapping from 3D to 2nd coordinates is ${\displaystyle (10',y')=(x/west,y/w)}$ . We tin can catechumen 2d points to homogeneous coordinates by defining them every bit ${\displaystyle (x,y,ane)}$ .

Assume that nosotros want to find intersection of two space lines in ii-dimensional space, defined equally ${\displaystyle a_{i}x+b_{1}y+c_{1}=0}$ and ${\displaystyle a_{2}x+b_{two}y+c_{ii}=0}$ . Nosotros can correspond these two lines in line coordinates every bit ${\displaystyle U_{ane}=(a_{one},b_{1},c_{1})}$ and ${\displaystyle U_{2}=(a_{two},b_{2},c_{2})}$ ,

The intersection ${\displaystyle P'}$ of two lines is then simply given by,^[three]

${\displaystyle P'=(a_{p},b_{p},c_{p})=U_{one}\times U_{2}=(b_{1}c_{two}-b_{2}c_{i},a_{2}c_{1}-a_{1}c_{2},a_{i}b_{2}-a_{two}b_{1})}$

If ${\displaystyle c_{p}=0}$ the lines practise not intersect.

More than two lines [edit]

The intersection of two lines tin be generalized to involve boosted lines. Existence of and expression for the n-line intersection trouble are equally follows.

In two dimensions [edit]

In 2 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 betoken, write the i-th equation (i = 1, …,northward) every bit ${\displaystyle {\brainstorm{bmatrix}a_{i1}&a_{i2}\end{bmatrix}}{\begin{bmatrix}10&y\end{bmatrix}}^{\mathsf {T}}=b_{i},}$ and stack these equations into matrix form as

${\displaystyle Aw=b,}$

where the i-th row of the n × ii matrix A is ${\displaystyle (a_{i1},a_{i2})}$ , w is the ii × 1 vector (10, y)^T, and the i-th element of the column vector b is b _i . If A has independent columns, its rank is 2. Then if and only if the rank of the augmented matrix [A | b] is likewise 2, there exists a solution of the matrix equation and thus an intersection betoken of the northward lines. The intersection point, if information technology exists, is given by

${\displaystyle due west=A^{1000}b=\left(A^{\mathsf {T}}A\correct)^{-1}A^{\mathsf {T}}b,}$

where ${\displaystyle A^{g}}$ is the Moore-Penrose generalized inverse of ${\displaystyle A}$ (which has the class shown because A has total column rank). Alternatively, the solution tin can exist found past jointly solving any two contained equations. But if the rank of A is only i, then if the rank of the augmented matrix is two at that place is no solution only if its rank is i and then all of the lines coincide with each other.

In three dimensions [edit]

The to a higher place approach tin can be readily extended to three dimensions. In iii or more than dimensions, even two lines near certainly do not intersect; pairs of non-parallel lines that do not intersect are called skew lines. But if an intersection does be it can be found, every bit follows.

In three dimensions a line is represented by the intersection of two planes, each of which has an equation of the form ${\displaystyle {\begin{bmatrix}a_{i1}&a_{i2}&a_{i3}\stop{bmatrix}}{\begin{bmatrix}ten&y&z\end{bmatrix}}^{\mathsf {T}}=b_{i}.}$ Thus a set of n lines tin exist represented past iin equations in the three-dimensional coordinate vector west = (10, y, z)^T:

${\displaystyle Aw=b}$

where at present A is 2n × 3 and b is iin × i. As earlier there is a unique intersection point if and merely if A has full column rank and the augmented matrix [A | b] does not, and the unique intersection if information technology exists is given by

${\displaystyle west=\left(A^{\mathsf {T}}A\right)^{-ane}A^{\mathsf {T}}b.}$

Nearest points to skew lines [edit]

In two or more than dimensions, we can usually find a indicate that is mutually closest to 2 or more lines in a least-squares sense.

In two dimensions [edit]

In the two-dimensional instance, first, stand for line i as a point, ${\displaystyle p_{i}}$ , on the line and a unit normal vector, ${\displaystyle {\lid {north}}_{i}}$ , perpendicular to that line. That is, if ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ are points on line 1, so let ${\displaystyle p_{1}=x_{1}}$ and let

${\displaystyle {\hat {n}}_{one}:={\begin{bmatrix}0&-1\\ane&0\end{bmatrix}}{\frac {x_{2}-x_{ane}}{\|x_{2}-x_{one}\|}}}$

which is the unit vector along the line, rotated by 90 degrees.

Note that the altitude from a betoken, x to the line ${\displaystyle (p,{\chapeau {north}})}$ is given by

${\displaystyle d(x,(p,n))=|(10-p)\cdot {\hat {north}}|=\left|(ten-p)^{\acme }{\hat {north}}\right|=\left|{\lid {north}}^{\superlative }(10-p)\right|={\sqrt {(x-p)^{\height }{\lid {north}}{\hat {n}}^{\top }(x-p)}}.}$

And and then the squared distance from a point, x, to a line is

${\displaystyle d(x,(p,due north))^{ii}=(ten-p)^{\top }\left({\chapeau {n}}{\hat {n}}^{\peak }\correct)(ten-p).}$

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

${\displaystyle Due east(x)=\sum _{i}(x-p_{i})^{\top }\left({\chapeau {n}}_{i}{\hat {n}}_{i}^{\superlative }\right)(10-p_{i}).}$

This can be rearranged:

${\displaystyle {\begin{aligned}East(ten)&=\sum _{i}x^{\top }{\hat {n}}_{i}{\lid {n}}_{i}^{\top }10-x^{\summit }{\lid {northward}}_{i}{\hat {n}}_{i}^{\acme }p_{i}-p_{i}^{\top }{\hat {n}}_{i}{\hat {northward}}_{i}^{\top }ten+p_{i}^{\acme }{\hat {n}}_{i}{\lid {n}}_{i}^{\top }p_{i}\\&=x^{\top }\left(\sum _{i}{\hat {n}}_{i}{\chapeau {due north}}_{i}^{\top }\right)x-2x^{\top }\left(\sum _{i}{\hat {due north}}_{i}{\hat {due north}}_{i}^{\summit }p_{i}\right)+\sum _{i}p_{i}^{\top }{\hat {n}}_{i}{\hat {n}}_{i}^{\top }p_{i}.\stop{aligned}}}$

To find the minimum, we differentiate with respect to ten and set the upshot equal to the nothing vector:

${\displaystyle {\frac {\partial E(x)}{\fractional x}}=0=ii\left(\sum _{i}{\lid {north}}_{i}{\hat {northward}}_{i}^{\top }\right)ten-2\left(\sum _{i}{\hat {n}}_{i}{\hat {northward}}_{i}^{\top }p_{i}\correct)}$

and then

${\displaystyle \left(\sum _{i}{\hat {n}}_{i}{\hat {n}}_{i}^{\height }\correct)x=\sum _{i}{\hat {north}}_{i}{\chapeau {n}}_{i}^{\top }p_{i}}$

and so

${\displaystyle x=\left(\sum _{i}{\chapeau {due north}}_{i}{\lid {due north}}_{i}^{\top }\right)^{-1}\left(\sum _{i}{\hat {n}}_{i}{\lid {due north}}_{i}^{\acme }p_{i}\right).}$

In more than ii dimensions [edit]

While ${\displaystyle {\hat {n}}_{i}}$ is non well-defined in more than two dimensions, this can be generalized to any number of dimensions past noting that ${\displaystyle {\lid {northward}}_{i}{\chapeau {n}}_{i}^{\top }}$ 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 ${\displaystyle p_{i}}$ and another point giving the altitude to the line. In any number of dimensions, if ${\displaystyle {\hat {v}}_{i}}$ is a unit vector along the i-thursday line, then

${\displaystyle {\hat {north}}_{i}{\hat {n}}_{i}^{\top }}$ becomes ${\displaystyle I-{\chapeau {v}}_{i}{\hat {v}}_{i}^{\top }}$

where I is the identity matrix, and and then^[4]

${\displaystyle x=\left(\sum _{i}I-{\hat {5}}_{i}{\hat {v}}_{i}^{\top }\right)^{-1}\left(\sum _{i}\left(I-{\hat {v}}_{i}{\chapeau {v}}_{i}^{\top }\right)p_{i}\right).}$

General derivation [edit]

In order to find the intersection point of a set of lines, we calculate the bespeak with minimum distance to them. Each line is defined past an origin ${\displaystyle a_{i}}$ and a unit of measurement management vector, ${\displaystyle n_{i}}$ . The square of the distance from a signal ${\displaystyle p}$ to one of the lines is given from Pythagoras:

${\displaystyle d_{i}^{ii}=\left\|p-a_{i}\correct\|^{2}-\left[\left(p-a_{i}\right)^{\mathsf {T}}*n_{i}\correct]^{ii}=\left(p-a_{i}\right)^{\mathsf {T}}*\left(p-a_{i}\right)-\left[\left(p-a_{i}\correct)^{\mathsf {T}}*n_{i}\right]^{ii}}$

Where : ${\displaystyle {\left(p-a_{i}\right)}^{\mathsf {T}}*n_{i}}$ is the project of ${\displaystyle \left(p-{{a}_{i}}\right)}$ on the line ${\displaystyle i}$ . The sum of distances to the square to all lines is:

${\displaystyle \sum _{i}d_{i}^{2}=\sum _{i}\left[{\left(p-a_{i}\right)^{\mathsf {T}}}*\left(p-a_{i}\right)-{\left[\left(p-a_{i}\right)^{\mathsf {T}}*n_{i}\right]^{2}}\correct]}$

To minimize this expression, nosotros differentiate it with respect to ${\displaystyle p}$ .

${\displaystyle \sum _{i}\left[2*\left(p-a_{i}\right)-2*\right[{{\left(p-a_{i}\right)}^{\mathsf {T}}}*n_{i}]*n_{i}]=0}$

${\displaystyle \sum _{i}\left(p-a_{i}\right)=\sum _{i}\left[n_{i}*n_{i}^{\mathsf {T}}\right]*\left(p-a_{i}\right)}$

Information technology results:

${\displaystyle \left[\sum _{i}\left[I-{n_{i}}*{n_{i}}^{\mathsf {T}}\correct]\right]*p=\sum _{i}\left[I-{n_{i}}*{n_{i}}^{\mathsf {T}}\right]*{a_{i}}}$

Where ${\displaystyle I}$ is the identity matrix. This is a matrix ${\displaystyle South*p=C}$ , with solution ${\displaystyle p={S^{+}}*C}$ , where ${\displaystyle {S}^{+}}$ is the pseudo-inverse of ${\displaystyle S}$ .

See also [edit]

• Line segment intersection
• Line intersection in projective space
• Distance between two parallel lines
• Altitude from a betoken to a line
• Line–plane intersection
• Parallel postulate
• Triangulation (calculator vision)
• Intersection (Euclidean geometry) § Ii line segments

References [edit]

1. ^ "Weisstein, Eric W. "Line-Line Intersection." From MathWorld". A Wolfram Web Resources . Retrieved 2008-01-10 .
2. ^ Antonio, Franklin (1992). "Chapter IV.6: Faster Line Segment Intersection". In Kirk, David (ed.). Graphics Gems Iii. Bookish Printing, Inc. pp. 199–202. ISBN0-12-059756-Ten.
3. ^ "Homogeneous coordinates". robotics.stanford.edu . Retrieved 2015-08-xviii .
4. ^ Traa, Johannes. "Least-Squares Intersection of Lines" (PDF) . Retrieved 30 Baronial 2022.

External links [edit]

• Distance between Lines and Segments with their Closest Bespeak of Arroyo, applicable to ii, 3, or more dimensions.

Source: https://en.wikipedia.org/wiki/Line%E2%80%93line_intersection

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