# How do I compute the intersection point of two lines?

## Question:

I have two lines that intersect at a point. I know the endpoints of the two lines. How do I compute the intersection point in Python?

```
# Given these endpoints
#line 1
A = [X, Y]
B = [X, Y]
#line 2
C = [X, Y]
D = [X, Y]
# Compute this:
point_of_intersection = [X, Y]
```

## Answers:

Unlike other suggestions, this is short and doesn’t use external libraries like `numpy`

. (Not that using other libraries is bad…it’s nice not need to, especially for such a simple problem.)

```
def line_intersection(line1, line2):
xdiff = (line1[0][0] - line1[1][0], line2[0][0] - line2[1][0])
ydiff = (line1[0][1] - line1[1][1], line2[0][1] - line2[1][1])
def det(a, b):
return a[0] * b[1] - a[1] * b[0]
div = det(xdiff, ydiff)
if div == 0:
raise Exception('lines do not intersect')
d = (det(*line1), det(*line2))
x = det(d, xdiff) / div
y = det(d, ydiff) / div
return x, y
print line_intersection((A, B), (C, D))
```

And FYI, I would use tuples instead of lists for your points. E.g.

```
A = (X, Y)
```

EDIT: Initially there was a typo. That was fixed Sept 2014 thanks to @zidik.

This is simply the Python transliteration of the following formula, where the lines are (**a1**, **a2**) and (**b1**, **b2**) and the intersection is **p**. (If the denominator is zero, the lines have no unique intersection.)

Can’t stand aside,

So we have linear system:

A

_{1}* x + B_{1}* y = C_{1}

A_{2}* x + B_{2}* y = C_{2}

let’s do it with Cramer’s rule, so solution can be found in determinants:

x = D

_{x}/D

y = D_{y}/D

where *D* is main determinant of the system:

A

_{1}B_{1}

A_{2}B_{2}

and *D _{x}* and

*D*can be found from matricies:

_{y}C

_{1}B_{1}

C_{2}B_{2}

and

A

_{1}C_{1}

A_{2}C_{2}

(notice, as *C* column consequently substitues the coef. columns of *x* and *y*)

So now the python, for clarity for us, to not mess things up let’s do mapping between math and python. We will use array `L`

for storing our coefs *A*, *B*, *C* of the line equations and intestead of pretty `x`

, `y`

we’ll have `[0]`

, `[1]`

, but anyway. Thus, what I wrote above will have the following form further in the code:

for *D*

L1[0] L1[1]

L2[0] L2[1]

for *D _{x}*

L1[2] L1[1]

L2[2] L2[1]

for *D _{y}*

L1[0] L1[2]

L2[0] L2[2]

Now go for coding:

`line`

– produces coefs *A*, *B*, *C* of line equation by two points provided,

`intersection`

– finds intersection point (if any) of two lines provided by coefs.

```
from __future__ import division
def line(p1, p2):
A = (p1[1] - p2[1])
B = (p2[0] - p1[0])
C = (p1[0]*p2[1] - p2[0]*p1[1])
return A, B, -C
def intersection(L1, L2):
D = L1[0] * L2[1] - L1[1] * L2[0]
Dx = L1[2] * L2[1] - L1[1] * L2[2]
Dy = L1[0] * L2[2] - L1[2] * L2[0]
if D != 0:
x = Dx / D
y = Dy / D
return x,y
else:
return False
```

Usage example:

```
L1 = line([0,1], [2,3])
L2 = line([2,3], [0,4])
R = intersection(L1, L2)
if R:
print "Intersection detected:", R
else:
print "No single intersection point detected"
```

I didn’t find an intuitive explanation on the web, so now that I worked it out, here’s my solution. This is for infinite lines (what I needed), not segments.

Some terms you might remember:

A line is defined as y = mx + b OR y = slope * x + y-intercept

Slope = rise over run = dy / dx = height / distance

Y-intercept is where the line crosses the Y axis, where X = 0

Given those definitions, here are some functions:

```
def slope(P1, P2):
# dy/dx
# (y2 - y1) / (x2 - x1)
return(P2[1] - P1[1]) / (P2[0] - P1[0])
def y_intercept(P1, slope):
# y = mx + b
# b = y - mx
# b = P1[1] - slope * P1[0]
return P1[1] - slope * P1[0]
def line_intersect(m1, b1, m2, b2):
if m1 == m2:
print ("These lines are parallel!!!")
return None
# y = mx + b
# Set both lines equal to find the intersection point in the x direction
# m1 * x + b1 = m2 * x + b2
# m1 * x - m2 * x = b2 - b1
# x * (m1 - m2) = b2 - b1
# x = (b2 - b1) / (m1 - m2)
x = (b2 - b1) / (m1 - m2)
# Now solve for y -- use either line, because they are equal here
# y = mx + b
y = m1 * x + b1
return x,y
```

Here’s a simple test between two (infinite) lines:

```
A1 = [1,1]
A2 = [3,3]
B1 = [1,3]
B2 = [3,1]
slope_A = slope(A1, A2)
slope_B = slope(B1, B2)
y_int_A = y_intercept(A1, slope_A)
y_int_B = y_intercept(B1, slope_B)
print(line_intersect(slope_A, y_int_A, slope_B, y_int_B))
```

Output:

```
(2.0, 2.0)
```

Using formula from:

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

```
def findIntersection(x1,y1,x2,y2,x3,y3,x4,y4):
px= ( (x1*y2-y1*x2)*(x3-x4)-(x1-x2)*(x3*y4-y3*x4) ) / ( (x1-x2)*(y3-y4)-(y1-y2)*(x3-x4) )
py= ( (x1*y2-y1*x2)*(y3-y4)-(y1-y2)*(x3*y4-y3*x4) ) / ( (x1-x2)*(y3-y4)-(y1-y2)*(x3-x4) )
return [px, py]
```

If your lines are multiple points instead, you can use this version.

```
import numpy as np
import matplotlib.pyplot as plt
"""
Sukhbinder
5 April 2017
Based on:
"""
def _rect_inter_inner(x1,x2):
n1=x1.shape[0]-1
n2=x2.shape[0]-1
X1=np.c_[x1[:-1],x1[1:]]
X2=np.c_[x2[:-1],x2[1:]]
S1=np.tile(X1.min(axis=1),(n2,1)).T
S2=np.tile(X2.max(axis=1),(n1,1))
S3=np.tile(X1.max(axis=1),(n2,1)).T
S4=np.tile(X2.min(axis=1),(n1,1))
return S1,S2,S3,S4
def _rectangle_intersection_(x1,y1,x2,y2):
S1,S2,S3,S4=_rect_inter_inner(x1,x2)
S5,S6,S7,S8=_rect_inter_inner(y1,y2)
C1=np.less_equal(S1,S2)
C2=np.greater_equal(S3,S4)
C3=np.less_equal(S5,S6)
C4=np.greater_equal(S7,S8)
ii,jj=np.nonzero(C1 & C2 & C3 & C4)
return ii,jj
def intersection(x1,y1,x2,y2):
"""
INTERSECTIONS Intersections of curves.
Computes the (x,y) locations where two curves intersect. The curves
can be broken with NaNs or have vertical segments.
usage:
x,y=intersection(x1,y1,x2,y2)
Example:
a, b = 1, 2
phi = np.linspace(3, 10, 100)
x1 = a*phi - b*np.sin(phi)
y1 = a - b*np.cos(phi)
x2=phi
y2=np.sin(phi)+2
x,y=intersection(x1,y1,x2,y2)
plt.plot(x1,y1,c='r')
plt.plot(x2,y2,c='g')
plt.plot(x,y,'*k')
plt.show()
"""
ii,jj=_rectangle_intersection_(x1,y1,x2,y2)
n=len(ii)
dxy1=np.diff(np.c_[x1,y1],axis=0)
dxy2=np.diff(np.c_[x2,y2],axis=0)
T=np.zeros((4,n))
AA=np.zeros((4,4,n))
AA[0:2,2,:]=-1
AA[2:4,3,:]=-1
AA[0::2,0,:]=dxy1[ii,:].T
AA[1::2,1,:]=dxy2[jj,:].T
BB=np.zeros((4,n))
BB[0,:]=-x1[ii].ravel()
BB[1,:]=-x2[jj].ravel()
BB[2,:]=-y1[ii].ravel()
BB[3,:]=-y2[jj].ravel()
for i in range(n):
try:
T[:,i]=np.linalg.solve(AA[:,:,i],BB[:,i])
except:
T[:,i]=np.NaN
in_range= (T[0,:] >=0) & (T[1,:] >=0) & (T[0,:] <=1) & (T[1,:] <=1)
xy0=T[2:,in_range]
xy0=xy0.T
return xy0[:,0],xy0[:,1]
if __name__ == '__main__':
# a piece of a prolate cycloid, and am going to find
a, b = 1, 2
phi = np.linspace(3, 10, 100)
x1 = a*phi - b*np.sin(phi)
y1 = a - b*np.cos(phi)
x2=phi
y2=np.sin(phi)+2
x,y=intersection(x1,y1,x2,y2)
plt.plot(x1,y1,c='r')
plt.plot(x2,y2,c='g')
plt.plot(x,y,'*k')
plt.show()
```

Here is a solution using the Shapely library. Shapely is often used for GIS work, but is built to be useful for computational geometry. I changed your inputs from lists to tuples.

# Problem

```
# Given these endpoints
#line 1
A = (X, Y)
B = (X, Y)
#line 2
C = (X, Y)
D = (X, Y)
# Compute this:
point_of_intersection = (X, Y)
```

# Solution

```
import shapely
from shapely.geometry import LineString, Point
line1 = LineString([A, B])
line2 = LineString([C, D])
int_pt = line1.intersection(line2)
point_of_intersection = int_pt.x, int_pt.y
print(point_of_intersection)
```

img And You can use this kode

```
class Nokta:
def __init__(self,x,y):
self.x=x
self.y=y
class Dogru:
def __init__(self,a,b):
self.a=a
self.b=b
def Kesisim(self,Dogru_b):
x1= self.a.x
x2=self.b.x
x3=Dogru_b.a.x
x4=Dogru_b.b.x
y1= self.a.y
y2=self.b.y
y3=Dogru_b.a.y
y4=Dogru_b.b.y
#Notlardaki denklemleri kullandım
pay1=((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3))
pay2=((x2-x1) * (y1 - y3) - (y2 - y1) * (x1 - x3))
payda=((y4 - y3) *(x2-x1)-(x4 - x3)*(y2 - y1))
if pay1==0 and pay2==0 and payda==0:
print("DOĞRULAR BİRBİRİNE ÇAKIŞIKTIR")
elif payda==0:
print("DOĞRULAR BİRBİRNE PARALELDİR")
else:
ua=pay1/payda if payda else 0
ub=pay2/payda if payda else 0
#x ve y buldum
x=x1+ua*(x2-x1)
y=y1+ua*(y2-y1)
print("DOĞRULAR {},{} NOKTASINDA KESİŞTİ".format(x,y))
```

The most concise solution I have found uses Sympy: https://www.geeksforgeeks.org/python-sympy-line-intersection-method/

```
# import sympy and Point, Line
from sympy import Point, Line
p1, p2, p3 = Point(0, 0), Point(1, 1), Point(7, 7)
l1 = Line(p1, p2)
# using intersection() method
showIntersection = l1.intersection(p3)
print(showIntersection)
```

With the scikit-spatial library you can easily do it in the following way:

```
import matplotlib.pyplot as plt
from skspatial.objects import Line
# Define the two lines.
line_1 = Line.from_points([3, -2], [5, 4])
line_2 = Line.from_points([-1, 0], [3, 2])
# Compute the intersection point
intersection_point = line_1.intersect_line(line_2)
# Plot
_, ax = plt.subplots()
line_1.plot_2d(ax, t_1=-2, t_2=3, c="k")
line_2.plot_2d(ax, t_1=-2, t_2=3, c="k")
intersection_point.plot_2d(ax, c="r", s=100)
grid = ax.grid()
```

there is already an answer that uses formula from Wikipedia but that doesn’t have any check point to check if line segments actually intersect so here you go

```
def line_intersection(a, b, c, d):
t = ((a[0] - c[0]) * (c[1] - d[1]) - (a[1] - c[1]) * (c[0] - d[0])) / ((a[0] - b[0]) * (c[1] - d[1]) - (a[1] - b[1]) * (c[0] - d[0]))
u = ((a[0] - c[0]) * (a[1] - b[1]) - (a[1] - c[1]) * (a[0] - b[0])) / ((a[0] - b[0]) * (c[1] - d[1]) - (a[1] - b[1]) * (c[0] - d[0]))
# check if line actually intersect
if (0 <= t and t <= 1 and 0 <= u and u <= 1):
return [a[0] + t * (b[0] - a[0]), a[1] + t * (b[1] - a[1])]
else:
return False
#usage
print(line_intersection([0,0], [10, 10], [0, 10], [10,0]))
#result [5.0, 5.0]
```