how to find where two lines intersect
Find the intersection of two lines
You are encouraged to solve this task according to the task description, using any language you may know.
Finding the intersection of two lines that are in the same plane is an important topic in collision detection.[1]
- Task
Find the point of intersection of two lines in 2D.
The 1st line passes though
(4,0)and
(6,10) .
The 2nd line passes though
and
(10,7).
Contents
- 1 11l
- 2 360 Assembly
- 3 Ada
- 4 ALGOL 68
- 5 APL
- 6 AutoHotkey
- 7 AWK
- 8 BASIC
- 8.1 Sinclair ZX81 BASIC
- 9 C
- 10 C#
- 11 C++
- 12 Clojure
- 13 D
- 14 F#
- 15 Factor
- 16 Fortran
- 17 FreeBASIC
- 18 Go
- 19 Groovy
- 20 Haskell
- 21 J
- 22 Java
- 23 JavaScript
- 23.1 ES6
- 24 jq
- 25 Julia
- 25.1 GeometryTypes library version
- 26 Kotlin
- 27 Lua
- 28 M2000 Interpreter
- 29 Maple
- 30 Mathematica/Wolfram Language
- 31 MATLAB
- 32 Modula-2
- 33 Nim
- 34 Perl
- 35 Phix
- 36 Processing
- 36.1 Processing Python mode
- 37 Python
- 38 Racket
- 39 Raku
- 40 REXX
- 40.1 version 1
- 40.2 version 2
- 41 Ring
- 42 Ruby
- 43 Rust
- 44 Scala
- 45 Sidef
- 46 Swift
- 47 TI-83 BASIC
- 48 Visual Basic
- 49 Visual Basic .NET
- 50 Wren
- 51 zkl
11l [edit]
Translation of: Python
F line_intersect(Ax1, Ay1, Ax2, Ay2, Bx1, By1, Bx2, By2)
V d = (By2 - By1) * (Ax2 - Ax1) - (Bx2 - Bx1) * (Ay2 - Ay1)
I d == 0
R (Float.infinity, Float.infinity)V uA = ((Bx2 - Bx1) * (Ay1 - By1) - (By2 - By1) * (Ax1 - Bx1)) / d
V uB = ((Ax2 - Ax1) * (Ay1 - By1) - (Ay2 - Ay1) * (Ax1 - Bx1)) / dI !(uA C 0.0..1.0 & uB C 0.0..1.0)
R (Float.infinity, Float.infinity)
V x = Ax1 + uA * (Ax2 - Ax1)
V y = Ay1 + uA * (Ay2 - Ay1)R (x, y)
V (a, b, c, d) = (4.0, 0.0, 6.0, 10.0)
V (e, f, g, h) = (0.0, 3.0, 10.0, 7.0)
V pt = line_intersect(a, b, c, d, e, f, g, h)
print(pt)
(5, 5)
360 Assembly [edit]
Translation of: Rexx
* Intersection of two lines 01/03/2019
INTERSEC CSECT
USING INTERSEC,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
SAVE (14,12) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
LE F0,XA xa
IF CE,F0,EQ,XB THEN if xa=xb then
STE F0,X1 x1=xa
LE F0,YA
IF CE,F0,EQ,YB THEN if ya=yb then
MVI MSG,C'=' msg='='
ENDIF , endif
ELSE , else
MVI FK1,X'01' fk1=true
LE F0,YB
SE F0,YA yb-ya
LE F2,XB
SE F2,XA xb-xa
DER F0,F2 /
STE F0,K1 k1=(yb-ya)/(xb-xa)
ME F0,XA k1*xa
LE F2,YA ya
SER F2,F0 -
STE F2,D1 d1=ya-k1*xa
ENDIF , endif
LE F0,XC
IF CE,F0,EQ,XD THEN if xc=xd then
STE F0,X2 x2=xc
LE F4,YC yc
IF CE,F4,EQ,YD THEN if yc=yd then
MVI MSG,C'=' msg='='
ENDIF , endif
ELSE , else
MVI FK2,X'01' fk2=true
LE F0,YD
SE F0,YC yd-yc
LE F2,XD
SE F2,XC xd-xc
DER F0,F2 /
STE F0,K2 k2=(yd-yc)/(xd-xc)
ME F0,XC k2*xc
LE F2,YC yc
SER F2,F0 -
STE F2,D2 d2=yc-k2*xc
ENDIF , endif
IF CLI,MSG,EQ,C' ' THEN if msg=' ' then
IF CLI,FK1,EQ,X'00' THEN if not fk1 then
IF CLI,FK2,EQ,X'00' THEN if not fk2 then
LE F4,X1
IF CE,F4,EQ,X2 if x1=x2 then
MVI MSG,C'=' msg='='
ELSE , else
MVI MSG,C'/' msg='/'
ENDIF , endif
ELSE , else
LE F0,X1
STE F0,X x=x1
LE F0,K2 k2
ME F0,X *x
AE F0,D2 +d2
STE F0,Y y=k2*x+d2
ENDIF , endif
ELSE , else
IF CLI,FK2,EQ,X'00' THEN if not fk2 then
LE F0,X2
STE F0,X x=x2
LE F0,K1 k1
ME F0,X *x
AE F0,D1 +d1
STE F0,Y y=k1*x+d1
ELSE , else
LE F4,K1
IF CE,F4,EQ,K2 THEN if k1=k2 then
LE F4,D1
IF CE,F4,EQ,D2 THEN if d1=d2 then
MVI MSG,C'=' msg='=';
ELSE , else
MVI MSG,C'/' msg='/';
ENDIF , endif
ELSE , else
LE F0,D2 d2
SE F0,D1 -d1
LE F2,K1 k1
SE F2,K2 -k2
DER F0,F2 /
STE F0,X x=(d2-d1)/(k1-k2)
LE F0,K1 k1
ME F0,X *x
AE F0,D1 +d1
STE F0,Y y=k1*x+d1
ENDIF , endif
ENDIF , endif
ENDIF , endif
ENDIF , endif
IF CLI,MSG,EQ,C' ' THEN if msg=' ' then
LE F0,X x
LA R0,3 decimal=3
BAL R14,FORMATF format x
MVC PG+0(13),0(R1) output x
LE F0,Y y
LA R0,3 decimal=3
BAL R14,FORMATF format y
MVC PG+13(13),0(R1) output y
ENDIF , endif
MVC PG+28(1),MSG output msg
XPRNT PG,L'PG print buffer
L R13,4(0,R13) restore previous savearea pointer
RETURN (14,12),RC=0 restore registers from calling sav
COPY plig\$_FORMATF.MLC
XA DC E'4.0' point A
YA DC E'0.0'
XB DC E'6.0' point B
YB DC E'10.0'
XC DC E'0.0' point C
YC DC E'3.0'
XD DC E'10.0' point D
YD DC E'7.0'
X DS E
Y DS E
X1 DS E
X2 DS E
K1 DS E
K2 DS E
D1 DS E
D2 DS E
FK1 DC X'00'
FK2 DC X'00'
MSG DC C' '
PG DC CL80' '
REGEQU
END INTERSEC
5.000 5.000
Ada [edit]
Works with: Ada version 2005
with Ada.Text_IO;procedure Intersection_Of_Two_Lines
is
Do_Not_Intersect : exception;type Line is record
a : Float;
b : Float;
end record;type Point is record
x : Float;
y : Float;
end record;function To_Line(p1, p2 : in Point) return Line
is
a : constant Float := (p1.y - p2.y ) / (p1.x - p2.x );
b : constant Float := p1.y - (a * p1.x );
begin
return (a,b);
end To_Line;function Intersection(Left, Right : in Line) return Point is
begin
if Left.a = Right.a then
raise Do_Not_Intersect with "The two lines do not intersect.";
end if;declare
b : constant Float := (Right.b - Left.b ) / (Left.a - Right.a );
begin
return (b, Left.a * b + Left.b );
end;
end Intersection;A1 : constant Line := To_Line( ( 4.0, 0.0 ), ( 6.0, 10.0 ) );
A2 : constant Line := To_Line( ( 0.0, 3.0 ), ( 10.0, 7.0 ) );
p : constant Point := Intersection(A1, A2);
begin
Ada.Text_IO.Put (p.x'Img);
Ada.Text_IO.Put_Line (p.y'Img);
end Intersection_Of_Two_Lines;
5.00000E+00 5.00000E+00
ALGOL 68 [edit]
Using "school maths".
BEGIN
# mode to hold a point #
MODE POINT = STRUCT ( REAL x, y ) ;
# mode to hold a line expressed as y = mx + c #
MODE LINE = STRUCT ( REAL m, c ) ;
# returns the line that passes through p1 and p2 #
PROC find line = ( POINT p1, p2 )LINE:
IF x OF p1 = x OF p2 THEN
# the line is vertical #
LINE( 0, x OF p1 )
ELSE
# the line is not vertical #
REAL m = ( y OF p1 - y OF p2 ) / ( x OF p1 - x OF p2 ) ;
LINE( m, y OF p1 - ( m * x OF p1 ) )
FI # find line # ;# returns the intersection of two lines - the lines must be distinct and not parallel #
PRIO INTERSECTION = 5;
OP INTERSECTION = ( LINE l1, l2 )POINT:
BEGIN
REAL x = ( c OF l2 - c OF l1 ) / ( m OF l1 - m OF l2 ) ;
POINT( x, ( m OF l1 * x ) + c OF l1 )
END # INTERSECTION # ;# find the intersection of the lines as per the task #
POINT i = find line( POINT( 4.0 , 0.0 ) , POINT( 6.0 , 10.0 ) )
INTERSECTION find line( ( 0.0 , 3.0 ) , ( 10.0 , 7.0 ) ) ;
print ( ( fixed ( x OF i, -8, 4 ) , fixed ( y OF i, -8, 4 ) , newline ) )END
5.0000 5.0000
APL [edit]
⍝ APL has a powerful operator the « dyadic domino » to solve a system of N linear equations with N unknowns
⍝ We use it first to solve the a and b, defining the 2 lines as y = ax + b, with the x and y of the given points
⍝ The system of equations for first line will be:
⍝ 0 = 4a + b
⍝ 10 = 6a + b
⍝ The two arguments to be passed to the dyadic domino are:
⍝ The (0, 10) vector as the left argument
⍝ The ( 4 1 ) matrix as the right argument.
⍝ ( 6 1 )
⍝ We will define a solver that will take the matrix of coordinates, one point per row, then massage the argument to extract x,y
⍝ and inject 1, where needed, and return a pair (a, b) of resolved unknowns.
⍝ Applied twice, we will have a, b and a', b' defining the two lines, we need to resolve it in x and y, in order to determine
⍝ their intersection
⍝ y = ax + b
⍝ y = a'x + b'
⍝ In order to reuse the same solver, we need to format a little bit the arguments, and change the sign of a and a', therefore
⍝ multiply (a,b) and (a', b') by (-1, 1):
⍝ b = -ax + y
⍝ b' = -a'x + y
A ← 4 0
B ← 6 10
C ← 0 3
D ← 10 7
solver ← {(,2 ¯1↑⍵)⌹(2 1↑⍵),1}
I ← solver 2 2⍴((¯1 1)×solver 2 2⍴A,B),(¯1 1)×solver 2 2⍴C,D
I 5 5
AutoHotkey [edit]
LineIntersectionByPoints(L1, L2) {
x1 := L1[ 1 , 1 ] , y1 := L1[ 1 , 2 ]
x2 := L1[ 2 , 1 ] , y2 := L1[ 2 , 2 ]
x3 := L2[ 1 , 1 ] , y3 := L2[ 1 , 2 ]
x4 := L2[ 2 , 1 ] , y4 := L2[ 2 , 2 ]
return ( (x1*y2-y1*x2) * (x3-x4) - (x1-x2) * (x3*y4-y3*x4) ) / ( (x1-x2) * (y3-y4) - (y1-y2) * (x3-x4) ) ", "
. ( (x1*y2-y1*x2) * (y3-y4) - (y1-y2) * (x3*y4-y3*x4) ) / ( (x1-x2) * (y3-y4) - (y1-y2) * (x3-x4) )
}
Examples:
L1 := [ [ 4 , 0 ] , [ 6 , 10 ] ]
L2 := [ [ 0 , 3 ] , [ 10 , 7 ] ]
MsgBox % LineIntersectionByPoints(L1, L2)
Outputs:
5.000000, 5.000000
AWK [edit]
# syntax: GAWK -f FIND_THE_INTERSECTION_OF_TWO_LINES.AWK
# converted from Ring
BEGIN {
intersect( 4,0,6,10,0,3,10,7 )
exit( 0 )
}
function intersect(xa,ya,xb,yb,xc,yc,xd,yd, errors,x,y) {
printf ( "the 1st line passes through (%g,%g) and (%g,%g)\n",xa,ya,xb,yb)
printf ( "the 2nd line passes through (%g,%g) and (%g,%g)\n",xc,yc,xd,yd)
if (xb-xa == 0 ) { print ( "error: xb-xa=0" ) ; errors++ }
if (xd-xc == 0 ) { print ( "error: xd-xc=0" ) ; errors++ }
if (errors > 0 ) {
print ( "" )
return ( 0 )
}
printf ( "the two lines are:\n" )
printf ( "yab=%g+x*%g\n",ya-xa* ( (yb-ya) / (xb-xa) ),(yb-ya) / (xb-xa) )
printf ( "ycd=%g+x*%g\n",yc-xc* ( (yd-yc) / (xd-xc) ),(yd-yc) / (xd-xc) )
x = ( (yc-xc* ( (yd-yc) / (xd-xc) ) ) - (ya-xa* ( (yb-ya) / (xb-xa) ) ) ) / ( ( (yb-ya) / (xb-xa) ) - ( (yd-yc) / (xd-xc) ) )
printf ( "x=%g\n",x)
y = ya-xa* ( (yb-ya) / (xb-xa) ) +x* ( (yb-ya) / (xb-xa) )
printf ( "yab=%g\n",y)
printf ( "ycd=%g\n",yc-xc* ( (yd-yc) / (xd-xc) ) +x* ( (yd-yc) / (xd-xc) ) )
printf ( "intersection: %g,%g\n \n",x,y)
return ( 1 )
}
the 1st line passes through (4,0) and (6,10) the 2nd line passes through (0,3) and (10,7) the two lines are: yab=-20+x*5 ycd=3+x*0.4 x=5 yab=5 ycd=5 intersection: 5,5
BASIC [edit]
Sinclair ZX81 BASIC [edit]
Translation of: REXX
(version 1)
Works with 1k of RAM.
10 LET XA=4
20 LET YA=0
30 LET XB=6
40 LET YB=10
50 LET XC=0
60 LET YC=3
70 LET XD=10
80 LET YD=0
90 PRINT "THE TWO LINES ARE:"
100 PRINT "YAB=";YA-XA*((YB-YA)/(XB-XA));"+X*";((YB-YA)/(XB-XA))
110 PRINT "YCD=";YC-XC*((YD-YC)/(XD-XC));"+X*";((YD-YC)/(XD-XC))
120 LET X=((YC-XC*((YD-YC)/(XD-XC)))-(YA-XA*((YB-YA)/(XB-XA))))/(((YB-YA)/(XB-XA))-((YD-YC)/(XD-XC)))
130 PRINT "X=";X
140 LET Y=YA-XA*((YB-YA)/(XB-XA))+X*((YB-YA)/(XB-XA))
150 PRINT "YAB=";Y
160 PRINT "YCD=";YC-XC*((YD-YC)/(XD-XC))+X*((YD-YC)/(XD-XC))
170 PRINT "INTERSECTION: ";X;",";Y
THE TWO LINES ARE: YAB=-20+X*5 YCD=3+X*0.4 X=5 YAB=5 YCD=5 INTERSECTION: 5,5
C [edit]
This implementation is generic and considers any two lines in the XY plane and not just the specified example. Usage is printed on incorrect invocation.
#include<stdlib.h>
#include<stdio.h>
#include<math.h>typedef struct {
double x,y;
}point;double lineSlope(point a,point b) {
if (a.x -b.x == 0.0 )
return NAN;
else
return (a.y -b.y ) / (a.x -b.x ) ;
}point extractPoint( char * str) {
int i,j,start,end,length;
char * holder;
point c;for (i= 0 ;str[i] != 00 ;i++ ) {
if (str[i] == '(' )
start = i;
if (str[i] == ',' ||str[i] == ')' )
{
end = i;length = end - start;
holder = ( char * ) malloc (length* sizeof ( char ) ) ;
for (j= 0 ;j<length- 1 ;j++ )
holder[j] = str[start + j + 1 ] ;
holder[j] = 00 ;if (str[i] == ',' ) {
start = i;
c.x = atof (holder) ;
}
else
c.y = atof (holder) ;
}
}return c;
}point intersectionPoint(point a1,point a2,point b1,point b2) {
point c;double slopeA = lineSlope(a1,a2) , slopeB = lineSlope(b1,b2) ;
if (slopeA==slopeB) {
c.x = NAN;
c.y = NAN;
}
else if (isnan(slopeA) && !isnan(slopeB) ) {
c.x = a1.x ;
c.y = (a1.x -b1.x ) *slopeB + b1.y ;
}
else if (isnan(slopeB) && !isnan(slopeA) ) {
c.x = b1.x ;
c.y = (b1.x -a1.x ) *slopeA + a1.y ;
}
else {
c.x = (slopeA*a1.x - slopeB*b1.x + b1.y - a1.y ) / (slopeA - slopeB) ;
c.y = slopeB* (c.x - b1.x ) + b1.y ;
}return c;
}int main( int argC, char * argV[ ] )
{
point c;if (argC < 5 )
printf ( "Usage : %s <four points specified as (x,y) separated by a space>" ,argV[ 0 ] ) ;
else {
c = intersectionPoint(extractPoint(argV[ 1 ] ) ,extractPoint(argV[ 2 ] ) ,extractPoint(argV[ 3 ] ) ,extractPoint(argV[ 4 ] ) ) ;if (isnan(c.x ) )
printf ( "The lines do not intersect, they are either parallel or co-incident." ) ;
else
printf ( "Point of intersection : (%lf,%lf)" ,c.x ,c.y ) ;
}return 0 ;
}
Invocation and output:
C:\rosettaCode>lineIntersect.exe (4,0) (6,10) (0,3) (10,7) Point of intersection : (5.000000,5.000000)
C# [edit]
using System ;
using System.Drawing ;
public class Program
{
static PointF FindIntersection(PointF s1, PointF e1, PointF s2, PointF e2) {
float a1 = e1. Y - s1. Y ;
float b1 = s1. X - e1. X ;
float c1 = a1 * s1. X + b1 * s1. Y ;float a2 = e2. Y - s2. Y ;
float b2 = s2. X - e2. X ;
float c2 = a2 * s2. X + b2 * s2. Y ;float delta = a1 * b2 - a2 * b1;
//If lines are parallel, the result will be (NaN, NaN).
return delta == 0 ? new PointF( float . NaN, float . NaN )
: new PointF( (b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta) ;
}static void Main( ) {
Func< float, float, PointF> p = (x, y) => new PointF(x, y) ;
Console. WriteLine (FindIntersection(p(4f, 0f), p(6f, 10f), p(0f, 3f), p(10f, 7f) ) ) ;
Console. WriteLine (FindIntersection(p(0f, 0f), p(1f, 1f), p(1f, 2f), p(4f, 5f) ) ) ;
}
}
{X=5, Y=5} {X=NaN, Y=NaN} C++ [edit]
#include <iostream>
#include <cmath>
#include <cassert>
using namespace std;/** Calculate determinant of matrix:
[a b]
[c d]
*/
inline double Det( double a, double b, double c, double d)
{
return a*d - b*c;
}/// Calculate intersection of two lines.
///\return true if found, false if not found or error
bool LineLineIntersect( double x1, double y1, // Line 1 start
double x2, double y2, // Line 1 end
double x3, double y3, // Line 2 start
double x4, double y4, // Line 2 end
double &ixOut, double &iyOut) // Output
{
double detL1 = Det(x1, y1, x2, y2) ;
double detL2 = Det(x3, y3, x4, y4) ;
double x1mx2 = x1 - x2;
double x3mx4 = x3 - x4;
double y1my2 = y1 - y2;
double y3my4 = y3 - y4;double denom = Det(x1mx2, y1my2, x3mx4, y3my4) ;
if (denom == 0.0 ) // Lines don't seem to cross
{
ixOut = NAN;
iyOut = NAN;
return false ;
}double xnom = Det(detL1, x1mx2, detL2, x3mx4) ;
double ynom = Det(detL1, y1my2, detL2, y3my4) ;
ixOut = xnom / denom;
iyOut = ynom / denom;
if ( !isfinite(ixOut) || !isfinite(iyOut) ) // Probably a numerical issue
return false ;return true ; //All OK
}int main( )
{
// **Simple crossing diagonal lines**// Line 1
double x1= 4.0, y1= 0.0 ;
double x2= 6.0, y2= 10.0 ;// Line 2
double x3= 0.0, y3= 3.0 ;
double x4= 10.0, y4= 7.0 ;double ix = - 1.0, iy = - 1.0 ;
bool result = LineLineIntersect(x1, y1, x2, y2, x3, y3, x4, y4, ix, iy) ;
cout << "result " << result << "," << ix << "," << iy << endl;double eps = 1e-6 ;
assert (result == true ) ;
assert ( fabs (ix - 5.0 ) < eps) ;
assert ( fabs (iy - 5.0 ) < eps) ;
return 0 ;
}
result 1,5,5
Clojure [edit]
;; Point is [x y] tuple
( defn compute-line [pt1 pt2]
( let [ [x1 y1] pt1
[x2 y2] pt2
m ( / ( - y2 y1) ( - x2 x1) ) ]
{:slope m
:offset ( - y1 ( * m x1) ) } ) )( defn intercept [line1 line2]
( let [x ( / ( - (:offset line1) (:offset line2) )
( - (:slope line2) (:slope line1) ) ) ]
{:x x
:y ( + ( * (:slope line1) x)
(:offset line1) ) } ) )
(def line1 (compute-line [4 0] [6 10])) (def line2 (compute-line [0 3] [10 7])) line1 ; {:slope 5, :offset -20} line2 ; {:slope 2/5, :offset 3} (intercept line1 line2) ; {:x 5, :y 5} D [edit]
Translation of: Kotlin
import std.stdio ;struct Point {
real x, y;void toString( scope void delegate ( const ( char ) [ ] ) sink) const {
import std.format ;
sink( "{" ) ;
sink.formattedWrite ! "%f" (x) ;
sink( ", " ) ;
sink.formattedWrite ! "%f" (y) ;
sink( "}" ) ;
}
}struct Line {
Point s, e;
}Point findIntersection(Line l1, Line l2) {
auto a1 = l1.e.y - l1.s.y ;
auto b1 = l1.s.x - l1.e.x ;
auto c1 = a1 * l1.s.x + b1 * l1.s.y ;auto a2 = l2.e.y - l2.s.y ;
auto b2 = l2.s.x - l2.e.x ;
auto c2 = a2 * l2.s.x + b2 * l2.s.y ;auto delta = a1 * b2 - a2 * b1;
// If lines are parallel, intersection point will contain infinite values
return Point( (b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta) ;
}void main( ) {
auto l1 = Line(Point( 4.0 , 0.0 ) , Point( 6.0 , 10.0 ) ) ;
auto l2 = Line(Point( 0f , 3f ) , Point( 10f , 7f ) ) ;
writeln(findIntersection(l1, l2) ) ;
l1 = Line(Point( 0.0 , 0.0 ) , Point( 1.0 , 1.0 ) ) ;
l2 = Line(Point( 1.0 , 2.0 ) , Point( 4.0 , 5.0 ) ) ;
writeln(findIntersection(l1, l2) ) ;
}
{5.000000, 5.000000} {-inf, -inf} F# [edit]
(*
Find the point of intersection of 2 lines.
Nigel Galloway May 20th., 2017
*)
type Line= {a:float;b:float;c:float } member N.toS =sprintf "%.2fx + %.2fy = %.2f" N.a N.b N.c
let intersect (n:Line) g = match (n.a *g.b -g.a *n.b ) with
|0.0 ->printfn "%s does not intersect %s" n.toS g.toS
|ng ->printfn "%s intersects %s at x=%.2f y=%.2f" n.toS g.toS ( (g.b *n.c -n.b *g.c ) /ng) ( (n.a *g.c -g.a *n.c ) /ng)
let fn (i,g) (e,l) = {a=g-l;b=e-i;c= (e-i) *g+ (g-l) *i}
intersect (fn ( 4.0,0.0 ) ( 6.0,10.0 ) ) (fn ( 0.0,3.0 ) ( 10.0,7.0 ) )
intersect {a= 3.18 ;b= 4.23 ;c= 7.13 } {a= 6.36 ;b= 8.46 ;c= 9.75 }
-10.00x + 2.00y = -40.00 intersects -4.00x + 10.00y = 30.00 at x=5.00 y=5.00 3.18x + 4.23y = 7.13 does not intersect 6.36x + 8.46y = 9.75
Factor [edit]
Works with: Factor version 0.99 2020-01-23
USING: arrays combinators.extras kernel math
math.matrices.laplace math.vectors prettyprint sequences ;: det ( pt pt -- x ) 2array determinant ;
: numerator ( x y pt pt quot -- z )
[email protected] swapd [ 2array ] [email protected] det ; inline: intersection ( pt pt pt pt -- pt )
[ [ det ] [email protected] ]
[ [ v- ] [email protected] ] 4bi
[ [ first ] numerator ]
[ [ second ] numerator ]
[ det 2nip ] 4tri
dup zero? [ 3drop { 0/0. 0/0. } ]
[ tuck [ / ] [email protected] 2array ] if ;{ 4 0 } { 6 10 } { 0 3 } { 10 7 } intersection .
{ 4 0 } { 6 10 } { 0 3 } { 10 7+1/10 } intersection .
{ 0 0 } { 1 1 } { 1 2 } { 4 5 } intersection .
{ 5 5 } { 5+5/459 5+25/459 } { NAN: 8000000000000 NAN: 8000000000000 } Fortran [edit]
Works with: Fortran version 90 and later
program intersect_two_lines
implicit nonetype point
real :: x,y
end type pointinteger, parameter :: n = 4
type (point) :: p (n)p( 1 ) %x = 4; p( 1 ) %y = 0; p( 2 ) %x = 6; p( 2 ) %y = 10 ! fist line
p( 3 ) %x = 0; p( 3 ) %y = 3; p( 4 ) %x = 10; p( 4 ) %y = 7 ! second linecall intersect(p, n)
contains
subroutine intersect(p,m)
integer, intent ( in ) :: m
type (point), intent ( in ) :: p (m)
integer :: i
real :: a ( 2 ), b( 2 ) ! y = a*x + b, for each line
real :: x, y ! intersect point
real :: dx,dy ! working variablesdo i = 1, 2
dx = p( 2 *i- 1 ) %x - p( 2 *i) %x
dy = p( 2 *i- 1 ) %y - p( 2 *i) %y
if ( dx == 0.) then ! in case this line is of the form y = b
a(i) = 0.
b (i) = p( 2 *i- 1 ) %y
else
a(i) = dy / dx
b(i) = p( 2 *i- 1 ) %y - a(i) *p( 2 *i- 1 ) %x
endif
enddoif ( a( 1 ) - a( 2 ) == 0. ) then
write( *,* ) "lines are not intersecting"
return
endifx = ( b( 2 ) - b( 1 ) ) / ( a( 1 ) - a( 2 ) )
y = a( 1 ) * x + b( 1 )
write( *,* )x,y
end subroutine intersect
end program intersect_two_lines
5.00000000 5.00000000
FreeBASIC [edit]
' version 16-08-2017
' compile with: fbc -s console
#Define NaN 0 / 0 ' FreeBASIC returns -1.#INDType _point_
As Double x, y
End TypeFunction l_l_intersect(s1 As _point_, e1 As _point_, s2 As _point_, e2 As _point_) As _point_
Dim As Double a1 = e1.y - s1.y
Dim As Double b1 = s1.x - e1.x
Dim As Double c1 = a1 * s1.x + b1 * s1.y
Dim As Double a2 = e2.y - s2.y
Dim As Double b2 = s2.x - e2.x
Dim As Double c2 = a2 * s2.x + b2 * s2.y
Dim As Double det = a1 * b2 - a2 * b1If det = 0 Then
Return Type (NaN, NaN)
Else
Return Type ( (b2 * c1 - b1 * c2) / det, (a1 * c2 - a2 * c1) / det)
End IfEnd Function
' ------=< MAIN >=------
Dim As _point_ s1, e1, s2, e2, answer
s1.x = 4.0 : s1.y = 0.0 : e1.x = 6.0 : e1.y = 10.0 ' start and end of first line
s2.x = 0.0 : s2.y = 3.0 : e2.x = 10.0 : e2.y = 7.0 ' start and end of second line
answer = l_l_intersect(s1, e1, s2, e2)
Print answer.x, answer.ys1.x = 0.0 : s1.y = 0.0 : e1.x = 0.0 : e1.y = 0.0 ' start and end of first line
s2.x = 0.0 : s2.y = 3.0 : e2.x = 10.0 : e2.y = 7.0 ' start and end of second line
answer = l_l_intersect(s1, e1, s2, e2)
Print answer.x, answer.y' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
5 5 -1.#IND -1.#IND
Go [edit]
package mainimport (
"fmt"
"errors"
)type Point struct {
x float64
y float64
}type Line struct {
slope float64
yint float64
}func CreateLine (a, b Point) Line {
slope := (b.y-a.y) / (b.x-a.x)
yint := a.y - slope*a.x
return Line{slope, yint}
}func EvalX (l Line, x float64 ) float64 {
return l.slope*x + l.yint
}func Intersection (l1, l2 Line) (Point, error) {
if l1.slope == l2.slope {
return Point{}, errors.New( "The lines do not intersect" )
}
x := (l2.yint-l1.yint) / (l1.slope-l2.slope)
y := EvalX(l1, x)
return Point{x, y}, nil
}func main() {
l1 := CreateLine(Point{ 4 , 0 }, Point{ 6 , 10 })
l2 := CreateLine(Point{ 0 , 3 }, Point{ 10 , 7 })
if result, err := Intersection(l1, l2); err == nil {
fmt.Println(result)
} else {
fmt.Println( "The lines do not intersect" )
}
}
{5 5} Groovy [edit]
Translation of: Java
class Intersection {
private static class Point {
double x, yPoint ( double x, double y) {
this.x = x
this.y = y
}@Override
String toString( ) {
return "($x, $y)"
}
}private static class Line {
Point s, eLine ( Point s, Point e) {
this.s = s
this.e = e
}
}private static Point findIntersection( Line l1, Line l2) {
double a1 = l1.e.y - l1.s.y
double b1 = l1.s.x - l1.e.x
double c1 = a1 * l1.s.x + b1 * l1.s.ydouble a2 = l2.e.y - l2.s.y
double b2 = l2.s.x - l2.e.x
double c2 = a2 * l2.s.x + b2 * l2.s.ydouble delta = a1 * b2 - a2 * b1
return new Point ( (b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta)
}static void main( String [ ] args) {
Line l1 = new Line ( new Point ( 4, 0 ), new Point ( 6, 10 ) )
Line l2 = new Line ( new Point ( 0, 3 ), new Point ( 10, 7 ) )
println (findIntersection(l1, l2) )l1 = new Line ( new Point ( 0, 0 ), new Point ( 1, 1 ) )
l2 = new Line ( new Point ( 1, 2 ), new Point ( 4, 5 ) )
println (findIntersection(l1, l2) )
}
}
(5.0, 5.0) (-Infinity, -Infinity)
Haskell [edit]
type Line = (Point, Point)type Point = ( Float , Float )
intersection :: Line -> Line -> Either String Point
intersection ab pq =
case determinant of
0 -> Left "(Parallel lines – no intersection)"
_ ->
let [abD, pqD] = (\(a, b) -> diff ( [ fst , snd ] <*> [a, b] ) ) <$> [ab, pq]
[ix, iy] =
[\(ab, pq) -> diff [abD, ab, pqD, pq] / determinant] <*>
[ (abDX, pqDX) , (abDY, pqDY) ]
in Right (ix, iy)
where
delta f x = f ( fst x) - f ( snd x)
diff [a, b, c, d] = a * d - b * c
[abDX, pqDX, abDY, pqDY] = [delta fst , delta snd ] <*> [ab, pq]
determinant = diff [abDX, abDY, pqDX, pqDY]-- TEST ----------------------------------------------------------------
ab :: Line
ab = ( ( 4.0 , 0.0 ) , ( 6.0 , 10.0 ) )pq :: Line
pq = ( ( 0.0 , 3.0 ) , ( 10.0 , 7.0 ) )interSection :: Either String Point
interSection = intersection ab pqmain :: IO ( )
main =
putStrLn $
case interSection of
Left x -> x
Right x -> show x
(5.0,5.0)
J [edit]
Translation of: C++
Solution:
det=: -/ .* NB. calculate determinant
findIntersection=: (det ,."1 [: |: -/"2 ) %&det -/"2
Examples:
line1=: 4 0 ,: 6 10
line2=: 0 3 ,: 10 7
line3=: 0 3 ,: 10 7.1
line4=: 0 0 ,: 1 1
line5=: 1 2 ,: 4 5
line6=: 1 _1 ,: 4 4
line7=: 2 5 ,: 3 _2findIntersection line1 ,: line2
5 5
findIntersection line1 ,: line3
5.01089 5.05447
findIntersection line4 ,: line5
__ __
findIntersection line6 ,: line7
2.5 1.5
Java [edit]
Translation of: Kotlin
public class Intersection {
private static class Point {
double x, y;Point ( double x, double y) {
this.x = x;
this.y = y;
}@Override
public String toString( ) {
return String.format ( "{%f, %f}", x, y) ;
}
}private static class Line {
Point s, e;Line ( Point s, Point e) {
this.s = s;
this.e = e;
}
}private static Point findIntersection( Line l1, Line l2) {
double a1 = l1.e.y - l1.s.y ;
double b1 = l1.s.x - l1.e.x ;
double c1 = a1 * l1.s.x + b1 * l1.s.y ;double a2 = l2.e.y - l2.s.y ;
double b2 = l2.s.x - l2.e.x ;
double c2 = a2 * l2.s.x + b2 * l2.s.y ;double delta = a1 * b2 - a2 * b1;
return new Point ( (b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta) ;
}public static void main( String [ ] args) {
Line l1 = new Line ( new Point ( 4, 0 ), new Point ( 6, 10 ) ) ;
Line l2 = new Line ( new Point ( 0, 3 ), new Point ( 10, 7 ) ) ;
System.out.println (findIntersection(l1, l2) ) ;l1 = new Line ( new Point ( 0, 0 ), new Point ( 1, 1 ) ) ;
l2 = new Line ( new Point ( 1, 2 ), new Point ( 4, 5 ) ) ;
System.out.println (findIntersection(l1, l2) ) ;
}
}
{5.000000, 5.000000} {-Infinity, -Infinity} JavaScript [edit]
Translation of: Haskell
ES6 [edit]
( ( ) => {
'use strict' ;
// INTERSECTION OF TWO LINES ----------------------------------------------// intersection :: Line -> Line -> Either String (Float, Float)
const intersection = (ab, pq) => {
const
delta = f => x => f(fst(x) ) - f(snd(x) ) ,
[abDX, pqDX, abDY, pqDY] = apList(
[delta(fst) , delta(snd) ] , [ab, pq]
) ,
determinant = abDX * pqDY - abDY * pqDX;return determinant !== 0 ? Right( ( ( ) => {
const [abD, pqD] = map(
( [a, b] ) => fst(a) * snd(b) - fst(b) * snd(a) ,
[ab, pq]
) ;
return apList(
[ ( [pq, ab] ) =>
(abD * pq - ab * pqD) / determinant
] , [
[pqDX, abDX] ,
[pqDY, abDY]
]
) ;
} ) ( ) ) : Left( '(Parallel lines – no intersection)' ) ;
} ;// GENERIC FUNCTIONS ------------------------------------------------------
// Left :: a -> Either a b
const Left = x => ( {
type: 'Either' ,
Left: x
} ) ;// Right :: b -> Either a b
const Right = x => ( {
type: 'Either' ,
Right: x
} ) ;// A list of functions applied to a list of arguments
// <*> :: [(a -> b)] -> [a] -> [b]
const apList = (fs, xs) => //
[ ].concat.apply ( [ ] , fs.map (f => //
[ ].concat.apply ( [ ] , xs.map (x => [f(x) ] ) ) ) ) ;// fst :: (a, b) -> a
const fst = tpl => tpl[ 0 ] ;// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map (f) ;// snd :: (a, b) -> b
const snd = tpl => tpl[ 1 ] ;// show :: a -> String
const show = x => JSON.stringify (x) ; //, null, 2);// TEST --------------------------------------------------
// lrIntersection ::Either String Point
const lrIntersection = intersection( [
[ 4.0 , 0.0 ] ,
[ 6.0 , 10.0 ]
] , [
[ 0.0 , 3.0 ] ,
[ 10.0 , 7.0 ]
] ) ;
return show(lrIntersection.Left || lrIntersection.Right ) ;
} ) ( ) ;
[5,5]
jq [edit]
The implementation closely follows the zkl entry but uses the JSON array [x,y] to represent the point (x,y), and an array [P1,P2] to represent the line through points P1 and P2. Array destructuring is used for simplicity.
# determinant of 2x2 matrix
def det(a;b;c;d): a*d - b*c ;# Input: an array representing a line (L1)
# Output: the intersection of L1 and L2 unless the lines are judged to be parallel
# This implementation uses "destructuring" to assign local variables
def lineIntersection(L2):
. as [[$ax,$ay], [$bx,$by]]
| L2 as [[$cx,$cy], [$dx,$dy]]
| {detAB: det($ax;$ay; $bx;$by),
detCD: det($cx;$cy; $dx;$dy),
abDx: ($ax - $bx),
cdDx: ($cx - $dx),
abDy: ($ay - $by),
cdDy: ($cy - $dy)}
| . + {xnom: det(.detAB;.abDx;.detCD;.cdDx),
ynom: det(.detAB;.abDy;.detCD;.cdDy),
denom: det(.abDx; .abDy;.cdDx; .cdDy) }
| if (.denom|length < 10e-6) # length/0 emits the absolute value
then error("lineIntersect: parallel lines")
else [.xnom/.denom, .ynom/.denom]
end ;
Example:
[[4.0, 0.0], [6.0,10.0]] | lineIntersection([[0.0, 3.0], [10.0, 7.0]])
[5,5]
Julia [edit]
Works with: Julia version 0.6
Translation of: Kotlin
struct Point{T}
x::T
y::T
end struct Line{T}
s::Point{T}
e::Point{T}
end
function intersection(l1::Line{T}, l2::Line{T}) where T<:Real
a1 = l1.e.y - l1.s.y
b1 = l1.s.x - l1.e.x
c1 = a1 * l1.s.x + b1 * l1.s.y
a2 = l2.e.y - l2.s.y
b2 = l2.s.x - l2.e.x
c2 = a2 * l2.s.x + b2 * l2.s.y
Δ = a1 * b2 - a2 * b1
# If lines are parallel, intersection point will contain infinite values
return Point((b2 * c1 - b1 * c2) / Δ, (a1 * c2 - a2 * c1) / Δ)
end
l1 = Line(Point{Float64}(4, 0), Point{Float64}(6, 10))
l2 = Line(Point{Float64}(0, 3), Point{Float64}(10, 7))
println(intersection(l1, l2))
l1 = Line(Point{Float64}(0, 0), Point{Float64}(1, 1))
l2 = Line(Point{Float64}(1, 2), Point{Float64}(4, 5))
println(intersection(l1, l2))
Point{Float64}(5.0, 5.0) Point{Float64}(-Inf, -Inf) GeometryTypes library version [edit]
using GeometryTypesa = LineSegment(Point2f0(4, 0), Point2f0(6, 10))
b = LineSegment(Point2f0(0, 3), Point2f0(10, 7))
@show intersects(a, b) # --> intersects(a, b) = (true, Float32[5.0, 5.0])
Kotlin [edit]
Translation of: C#
// version 1.1.2class PointF( val x: Float, val y: Float) {
override fun toString( ) = "{$x, $y}"
}class LineF( val s: PointF, val e: PointF)
fun findIntersection(l1: LineF, l2: LineF) : PointF {
val a1 = l1.e.y - l1.s.y
val b1 = l1.s.x - l1.e.x
val c1 = a1 * l1.s.x + b1 * l1.s.yval a2 = l2.e.y - l2.s.y
val b2 = l2.s.x - l2.e.x
val c2 = a2 * l2.s.x + b2 * l2.s.yval delta = a1 * b2 - a2 * b1
// If lines are parallel, intersection point will contain infinite values
return PointF( (b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta)
}fun main(args: Array<String> ) {
var l1 = LineF(PointF(4f, 0f), PointF(6f, 10f) )
var l2 = LineF(PointF(0f, 3f), PointF(10f, 7f) )
println(findIntersection(l1, l2) )
l1 = LineF(PointF(0f, 0f), PointF(1f, 1f) )
l2 = LineF(PointF(1f, 2f), PointF(4f, 5f) )
println(findIntersection(l1, l2) )
}
{5.0, 5.0} {-Infinity, -Infinity} Lua [edit]
Translation of: C#
function intersection (s1, e1, s2, e2)
local d = (s1.x - e1.x) * (s2.y - e2.y) - (s1.y - e1.y) * (s2.x - e2.x)
local a = s1.x * e1.y - s1.y * e1.x
local b = s2.x * e2.y - s2.y * e2.x
local x = (a * (s2.x - e2.x) - (s1.x - e1.x) * b) / d
local y = (a * (s2.y - e2.y) - (s1.y - e1.y) * b) / d
return x, y
endlocal line1start, line1end = {x = 4 , y = 0 } , {x = 6 , y = 10 }
local line2start, line2end = {x = 0 , y = 3 } , {x = 10 , y = 7 }
print (intersection(line1start, line1end, line2start, line2end) )
5 5
M2000 Interpreter [edit]
Module Lineintersection (lineAtuple, lineBtuple) {
class line {
private:
slop, k
public:
function f(x) {
=x*.slop-.k
}
function intersection(b as line) {
if b.slop==.slop then
=(,)
else
x1=(.k-b.k)/(.slop-b.slop)
=(x1, .f(x1))
end if
}
Class:
module line {
read x1, y1, x2, y2
if x1==x2 then error "wrong input"
if x1>x2 then swap x1,x2 : swap y1, y2
.slop<=(y1-y2)/(x1-x2)
.k<=x1*.slop-y1
}
}
M=line(!lineAtuple)
N=line(!lineBtuple)
Print M.intersection(N)
}
Lineintersection (4,0,6,10), (0,3,10,7) ' print 5 5
5 5
Maple [edit]
with(geometry):
line(L1, [point(A,[4,0]), point(B,[6,10])]):
line(L2, [point(C,[0,3]), point(E,[10,7])]):
coordinates(intersection(x,L1,L2));
[5, 5]
Mathematica/Wolfram Language [edit]
RegionIntersection[
InfiniteLine[{{4, 0}, {6, 10}}],
InfiniteLine[{{0, 3}, {10, 7}}]
]
Point[{5, 5}] MATLAB [edit]
function cross=intersection(line1,line2)
a=polyfit (line1(:,1 ),line1(:,2 ),1 );
b=polyfit (line2(:,1 ),line2(:,2 ),1 );
cross=[a( 1 ) -1; b( 1 ) -1 ]\[-a( 2 );-b( 2 ) ];
end
line1=[4 0; 6 10]; line2=[0 3; 10 7]; cross=intersection(line1,line2) cross = 5.0000 5.0000
Modula-2 [edit]
MODULE LineIntersection;
FROM RealStr IMPORT RealToStr;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;TYPE
Point = RECORD
x,y : REAL;
END;PROCEDURE PrintPoint(p : Point);
VAR buf : ARRAY [ 0..31 ] OF CHAR;
BEGIN
WriteString( "{" );
RealToStr(p.x, buf);
WriteString(buf);
WriteString( ", " );
RealToStr(p.y, buf);
WriteString(buf);
WriteString( "}" );
END PrintPoint;TYPE
Line = RECORD
s,e : Point;
END;PROCEDURE FindIntersection(l1,l2 : Line) : Point;
VAR a1,b1,c1,a2,b2,c2,delta : REAL;
BEGIN
a1 := l1.e.y - l1.s.y;
b1 := l1.s.x - l1.e.x;
c1 := a1 * l1.s.x + b1 * l1.s.y;a2 := l2.e.y - l2.s.y;
b2 := l2.s.x - l2.e.x;
c2 := a2 * l2.s.x + b2 * l2.s.y;delta := a1 * b2 - a2 * b1;
RETURN Point{ (b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta};
END FindIntersection;VAR
l1,l2 : Line;
result : Point;
BEGIN
l1 := Line{ { 4.0 , 0.0 } , { 6.0 , 10.0 } };
l2 := Line{ { 0.0 , 3.0 } , { 10.0 , 7.0 } };
PrintPoint(FindIntersection(l1,l2) );
WriteLn;l1 := Line{ { 0.0 , 0.0 } , { 1.0 , 1.0 } };
l2 := Line{ { 1.0 , 2.0 } , { 4.0 , 5.0 } };
PrintPoint(FindIntersection(l1,l2) );
WriteLn;ReadChar;
END LineIntersection.
Nim [edit]
Translation of: Go
type
Line = tuple
slope: float
yInt: float
Point = tuple
x: float
y: floatfunc createLine(a, b: Point): Line =
result.slope = (b.y - a.y) / (b.x - a.x)
result.yInt = a.y - result.slope * a.xfunc evalX(line: Line, x: float): float =
line.slope * x + line.yIntfunc intersection(line1, line2: Line): Point =
let x = (line2.yInt - line1.yInt) / (line1.slope - line2.slope)
let y = evalX(line1, x)
(x, y)var line1 = createLine((4.0, 0.0), (6.0, 10.0))
var line2 = createLine((0.0, 3.0), (10.0, 7.0))
echo intersection(line1, line2)
line1 = createLine((0.0, 0.0), (1.0, 1.0))
line2 = createLine((1.0, 2.0), (4.0, 5.0))
echo intersection(line1, line2)
(x: 5.0, y: 5.0) (x: inf, y: inf)
Perl [edit]
Translation of: C#
If warning are enabled the second print will issue a warning since we are trying to print out an undef
sub intersect {
my ( $x1 , $y1 , $x2 , $y2 , $x3 , $y3 , $x4 , $y4 ) = @_ ;
my $a1 = $y2 - $y1 ;
my $b1 = $x1 - $x2 ;
my $c1 = $a1 * $x1 + $b1 * $y1 ;
my $a2 = $y4 - $y3 ;
my $b2 = $x3 - $x4 ;
my $c2 = $a2 * $x3 + $b2 * $y3 ;
my $delta = $a1 * $b2 - $a2 * $b1 ;
return ( undef , undef ) if $delta == 0 ;
# If delta is 0, i.e. lines are parallel then the below will fail
my $ix = ( $b2 * $c1 - $b1 * $c2 ) / $delta ;
my $iy = ( $a1 * $c2 - $a2 * $c1 ) / $delta ;
return ( $ix , $iy ) ;
}my ( $ix , $iy ) = intersect( 4 , 0 , 6 , 10 , 0 , 3 , 10 , 7 ) ;
print "$ix $iy\n" ;
( $ix , $iy ) = intersect( 0 , 0 , 1 , 1 , 1 , 2 , 4 , 5 ) ;
print "$ix $iy\n" ;
Phix [edit]
enum X, Yfunction abc(sequence s,e)
-- yeilds {a,b,c}, corresponding to ax+by=c
atom a = e[Y]-s[Y], b = s[X]-e[X], c = a*s[X]+b*s[Y]
return {a,b,c}
end functionprocedure intersect(sequence s1, e1, s2, e2)
atom {a1,b1,c1} = abc(s1,e1),
{a2,b2,c2} = abc(s2,e2),
delta = a1*b2 - a2*b1,
x = b2*c1 - b1*c2,
y = a1*c2 - a2*c1
?iff(delta=0?"parallel lines/do not intersect"
:{x/delta, y/delta})
end procedureintersect({4,0},{6,10},{0,3},{10,7}) -- {5,5}
intersect({4,0},{6,10},{0,3},{10,7.1}) -- {5.010893246,5.054466231}
intersect({0,0},{0,0},{0,3},{10,7}) -- "parallel lines/do not intersect"
intersect({0,0},{1,1},{1,2},{4,5}) -- "parallel lines/do not intersect"
intersect({1,-1},{4,4},{2,5},{3,-2}) -- {2.5,1.5}
Processing [edit]
void setup( ) {
// test lineIntersect() with visual and textual output
float lineA[ ] = { 4, 0, 6, 10 } ; // try 4, 0, 6, 4
float lineB[ ] = { 0, 3, 10, 7 } ; // for non intersecting test
PVector pt = lineInstersect(lineA[ 0 ], lineA[ 1 ], lineA[ 2 ], lineA[ 3 ],
lineB[ 0 ], lineB[ 1 ], lineB[ 2 ], lineB[ 3 ] ) ;
scale( 9 ) ;
line(lineA[ 0 ], lineA[ 1 ], lineA[ 2 ], lineA[ 3 ] ) ;
line(lineB[ 0 ], lineB[ 1 ], lineB[ 2 ], lineB[ 3 ] ) ;
if (pt != null ) {
stroke( 255 ) ;
point(pt.x, pt.y ) ;
println(pt.x, pt.y ) ;
} else {
println( "No point" ) ;
}
}PVector lineInstersect( float Ax1, float Ay1, float Ax2, float Ay2,
float Bx1, float By1, float Bx2, float By2) {
// returns null if there is no intersection
float uA, uB;
float d = ( (By2 - By1) * (Ax2 - Ax1) - (Bx2 - Bx1) * (Ay2 - Ay1) ) ;
if (d != 0 ) {
uA = ( (Bx2 - Bx1) * (Ay1 - By1) - (By2 - By1) * (Ax1 - Bx1) ) / d;
uB = ( (Ax2 - Ax1) * (Ay1 - By1) - (Ay2 - Ay1) * (Ax1 - Bx1) ) / d;
} else {
return null ;
}
if ( 0 > uA || uA > 1 || 0 > uB || uB > 1 ) {
return null ;
}
float x = Ax1 + uA * (Ax2 - Ax1) ;
float y = Ay1 + uA * (Ay2 - Ay1) ;
return new PVector(x, y) ;
}
Processing Python mode [edit]
from __future__ import divisiondef setup( ):
""" test line_intersect() with visual and textual output """
(a, b) , (c, d) = ( 4 , 0 ) , ( 6 , 10 ) # try (4, 0), (6, 4)
(e, f) , (g, h) = ( 0 , 3 ) , ( 10 , 7 ) # for non intersecting test
pt = line_instersect(a, b, c, d, e, f, g, h)
scale( 9 )
line(a, b, c, d)
line(e, f, g, h)
if pt:
x, y = pt
stroke( 255 )
point(x, y)
println(pt) # prints x, y coordinates or 'None'def line_instersect(Ax1, Ay1, Ax2, Ay2, Bx1, By1, Bx2, By2):
""" returns a (x, y) tuple or None if there is no intersection """
d = (By2 - By1) * (Ax2 - Ax1) - (Bx2 - Bx1) * (Ay2 - Ay1)
if d:
uA = ( (Bx2 - Bx1) * (Ay1 - By1) - (By2 - By1) * (Ax1 - Bx1) ) / d
uB = ( (Ax2 - Ax1) * (Ay1 - By1) - (Ay2 - Ay1) * (Ax1 - Bx1) ) / d
else:
return
if not ( 0 <= uA <= 1 and 0 <= uB <= 1 ):
return
x = Ax1 + uA * (Ax2 - Ax1)
y = Ay1 + uA * (Ay2 - Ay1)
return x, y
Python [edit]
Find the intersection without importing third-party libraries.
def line_intersect(Ax1, Ay1, Ax2, Ay2, Bx1, By1, Bx2, By2):
""" returns a (x, y) tuple or None if there is no intersection """
d = (By2 - By1) * (Ax2 - Ax1) - (Bx2 - Bx1) * (Ay2 - Ay1)
if d:
uA = ( (Bx2 - Bx1) * (Ay1 - By1) - (By2 - By1) * (Ax1 - Bx1) ) / d
uB = ( (Ax2 - Ax1) * (Ay1 - By1) - (Ay2 - Ay1) * (Ax1 - Bx1) ) / d
else:
return
if not ( 0 <= uA <= 1 and 0 <= uB <= 1 ):
return
x = Ax1 + uA * (Ax2 - Ax1)
y = Ay1 + uA * (Ay2 - Ay1)return x, y
if __name__ == '__main__':
(a, b) , (c, d) = ( 4 , 0 ) , ( 6 , 10 ) # try (4, 0), (6, 4)
(e, f) , (g, h) = ( 0 , 3 ) , ( 10 , 7 ) # for non intersecting test
pt = line_intersect(a, b, c, d, e, f, g, h)
print (pt)
(5.0, 5.0)
Or, labelling the moving parts a little more, and returning a composable option value containing either a message (in the absence of an intersection), or a pair of coordinates:
Works with: Python version 3.7
'''The intersection of two lines.'''from itertools import product
# intersection :: Line -> Line -> Either String Point
def intersection(ab):
'''Either the point at which the lines ab and pq
intersect, or a message string indicating that
they are parallel and have no intersection.'''
def delta(f):
return lambda x: f(fst(x) ) - f(snd(x) )def prodDiff(abcd):
[a, b, c, d] = abcd
return (a * d) - (b * c)def go(pq):
[abDX, pqDX, abDY, pqDY] = apList(
[delta(fst) , delta(snd) ]
) ( [ab, pq] )
determinant = prodDiff( [abDX, abDY, pqDX, pqDY] )def point( ):
[abD, pqD] = map (
lambda xy: prodDiff(
apList( [fst, snd] ) ( [fst(xy) , snd(xy) ] )
) , [ab, pq]
)
return apList(
[ lambda abpq: prodDiff(
[abD, fst(abpq) , pqD, snd(abpq) ] ) / determinant]
) (
[ (abDX, pqDX) , (abDY, pqDY) ]
)
return Right(point( ) ) if 0 != determinant else Left(
'( Parallel lines - no intersection )'
)return lambda pq: bindLR(go(pq) ) (
lambda xs: Right( (fst(xs) , snd(xs) ) )
)# --------------------------TEST---------------------------
# main :: IO()
def main( ):
'''Test'''# Left(message - no intersection) or Right(point)
# lrPoint :: Either String Point
lrPoint = intersection(
( ( 4.0 , 0.0 ) , ( 6.0 , 10.0 ) )
) (
( ( 0.0 , 3.0 ) , ( 10.0 , 7.0 ) )
)
print (
lrPoint[ 'Left' ] or lrPoint[ 'Right' ]
)# --------------------GENERIC FUNCTIONS--------------------
# Left :: a -> Either a b
def Left(x):
'''Constructor for an empty Either (option type) value
with an associated string.'''
return { 'type': 'Either' , 'Right': None , 'Left': x}# Right :: b -> Either a b
def Right(x):
'''Constructor for a populated Either (option type) value'''
return { 'type': 'Either' , 'Left': None , 'Right': x}# apList (<*>) :: [(a -> b)] -> [a] -> [b]
def apList(fs):
'''The application of each of a list of functions,
to each of a list of values.
'''
def go(fx):
f, x = fx
return f(x)
return lambda xs: [
go(x) for x
in product(fs, xs)
]# bindLR (>>=) :: Either a -> (a -> Either b) -> Either b
def bindLR(m):
'''Either monad injection operator.
Two computations sequentially composed,
with any value produced by the first
passed as an argument to the second.'''
return lambda mf: (
mf(m.get ( 'Right' ) ) if None is m.get ( 'Left' ) else m
)# fst :: (a, b) -> a
def fst(tpl):
'''First member of a pair.'''
return tpl[ 0 ]# snd :: (a, b) -> b
def snd(tpl):
'''Second member of a pair.'''
return tpl[ 1 ]# MAIN ---
if __name__ == '__main__':
main( )
(5.0, 5.0)
Find the intersection by importing the external Shapely library.
from shapely.geometry import LineStringif __name__ == "__main__":
line1 = LineString( [ ( 4 , 0 ) , ( 6 , 10 ) ] )
line2 = LineString( [ ( 0 , 3 ) , ( 10 , 7 ) ] )
print (line1.intersection (line2) )
POINT (5 5)
Racket [edit]
Translation of: C++
#lang racket/base
(define (det a b c d) (- (* a d) (* b c))) ; determinant(define (line-intersect ax ay bx by cx cy dx dy) ; --> (values x y)
(let* ((det.ab (det ax ay bx by))
(det.cd (det cx cy dx dy))
(abΔx (- ax bx))
(cdΔx (- cx dx))
(abΔy (- ay by))
(cdΔy (- cy dy))
(xnom (det det.ab abΔx det.cd cdΔx))
(ynom (det det.ab abΔy det.cd cdΔy))
(denom (det abΔx abΔy cdΔx cdΔy)))
(when (zero? denom)
(error 'line-intersect "parallel lines"))
(values (/ xnom denom) (/ ynom denom))))(module+ test (line-intersect 4 0 6 10
0 3 10 7))
5 5
Raku [edit]
(formerly Perl 6)
Works with: Rakudo version 2016.11
Translation of: zkl
sub intersection (Real $ax , Real $ay , Real $bx , Real $by ,
Real $cx , Real $cy , Real $dx , Real $dy ) {sub term:<|AB|> { determinate( $ax , $ay , $bx , $by ) }
sub term:<|CD|> { determinate( $cx , $cy , $dx , $dy ) }my $ΔxAB = $ax - $bx ;
my $ΔyAB = $ay - $by ;
my $ΔxCD = $cx - $dx ;
my $ΔyCD = $cy - $dy ;my $x - numerator = determinate( |AB|, $ΔxAB, |CD|, $ΔxCD ) ;
my $y - numerator = determinate( |AB|, $ΔyAB, |CD|, $ΔyCD ) ;
my $denominator = determinate( $ΔxAB, $ΔyAB, $ΔxCD, $ΔyCD ) ;return 'Lines are parallel' if $denominator == 0 ;
( $x - numerator / $denominator , $y - numerator / $denominator ) ;
}sub determinate ( Real $a , Real $b , Real $c , Real $d ) { $a * $d - $b * $c }
# TESTING
say 'Intersection point: ' , intersection( 4 , 0 , 6 , 10 , 0 , 3 , 10 , 7 ) ;
say 'Intersection point: ' , intersection( 4 , 0 , 6 , 10 , 0 , 3 , 10 , 7.1 ) ;
say 'Intersection point: ' , intersection( 0 , 0 , 1 , 1 , 1 , 2 , 4 , 5 ) ;
Intersection point: (5 5) Intersection point: (5.010893 5.054466) Intersection point: Lines are parallel
REXX [edit]
version 1 [edit]
Naive implementation. To be improved for parallel lines and degenerate lines such as y=5 or x=8.
/* REXX */
Parse Value '(4.0,0.0)' With '(' xa ',' ya ')'
Parse Value '(6.0,10.0)' With '(' xb ',' yb ')'
Parse Value '(0.0,3.0)' With '(' xc ',' yc ')'
Parse Value '(10.0,7.0)' With '(' xd ',' yd ')'Say 'The two lines are:'
Say 'yab='ya-xa* ( (yb-ya) / (xb-xa) ) '+x*' || ( (yb-ya) / (xb-xa) )
Say 'ycd='yc-xc* ( (yd-yc) / (xd-xc) ) '+x*' || ( (yd-yc) / (xd-xc) )x=( (yc-xc* ( (yd-yc) / (xd-xc) ) )-(ya-xa* ( (yb-ya) / (xb-xa) ) ) ) /,
( ( (yb-ya) / (xb-xa) )-( (yd-yc) / (xd-xc) ) )
Say 'x=' ||x
y=ya-xa* ( (yb-ya) / (xb-xa) )+x* ( (yb-ya) / (xb-xa) )
Say 'yab='y
Say 'ycd='yc-xc* ( (yd-yc) / (xd-xc) )+x* ( (yd-yc) / (xd-xc) )
Say 'Intersection: (' ||x','y')'
The two lines are: yab=-20.0+x*5 ycd=3.0+x*0.4 x=5 yab=5.0 ycd=5.0 Intersection: (5,5.0)
version 2 [edit]
complete implementation taking care of all possibilities.
Variables are named after the Austrian notation for a straight line: y=k*x+d
say ggx1( '4.0 0.0 6.0 10.0 0.0 3.0 10.0 7.0' )
say ggx1( '0.0 0.0 0.0 10.0 0.0 3.0 10.0 7.0' )
say ggx1( '0.0 0.0 0.0 10.0 0.0 3.0 10.0 7.0' )
say ggx1( '0.0 0.0 0.0 1.0 1.0 0.0 1.0 7.0' )
say ggx1( '0.0 0.0 0.0 0.0 0.0 3.0 10.0 7.0' )
say ggx1( '0.0 0.0 3.0 3.0 0.0 0.0 6.0 6.0' )
say ggx1( '0.0 0.0 3.0 3.0 0.0 1.0 6.0 7.0' )
Exitggx1: Procedure
/*---------------------------------------------------------------------
* find the intersection of the lines AB and CD
*--------------------------------------------------------------------*/
Parse Arg xa ya xb yb xc yc xd yd
Say 'A=('xa'/'ya') B=(' ||xb'/'yb') C=(' ||xc'/'yc') D=(' ||xd'/'yd')'
res=''
If xa=xb Then Do /* AB is a vertical line */
k1='*' /* slope is infinite */
x1=xa /* intersection's x is xa */
If ya=yb Then /* coordinates are equal */
res='Points A and B are identical' /* special case */
End
Else Do /* AB is not a vertical line */
k1=(yb-ya) / (xb-xa) /* compute the slope of AB */
d1=ya-k1*xa /* and its intersection with the y-axis */
End
If xc=xd Then Do /* CD is a vertical line */
k2='*' /* slope is infinite */
x2=xc /* intersection's x is xc */
If yc=yd Then /* coordinates are equal */
res='Points C and D are identical' /* special case */
End
Else Do /* CD is not a vertical line */
k2=(yd-yc) / (xd-xc) /* compute the slope of CD */
d2=yc-k2*xc /* and its intersection with the y-axis */
EndIf res='' Then Do /* no special case so far */
If k1='*' Then Do /* AB is vertical */
If k2='*' Then Do /* CD is vertical */
If x1=x2 Then /* and they are identical */
res='Lines AB and CD are identical'
Else /* not identical */
res='Lines AB and CD are parallel'
End
Else Do
x=x1 /* x is taken from AB */
y=k2*x+d2 /* y is computed from CD */
End
End
Else Do /* AB is not verical */
If k2='*' Then Do /* CD is vertical */
x=x2 /* x is taken from CD */
y=k1*x+d1 /* y is computed from AB */
End
Else Do /* AB and CD are not vertical */
If k1=k2 Then Do /* identical slope */
If d1=d2 Then /* same intersection with x=0 */
res='Lines AB and CD are identical'
Else /* otherwise */
res='Lines AB and CD are parallel'
End
Else Do /* finally the normal case */
x=(d2-d1) / (k1-k2) /* compute x */
y=k1*x+d1 /* and y */
End
End
End
End
If res='' Then /* not any special case */
res='Intersection is (' ||x'/'y')' /* that's the result */
Return ' -->' res
A=(4.0/0.0) B=(6.0/10.0) C=(0.0/3.0) D=(10.0/7.0) --> Intersection is (5/5.0) A=(0.0/0.0) B=(0.0/10.0) C=(0.0/3.0) D=(10.0/7.0) --> Intersection is (0.0/3.0) A=(0.0/0.0) B=(0.0/10.0) C=(0.0/3.0) D=(10.0/7.0) --> Intersection is (0.0/3.0) A=(0.0/0.0) B=(0.0/1.0) C=(1.0/0.0) D=(1.0/7.0) --> Lines AB and CD are parallel A=(0.0/0.0) B=(0.0/0.0) C=(0.0/3.0) D=(10.0/7.0) --> Points A and B are identical A=(0.0/0.0) B=(3.0/3.0) C=(0.0/0.0) D=(6.0/6.0) --> Lines AB and CD are identical A=(0.0/0.0) B=(3.0/3.0) C=(0.0/1.0) D=(6.0/7.0) --> Lines AB and CD are parallel
Ring [edit]
# Project : Find the intersection of two linesxa=4
ya=0
xb=6
yb=10
xc=0
yc=3
xd=10
yd=7
see "the two lines are:" + nl
see "yab=" + (ya-xa*((yb-ya)/(xb-xa))) + "+x*" + ((yb-ya)/(xb-xa)) + nl
see "ycd=" + (yc-xc*((yd-yc)/(xd-xc))) + "+x*" + ((yd-yc)/(xd-xc)) + nl
x=((yc-xc*((yd-yc)/(xd-xc)))-(ya-xa*((yb-ya)/(xb-xa))))/(((yb-ya)/(xb-xa))-((yd-yc)/(xd-xc)))
see "x=" + x + nl
y=ya-xa*((yb-ya)/(xb-xa))+x*((yb-ya)/(xb-xa))
see "yab=" + y + nl
see "ycd=" + (yc-xc*((yd-yc)/(xd-xc))+x*((yd-yc)/(xd-xc))) + nl
see "intersection: " + x + "," + y + nl
Output:
the two lines are: yab=-20+x*5 ycd=3+x*0.4 x=5 yab=5 ycd=5 intersection: 5,5
Ruby [edit]
Point = Struct.new ( :x, :y )class Line
attr_reader :a, :bdef initialize(point1, point2)
@a = (point1.y - point2.y ).fdiv (point1.x - point2.x )
@b = point1.y - @a *point1.x
enddef intersect(other)
return nil if @a == other.a
x = (other.b - @b ).fdiv (@a - other.a )
y = @a *x + @b
Point.new (x,y)
enddef to_s
"y = #{@a}x + #{@b}"
endend
l1 = Line.new (Point.new ( 4, 0 ), Point.new ( 6, 10 ) )
l2 = Line.new (Point.new ( 0, 3 ), Point.new ( 10, 7 ) )puts "Line #{l1} intersects line #{l2} at #{l1.intersect(l2)}."
Line y = 5.0x + -20.0 intersects line y = 0.4x + 3.0 at #<struct Point x=5.0, y=5.0>.
Rust [edit]
#[derive(Copy, Clone, Debug)]
struct Point {
x: f64,
y: f64,
}impl Point {
pub fn new(x: f64, y: f64) -> Self {
Point { x, y }
}
}#[derive(Copy, Clone, Debug)]
struct Line(Point, Point);impl Line {
pub fn intersect(self, other: Self) -> Option<Point> {
let a1 = self.1.y - self.0.y;
let b1 = self.0.x - self.1.x;
let c1 = a1 * self.0.x + b1 * self.0.y;let a2 = other.1.y - other.0.y;
let b2 = other.0.x - other.1.x;
let c2 = a2 * other.0.x + b2 * other.0.y;let delta = a1 * b2 - a2 * b1;
if delta == 0.0 {
return None;
}Some(Point {
x: (b2 * c1 - b1 * c2) / delta,
y: (a1 * c2 - a2 * c1) / delta,
})
}
}fn main() {
let l1 = Line(Point::new(4.0, 0.0), Point::new(6.0, 10.0));
let l2 = Line(Point::new(0.0, 3.0), Point::new(10.0, 7.0));
println!("{:?}", l1.intersect(l2));let l1 = Line(Point::new(0.0, 0.0), Point::new(1.0, 1.0));
let l2 = Line(Point::new(1.0, 2.0), Point::new(4.0, 5.0));
println!("{:?}", l1.intersect(l2));
}
Some(Point { x: 5.0, y: 5.0 }) None Scala [edit]
object Intersection extends App {
val (l1, l2) = (LineF(PointF( 4, 0 ), PointF( 6, 10 ) ), LineF(PointF( 0, 3 ), PointF( 10, 7 ) ) )def findIntersection(l1: LineF, l2: LineF) : PointF = {
val a1 = l1.e.y - l1.s.y
val b1 = l1.s.x - l1.e.x
val c1 = a1 * l1.s.x + b1 * l1.s.yval a2 = l2.e.y - l2.s.y
val b2 = l2.s.x - l2.e.x
val c2 = a2 * l2.s.x + b2 * l2.s.yval delta = a1 * b2 - a2 * b1
// If lines are parallel, intersection point will contain infinite values
PointF( (b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta)
}def l01 = LineF(PointF(0f, 0f), PointF(1f, 1f) )
def l02 = LineF(PointF(1f, 2f), PointF(4f, 5f) )case class PointF(x: Float, y: Float) {
override def toString = s"{$x, $y}"
}case class LineF(s: PointF, e: PointF)
println(findIntersection(l1, l2) )
println(findIntersection(l01, l02) )}
See it in running in your browser by (JavaScript)
or by Scastie (JVM).
Sidef [edit]
Translation of: Raku
func det(a, b, c, d) { a*d - b*c }func intersection(ax, ay, bx, by,
cx, cy, dx, dy) {var detAB = det(ax,ay, bx,by)
var detCD = det(cx,cy, dx,dy)var ΔxAB = (ax - bx)
var ΔyAB = (ay - by)
var ΔxCD = (cx - dx)
var ΔyCD = (cy - dy)var x_numerator = det(detAB, ΔxAB, detCD, ΔxCD)
var y_numerator = det(detAB, ΔyAB, detCD, ΔyCD)
var denominator = det( ΔxAB, ΔyAB, ΔxCD, ΔyCD)denominator == 0 && return 'lines are parallel'
[x_numerator / denominator, y_numerator / denominator]
}say ( 'Intersection point: ', intersection( 4,0, 6,10, 0,3, 10,7 ) )
say ( 'Intersection point: ', intersection( 4,0, 6,10, 0,3, 10,7.1 ) )
say ( 'Intersection point: ', intersection( 0,0, 1,1, 1,2, 4,5 ) )
Intersection point: [5, 5] Intersection point: [2300/459, 2320/459] Intersection point: lines are parallel
Swift [edit]
struct Point {
var x: Double
var y: Double
} struct Line {
var p1: Point
var p2: Point
var slope: Double {
guard p1.x - p2.x != 0.0 else { return .nan }
return (p1.y-p2.y) / (p1.x-p2.x)
}
func intersection(of other: Line) -> Point? {
let ourSlope = slope
let theirSlope = other.slope
guard ourSlope != theirSlope else { return nil }
if ourSlope.isNaN && !theirSlope.isNaN {
return Point(x: p1.x, y: (p1.x - other.p1.x) * theirSlope + other.p1.y)
} else if theirSlope.isNaN && !ourSlope.isNaN {
return Point(x: other.p1.x, y: (other.p1.x - p1.x) * ourSlope + p1.y)
} else {
let x = (ourSlope*p1.x - theirSlope*other.p1.x + other.p1.y - p1.y) / (ourSlope - theirSlope)
return Point(x: x, y: theirSlope*(x - other.p1.x) + other.p1.y)
}
}
}
let l1 = Line(p1: Point(x: 4.0, y: 0.0), p2: Point(x: 6.0, y: 10.0))
let l2 = Line(p1: Point(x: 0.0, y: 3.0), p2: Point(x: 10.0, y: 7.0))
print("Intersection at : \(l1.intersection(of: l2)!)")
Intersection at : Point(x: 5.0, y: 5.0)
TI-83 BASIC [edit]
Works with: TI-83 BASIC version TI-84Plus 2.55MP
Translation of: Rexx
Simple version:
[[4,0][6,10][0,3][10,7]]→[A]
([A](2,2)-[A](1,2))/([A](2,1)-[A](1,1))→B
[A](1,2)-[A](1,1)*B→A
([A](4,2)-[A](3,2))/([A](4,1)-[A](3,1))→D
[A](3,2)-[A](3,1)*D→C
(C-A)/(B-D)→X
A+X*B→Y
C+X*D→Z
Disp {X,Y}
{5 5} Full version:
[[4,0][6,10][0,3][10,7]]→[A]
{4,2}→Dim([B])
0→M
If [A](1,1)=[A](2,1)
Then
[A](1,1)→[B](3,1)
If [A](1,2)=[A](2,2):1→M
Else
1→[B](4,1)
([A](2,2)-[A](1,2))/([A](2,1)-[A](1,1))→[B](1,1)
[A](1,2)-[B](1,1)*[A](1,1)→[B](2,1)
End
If [A](3,1)=[A](4,1)
Then
[A](3,1)→[B](3,2)
If [A](3,2)=[A](4,2):2→M
Else
1→[B](4,2)
([A](4,2)-[A](3,2))/([A](4,1)-[A](3,1))→[B](1,2)
[A](3,2)-[B](1,2)*[A](3,1)→[B](2,2)
End
If M=0 Then
If [B](4,1)=0
Then
If [B](4,2)=0
Then
If [B](3,1)=[B](3,2)
Then:3→M
Else:4→M
End
Else
[B](3,1)→X
[B](1,2)*X+[B](2,2)→Y
End
Else
If [B](4,2)=0
Then
[B](3,2)→X
[B](1,1)*X+[B](2,1)→Y
Else
If [B](1,1)=[B](1,2)
Then
If [B](2,1)=[B](2,2)
Then:5→M
Else:6→M
End
Else
([B](2,2)-[B](2,1))/([B](1,1)-[B](1,2))→X
[B](1,1)*X+[B](2,1)→Y
End
End
End
End
Disp {X,Y,M}
{5 5} Visual Basic [edit]
Works with: VBA version 6.5
Works with: VBA version 7.1
Option ExplicitPublic Type Point
x As Double
y As Double
invalid As Boolean
End TypePublic Type Line
s As Point
e As Point
End TypePublic Function GetIntersectionPoint(L1 As Line, L2 As Line) As Point
Dim a1 As Double
Dim b1 As Double
Dim c1 As Double
Dim a2 As Double
Dim b2 As Double
Dim c2 As Double
Dim det As Doublea1 = L1.e.y - L1.s.y
b1 = L1.s.x - L1.e.x
c1 = a1 * L1.s.x + b1 * L1.s.y
a2 = L2.e.y - L2.s.y
b2 = L2.s.x - L2.e.x
c2 = a2 * L2.s.x + b2 * L2.s.y
det = a1 * b2 - a2 * b1If det Then
With GetIntersectionPoint
.x = (b2 * c1 - b1 * c2) / det
.y = (a1 * c2 - a2 * c1) / det
End With
Else
GetIntersectionPoint.invalid = True
End If
End FunctionSub Main()
Dim ln1 As Line
Dim ln2 As Line
Dim ip As Pointln1.s.x = 4
ln1.s.y = 0
ln1.e.x = 6
ln1.e.y = 10
ln2.s.x = 0
ln2.s.y = 3
ln2.e.x = 10
ln2.e.y = 7
ip = GetIntersectionPoint(ln1, ln2)
Debug.Assert Not ip.invalid
Debug.Assert ip.x = 5 And ip.y = 5LSet ln2.s = ln2.e
ip = GetIntersectionPoint(ln1, ln2)
Debug.Assert ip.invalidLSet ln2 = ln1
ip = GetIntersectionPoint(ln1, ln2)
Debug.Assert ip.invalidEnd Sub
Visual Basic .NET [edit]
Translation of: C#
Imports System. DrawingModule Module1
Function FindIntersection(s1 As PointF, e1 As PointF, s2 As PointF, e2 As PointF) As PointF
Dim a1 = e1. Y - s1. Y
Dim b1 = s1. X - e1. X
Dim c1 = a1 * s1. X + b1 * s1. YDim a2 = e2. Y - s2. Y
Dim b2 = s2. X - e2. X
Dim c2 = a2 * s2. X + b2 * s2. YDim delta = a1 * b2 - a2 * b1
'If lines are parallel, the result will be (NaN, NaN).
Return If (delta = 0, New PointF( Single . NaN, Single . NaN ), New PointF( (b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta) )
End FunctionSub Main( )
Dim p = Function (x As Single, y As Single ) New PointF(x, y)
Console. WriteLine (FindIntersection(p(4.0F, 0F), p(6.0F, 10.0F), p(0F, 3.0F), p(10.0F, 7.0F) ) )
Console. WriteLine (FindIntersection(p(0F, 0F), p(1.0F, 1.0F), p(1.0F, 2.0F), p(4.0F, 5.0F) ) )
End SubEnd Module
{X=5, Y=5} {X=NaN, Y=NaN} Wren [edit]
Translation of: Kotlin
class Point {
construct new (x, y) {
_x = x
_y = y
}x { _x }
y { _y }toString { "(%(_x), %(_y))" }
}class Line {
construct new (s, e) {
_s = s
_e = e
}s { _s }
e { _e }
}var findIntersection = Fn. new { |l1, l2|
var a1 = l1. e . y - l1. s . y
var b1 = l1. s . x - l1. e . x
var c1 = a1*l1. s . x + b1*l1. s . yvar a2 = l2. e . y - l2. s . y
var b2 = l2. s . x - l2. e . x
var c2 = a2*l2. s . x + b2*l2. s . yvar delta = a1*b2 - a2*b1
// if lines are parallel, intersection point will contain infinite values
return Point. new ( (b2*c1 - b1*c2) /delta, (a1*c2 - a2*c1) /delta)
}var l1 = Line. new (Point. new ( 4 , 0 ) , Point. new ( 6 , 10 ) )
var l2 = Line. new (Point. new ( 0 , 3 ) , Point. new ( 10 , 7 ) )
System. print (findIntersection. call (l1, l2) )
l1 = Line. new (Point. new ( 0 , 0 ) , Point. new ( 1 , 1 ) )
l2 = Line. new (Point. new ( 1 , 2 ) , Point. new ( 4 , 5 ) )
System. print (findIntersection. call (l1, l2) )
(5, 5) (-infinity, -infinity)
zkl [edit]
Translation of: C++
fcn lineIntersect(ax,ay, bx,by, cx,cy, dx,dy){ // --> (x,y)
detAB,detCD := det(ax,ay, bx,by), det(cx,cy, dx,dy);
abDx,cdDx := ax - bx, cx - dx; // delta x
abDy,cdDy := ay - by, cy - dy; // delta y xnom,ynom := det(detAB,abDx, detCD,cdDx), det(detAB,abDy, detCD,cdDy);
denom := det(abDx,abDy, cdDx,cdDy);
if(denom.closeTo(0.0, 0.0001))
throw(Exception.MathError("lineIntersect: Parallel lines"));
return(xnom/denom, ynom/denom);
}
fcn det(a,b,c,d){ a*d - b*c } // determinant
lineIntersect(4.0,0.0, 6.0,10.0, 0.0,3.0, 10.0,7.0).println();
L(5,5)
References
- ↑ [1]
how to find where two lines intersect
Source: http://www.rosettacode.org/wiki/Find_the_intersection_of_two_lines
Posted by: mahonthised.blogspot.com
0 Response to "how to find where two lines intersect"
Post a Comment