Given points A and B corresponding to line AB and points P and Q corresponding to line PQ, find the point of intersection of these lines. The points are given in 2D Plane with their X and Y Coordinates. Examples:
Input : A = (1, 1), B = (4, 4)
C = (1, 8), D = (2, 4)
Output : The intersection of the given lines
AB and CD is: (2.4, 2.4)
Input : A = (0, 1), B = (0, 4)
C = (1, 8), D = (1, 4)
Output : The given lines AB and CD are parallel.
First of all, let us assume that we have two points (x1, y1) and (x2, y2). Now, we find the equation of line formed by these points. Let the given lines be :
- a1x + b1y = c1
- a2x + b2y = c2
We have to now solve these 2 equations to find the point of intersection. To solve, we multiply 1. by b2 and 2 by b1 This gives us, a1b2x + b1b2y = c1b2 a2b1x + b2b1y = c2b1 Subtracting these we get, (a1b2 – a2b1) x = c1b2 – c2b1 This gives us the value of x. Similarly, we can find the value of y. (x, y) gives us the point of intersection. Note: This gives the point of intersection of two lines, but if we are given line segments instead of lines, we have to also recheck that the point so computed actually lies on both the line segments. If the line segment is specified by points (x1, y1) and (x2, y2), then to check if (x, y) is on the segment we have to just check that
- min (x1, x2) <= x <= max (x1, x2)
- min (y1, y2) <= y <= max (y1, y2)
The pseudo code for the above implementation:
determinant = a1 b2 - a2 b1
if (determinant == 0)
{
// Lines are parallel
}
else
{
x = (c1b2 - c2b1)/determinant
y = (a1c2 - a2c1)/determinant
}
These can be derived by first getting the slope directly and then finding the intercept of the line.
C++
// C++ Implementation. To find the point of// intersection of two lines#include <bits/stdc++.h>using namespace std;// This pair is used to store the X and Y// coordinates of a point respectively#define pdd pair<double, double>// Function used to display X and Y coordinates// of a pointvoid displayPoint(pdd P){ cout << "(" << P.first << ", " << P.second << ")" << endl;}pdd lineLineIntersection(pdd A, pdd B, pdd C, pdd D){ // Line AB represented as a1x + b1y = c1 double a1 = B.second - A.second; double b1 = A.first - B.first; double c1 = a1*(A.first) + b1*(A.second); // Line CD represented as a2x + b2y = c2 double a2 = D.second - C.second; double b2 = C.first - D.first; double c2 = a2*(C.first)+ b2*(C.second); double determinant = a1*b2 - a2*b1; if (determinant == 0) { // The lines are parallel. This is simplified // by returning a pair of FLT_MAX return make_pair(FLT_MAX, FLT_MAX); } else { double x = (b2*c1 - b1*c2)/determinant; double y = (a1*c2 - a2*c1)/determinant; return make_pair(x, y); }}// Driver codeint main(){ pdd A = make_pair(1, 1); pdd B = make_pair(4, 4); pdd C = make_pair(1, 8); pdd D = make_pair(2, 4); pdd intersection = lineLineIntersection(A, B, C, D); if (intersection.first == FLT_MAX && intersection.second==FLT_MAX) { cout << "The given lines AB and CD are parallel.\n"; } else { // NOTE: Further check can be applied in case // of line segments. Here, we have considered AB // and CD as lines cout << "The intersection of the given lines AB " "and CD is: "; displayPoint(intersection); } return 0;} |
Java
// Java Implementation. To find the point of// intersection of two lines// Class used to used to store the X and Y// coordinates of a point respectivelyclass Point{ double x,y; public Point(double x, double y) { this.x = x; this.y = y; } // Method used to display X and Y coordinates // of a point static void displayPoint(Point p) { System.out.println("(" + p.x + ", " + p.y + ")"); }}class Test{ static Point lineLineIntersection(Point A, Point B, Point C, Point D) { // Line AB represented as a1x + b1y = c1 double a1 = B.y - A.y; double b1 = A.x - B.x; double c1 = a1*(A.x) + b1*(A.y); // Line CD represented as a2x + b2y = c2 double a2 = D.y - C.y; double b2 = C.x - D.x; double c2 = a2*(C.x)+ b2*(C.y); double determinant = a1*b2 - a2*b1; if (determinant == 0) { // The lines are parallel. This is simplified // by returning a pair of FLT_MAX return new Point(Double.MAX_VALUE, Double.MAX_VALUE); } else { double x = (b2*c1 - b1*c2)/determinant; double y = (a1*c2 - a2*c1)/determinant; return new Point(x, y); } } // Driver method public static void main(String args[]) { Point A = new Point(1, 1); Point B = new Point(4, 4); Point C = new Point(1, 8); Point D = new Point(2, 4); Point intersection = lineLineIntersection(A, B, C, D); if (intersection.x == Double.MAX_VALUE && intersection.y == Double.MAX_VALUE) { System.out.println("The given lines AB and CD are parallel."); } else { // NOTE: Further check can be applied in case // of line segments. Here, we have considered AB // and CD as lines System.out.print("The intersection of the given lines AB " + "and CD is: "); Point.displayPoint(intersection); } }} |
Python3
# Python program to find the point of# intersection of two lines# Class used to used to store the X and Y# coordinates of a point respectivelyclass Point: def __init__(self, x, y): self.x = x self.y = y # Method used to display X and Y coordinates # of a point def displayPoint(self, p): print(f"({p.x}, {p.y})")def lineLineIntersection(A, B, C, D): # Line AB represented as a1x + b1y = c1 a1 = B.y - A.y b1 = A.x - B.x c1 = a1*(A.x) + b1*(A.y) # Line CD represented as a2x + b2y = c2 a2 = D.y - C.y b2 = C.x - D.x c2 = a2*(C.x) + b2*(C.y) determinant = a1*b2 - a2*b1 if (determinant == 0): # The lines are parallel. This is simplified # by returning a pair of FLT_MAX return Point(10**9, 10**9) else: x = (b2*c1 - b1*c2)/determinant y = (a1*c2 - a2*c1)/determinant return Point(x, y)# Driver codeA = Point(1, 1)B = Point(4, 4)C = Point(1, 8)D = Point(2, 4)intersection = lineLineIntersection(A, B, C, D)if (intersection.x == 10**9 and intersection.y == 10**9): print("The given lines AB and CD are parallel.")else: # NOTE: Further check can be applied in case # of line segments. Here, we have considered AB # and CD as lines print("The intersection of the given lines AB " + "and CD is: ") intersection.displayPoint(intersection)# This code is contributed by Saurabh Jaiswal |
C#
using System;// C# Implementation. To find the point of // intersection of two lines // Class used to used to store the X and Y // coordinates of a point respectively public class Point{ public double x, y; public Point(double x, double y) { this.x = x; this.y = y; } // Method used to display X and Y coordinates // of a point public static void displayPoint(Point p) { Console.WriteLine("(" + p.x + ", " + p.y + ")"); }}public class Test{ public static Point lineLineIntersection(Point A, Point B, Point C, Point D) { // Line AB represented as a1x + b1y = c1 double a1 = B.y - A.y; double b1 = A.x - B.x; double c1 = a1 * (A.x) + b1 * (A.y); // Line CD represented as a2x + b2y = c2 double a2 = D.y - C.y; double b2 = C.x - D.x; double c2 = a2 * (C.x) + b2 * (C.y); double determinant = a1 * b2 - a2 * b1; if (determinant == 0) { // The lines are parallel. This is simplified // by returning a pair of FLT_MAX return new Point(double.MaxValue, double.MaxValue); } else { double x = (b2 * c1 - b1 * c2) / determinant; double y = (a1 * c2 - a2 * c1) / determinant; return new Point(x, y); } } // Driver method public static void Main(string[] args) { Point A = new Point(1, 1); Point B = new Point(4, 4); Point C = new Point(1, 8); Point D = new Point(2, 4); Point intersection = lineLineIntersection(A, B, C, D); if (intersection.x == double.MaxValue && intersection.y == double.MaxValue) { Console.WriteLine("The given lines AB and CD are parallel."); } else { // NOTE: Further check can be applied in case // of line segments. Here, we have considered AB // and CD as lines Console.Write("The intersection of the given lines AB " + "and CD is: "); Point.displayPoint(intersection); } }} // This code is contributed by Shrikant13 |
Javascript
<script> // Javascript program to find the point of // intersection of two lines // Class used to used to store the X and Y // coordinates of a point respectively class Point { constructor(x, y) { this.x = x; this.y = y; } // Method used to display X and Y coordinates // of a point displayPoint(p){ document.write("(" + p.x + ", " + p.y + ")"); } } function lineLineIntersection(A,B,C,D){ // Line AB represented as a1x + b1y = c1 var a1 = B.y - A.y; var b1 = A.x - B.x; var c1 = a1*(A.x) + b1*(A.y); // Line CD represented as a2x + b2y = c2 var a2 = D.y - C.y; var b2 = C.x - D.x; var c2 = a2*(C.x)+ b2*(C.y); var determinant = a1*b2 - a2*b1; if (determinant == 0) { // The lines are parallel. This is simplified // by returning a pair of FLT_MAX return new Point(Number.MAX_VALUE, Number.MAX_VALUE); } else { var x = (b2*c1 - b1*c2)/determinant; var y = (a1*c2 - a2*c1)/determinant; return new Point(x, y); } } // Driver code let A = new Point(1, 1); let B = new Point(4, 4); let C = new Point(1, 8); let D = new Point(2, 4); var intersection = lineLineIntersection(A, B, C, D); if (intersection.x == Number.MAX_VALUE && intersection.y == Number.MAX_VALUE){ document.write("The given lines AB and CD are parallel."); }else{ // NOTE: Further check can be applied in case // of line segments. Here, we have considered AB // and CD as lines document.write("The intersection of the given lines AB " + "and CD is: "); intersection.displayPoint(intersection); } // This code is contributed by shruti456rawal</script> |
Output:
The intersection of the given lines AB and CD is: (2.4, 2.4)
Time Complexity: O(1)
Auxiliary Space: O(1)
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