Program for Point of Intersection of Two Lines

2

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 :

  1. a1x + b1y = c1
  2. 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 point
void 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 code
int 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 respectively
class 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);
	    }
	}
}


Output:

The intersection of the given lines AB and 
CD is: (2.4, 2.4)

This article is contributed by Aanya Jindal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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