Find X and Y intercepts of a line passing through the given points

Given two points on a 2D plane, the task is to find the x – intercept and the y – intercept of a line passing through the given points.

Examples:

Input: points[][] = {{5, 2}, {2, 7}}
Output:
6.2
10.333333333333334

Input: points[][] = {{3, 2}, {2, 4}}
Output:
4.0
8.0

Approach:



  • Find the slope using the given points.
  • Put the value of the slope in the expression of the line i.e. y = mx + c.
  • Now find the value of c using the values of any of the given points in the equation y = mx + c
  • To find the x-intercept, put y = 0 in y = mx + c.
  • To find the y-intercept, put x = 0 in y = mx + c.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the X and Y intercepts
// of the line passing through
// the given points
void getXandYintercept(int P[], int Q[])
{
    int a = P[1] - Q[1];
    int b = P[0] - Q[0];
  
    // if line is parallel to y axis
    if (b == 0) {
        cout << P[0] << endl; // x - intercept will be p[0]
        cout << "infinity"; // y - intercept will be infinity
        return;
    }
  
    // if line is parallel to x axis
    if (a == 0) {
        cout << "infinity"; // x - intercept will be infinity
        cout << P[1] << endl; // y - intercept will be p[1]
        return;
    }
  
    // Slope of the line
    double m = a / (b * 1.0);
  
    // y = mx + c in where c is unknown
    // Use any of the given point to find c
    int x = P[0];
    int y = P[1];
    double c = y - m * x;
  
    // For finding the x-intercept put y = 0
    y = 0;
    double r = (y - c) / (m * 1.0);
    cout << r << endl;
  
    // For finding the y-intercept put x = 0
    x = 0;
    y = m * x + c;
    printf("%.8f", c);
}
  
// Driver code
int main()
{
    int p1[] = { 5, 2 };
    int p2[] = { 2, 7 };
    getXandYintercept(p1, p2);
    return 0;
}
  
// This code is contributed by Mohit Kumar

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Java

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// Java implementation of the approach
class GFG {
  
    // Function to find the X and Y intercepts
    // of the line passing through
    // the given points
    static void getXandYintercept(int P[],
                                  int Q[])
    {
        int a = P[1] - Q[1];
        int b = P[0] - Q[0];
  
        // if line is parallel to y axis
        if (b == 0) {
  
            // x - intercept will be p[0]
            System.out.println(P[0]); 
  
            // y - intercept will be infinity
            System.out.println("infinity");
            return;
        }
  
        // if line is parallel to x axis
        if (a == 0) {
  
            // x - intercept will be infinity  
            System.out.println("infinity"); 
  
            // y - intercept will be p[1]
            System.out.println(P[1]); 
            return;
        }
  
        // Slope of the line
        double m = a / (b * 1.0);
  
        // y = mx + c in where c is unknown
        // Use any of the given point to find c
        int x = P[0];
        int y = P[1];
        double c = y - m * x;
  
        // For finding the x-intercept put y = 0
        y = 0;
        double r = (y - c) / (m * 1.0);
        System.out.println(r);
  
        // For finding the y-intercept put x = 0
        x = 0;
        y = (int)(m * x + c);
        System.out.print(c);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int p1[] = { 5, 2 };
        int p2[] = { 2, 7 };
        getXandYintercept(p1, p2);
    }
}
  
// This code is contributed by kanugargng

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Python3

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# Python3 implementation of the approach
  
# Function to find the X and Y intercepts 
# of the line passing through 
# the given points
def getXandYintercept(P, Q):
  
    a = P[1] - Q[1
    b = P[0] - Q[0]
      
    # if line is parallel to y axis
    if b == 0:
        print(P[0])         # x - intercept will be p[0]
        print("infinity")   # y - intercept will be infinity
        return
      
    # if line is parallel to x axis
    if a == 0:
        print("infinity")     # x - intercept will be infinity
        print(P[1])           # y - intercept will be p[1]
        return
      
      
          
    # Slope of the line
    m = a / b
      
    # y = mx + c in where c is unknown 
    # Use any of the given point to find c
    x = P[0]
    y = P[1]
    c = y-m * x
      
    # For finding the x-intercept put y = 0
    y = 0
    x =(y-c)/m
    print(x)
      
    # For finding the y-intercept put x = 0
    x = 0
    y = m * x + c
    print(y) 
  
# Driver code
p1 = [5, 2
p2 = [7, 2]
getXandYintercept(p1, p2)

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C#

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// C# implementation of the approach
using System;
  
class GFG {
  
    // Function to find the X and Y intercepts
    // of the line passing through
    // the given points
    static void getXandYintercept(int[] P,
                                  int[] Q)
    {
        int a = P[1] - Q[1];
        int b = P[0] - Q[0];
  
        // if line is parallel to y axis
        if (b == 0) {
            Console.WriteLine(P[0]); // x - intercept will be p[0]
            Console.WriteLine("infinity"); // y - intercept will be infinity
            return;
        }
  
        // if line is parallel to x axis
        if (a == 0) {
            Console.WriteLine("infinity"); // x - intercept will be infinity
            Console.WriteLine(P[1]); // y - intercept will be p[1]
            return;
        }
  
        // Slope of the line
        double m = a / (b * 1.0);
  
        // y = mx + c in where c is unknown
        // Use any of the given point to find c
        int x = P[0];
        int y = P[1];
        double c = y - m * x;
  
        // For finding the x-intercept put y = 0
        y = 0;
        double r = (y - c) / (m * 1.0);
        Console.WriteLine(r);
  
        // For finding the y-intercept put x = 0
        x = 0;
        y = (int)(m * x + c);
        Console.WriteLine(c);
    }
  
    // Driver code
    public static void Main()
    {
        int[] p1 = { 5, 2 };
        int[] p2 = { 2, 7 };
        getXandYintercept(p1, p2);
    }
}
  
// This code is contributed by AnkitRai01

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Output:

6.2
10.33333333333

Time Complexity: O(1)

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