Equation of straight line passing through a given point which bisects it into two equal line segments

Last Updated : 08 Jun, 2022

Given a straight line which passes through a given point (x₀, y₀) such that this point bisects the line segment in two equal line segments. The task is to find the equation of this straight line.
Examples:

Input: x₀ = 4, y₀ = 3
Output: 3x + 4y = 24
Input: x₀ = 7, y₀ = 12
Output: 12x + 7y = 168

Approach:

Let PQ be the line and AB be the line segment between the axes. The x-intercept and y-intercept are a & b respectively.
Now, as C(x₀, y₀) bisects AB so,
x₀ = (a + 0) / 2 i.e. a = 2x₀
Similarly, y₀ = (0 + b) / 2 i.e. b = 2y₀
We know that the equation of a straight line in intercept form is,

x / a + y / b = 1
Here, a = 2x₀ & b = 2y₀
So, x / 2x₀ + y / 2y₀ = 1
or, x / x₀ + y / y₀ = 2
Therefore, x * y₀ + y * x₀ = 2 * x₀ * y₀

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to print the equation
// of the required line
void line(double x0, double y0)
{
    double c = 2 * y0 * x0;
    cout << y0 << "x"
         << " + " << x0 << "y = " << c;
}
 
// Driver code
int main()
{
    double x0 = 4, y0 = 3;
    line(x0, y0);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function to print the equation
// of the required line
static void line(double x0, double y0)
{
    double c = (int)(2 * y0 * x0);
    System.out.println(y0 + "x" + " + " + 
                       x0 + "y = " + c);
}
 
// Driver code
public static void main(String[] args)
{
    double x0 = 4, y0 = 3;
    line(x0, y0);
}
}
 
// This code is contributed 
// by Code_Mech

Python3

# Python 3 implementation of the approach
 
# Function to print the equation
# of the required line
def line(x0, y0):
    c = 2 * y0 * x0
    print(y0, "x", "+", x0, "y=", c)
 
# Driver code
if __name__ == '__main__':
    x0 = 4
    y0 = 3
    line(x0, y0)
     
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to print the equation
// of the required line
static void line(double x0, double y0)
{
    double c = (int)(2 * y0 * x0);
    Console.WriteLine(y0 + "x" + " + " + 
                    x0 + "y = " + c);
}
 
// Driver code
public static void Main(String[] args)
{
    double x0 = 4, y0 = 3;
    line(x0, y0);
}
}
 
/* This code contributed by PrinciRaj1992 */

PHP

<?php
// PHP implementation of the approach 
 
// Function to print the equation 
// of the required line 
function line($x0, $y0) 
{ 
    $c = 2 * $y0 * $x0; 
    echo $y0 , "x"," + ",
         $x0 , "y = " , $c; 
} 
 
// Driver code 
$x0 = 4; $y0 = 3; 
line($x0, $y0); 
 
// This code is contributed by Ryuga
?>

Javascript

<script>
 
// javascript implementation of the approach
 
     
// Function to print the equation
// of the required line
function line(x0 , y0)
{
    var c = parseInt(2 * y0 * x0);
    document.write(y0 + "x" + " + " + 
                       x0 + "y = " + c);
}
 
// Driver code
var x0 = 4, y0 = 3;
line(x0, y0);
 
 
// This code is contributed by Amit Katiyar 
 
</script>