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

  • Last Updated : 11 Nov, 2021

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 
 

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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>
Output: 
3x + 4y = 24

 




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