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++ 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 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 |
# 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# 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 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 ?> |
<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> |
3x + 4y = 24
Time Complexity: O(1)
Auxiliary Space: O(1)