Given the values of m and c for the equation of a line y = (m * x) + c, the task is to find whether the point (x, y) lies on the given line.
Examples:
Input: m = 3, c = 2, x = 1, y = 5
Output: Yes
m * x + c = 3 * 1 + 2 = 3 + 2 = 5 which is equal to y
Hence, the given point satisfies the line’s equationInput: m = 5, c = 2, x = 2, y = 5
Output: No
Approach: In order for the given point to lie on the line, it must satisfy the equation of the line. Check whether y = (m * x) + c holds true.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function that return true if // the given point lies on the given line bool pointIsOnLine( int m, int c, int x, int y)
{ // If (x, y) satisfies the equation of the line
if (y == ((m * x) + c))
return true ;
return false ;
} // Driver code int main()
{ int m = 3, c = 2;
int x = 1, y = 5;
if (pointIsOnLine(m, c, x, y))
cout << "Yes" ;
else
cout << "No" ;
} |
// Java implementation of the approach class GFG
{ // Function that return true if // the given point lies on the given line static boolean pointIsOnLine( int m, int c,
int x, int y)
{ // If (x, y) satisfies the equation
// of the line
if (y == ((m * x) + c))
return true ;
return false ;
} // Driver code public static void main(String[] args)
{ int m = 3 , c = 2 ;
int x = 1 , y = 5 ;
if (pointIsOnLine(m, c, x, y))
System.out.print( "Yes" );
else
System.out.print( "No" );
} } // This code has been contributed by 29AjayKumar |
# Python3 implementation of the approach # Function that return true if the # given point lies on the given line def pointIsOnLine(m, c, x, y):
# If (x, y) satisfies the
# equation of the line
if (y = = ((m * x) + c)):
return True ;
return False ;
# Driver code m = 3 ; c = 2 ;
x = 1 ; y = 5 ;
if (pointIsOnLine(m, c, x, y)):
print ( "Yes" );
else :
print ( "No" );
# This code is contributed by mits |
// C# implementation of the approach using System;
class GFG
{ // Function that return true if // the given point lies on the given line static bool pointIsOnLine( int m, int c,
int x, int y)
{ // If (x, y) satisfies the equation
// of the line
if (y == ((m * x) + c))
return true ;
return false ;
} // Driver code public static void Main()
{ int m = 3, c = 2;
int x = 1, y = 5;
if (pointIsOnLine(m, c, x, y))
Console.Write( "Yes" );
else
Console.Write( "No" );
} } // This code is contributed by Akanksha Rai |
<script> // Javascript implementation of the approach // Function that return true if // the given point lies on the given line function pointIsOnLine(m, c, x, y)
{ // If (x, y) satisfies the equation
// of the line
if (y == ((m * x) + c))
return true ;
return false ;
} // Driver code var m = 3, c = 2;
var x = 1, y = 5;
if (pointIsOnLine(m, c, x, y))
document.write( "Yes" );
else document.write( "No" );
// This code is contributed by Rajput-Ji </script> |
<?php // PHP implementation of the approach // Function that return true if the // given point lies on the given line function pointIsOnLine( $m , $c , $x , $y )
{ // If (x, y) satisfies the equation
// of the line
if ( $y == (( $m * $x ) + $c ))
return true;
return false;
} // Driver code $m = 3; $c = 2;
$x = 1; $y = 5;
if (pointIsOnLine( $m , $c , $x , $y ))
echo "Yes" ;
else echo "No" ;
// This code is contributed by Ryuga ?> |
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)