Given a point P in 2-D plane and equation of a line, the task is to find the foot of the perpendicular from P to the line.
Note: Equation of line is in form ax+by+c=0.
Examples:
Input : P=(1, 0), a = -1, b = 1, c = 0 Output : Q = (0.5, 0.5) The foot of perpendicular from point (1, 0) to line -x + y = 0 is (0.5, 0.5) Input : P=(3, 3), a = 0, b = 1, c = -2 Output : Q = (3, 2) The foot of perpendicular from point (3, 3) to line y-2 = 0 is (3, 2)
Since equation of the line is given to be of the form ax + by + c = 0. Equation of line passing through P and is perpendicular to line. Therefore equation of line passing through P and Q becomes ay – bx + d = 0. Also, P passes through line passing through P and Q, so we put coordinate of P in the above equation:
ay1 - bx1 + d = 0 or, d = bx1 - ay1
Also, Q is the intersection of the given line and the line passing through P and Q. So we can find the solution of:
ax + by + c = 0 and, ay - bx + (bx1-ay1) = 0
Since a, b, c, d all are known we can find x and y here as:
Below is the implementation of the above approach:
// C++ program for implementation of // the above approach #include <iostream> using namespace std;
// Function to find foot of perpendicular from // a point in 2 D plane to a Line pair< double , double > findFoot( double a, double b, double c,
double x1, double y1)
{ double temp = -1 * (a * x1 + b * y1 + c) / (a * a + b * b);
double x = temp * a + x1;
double y = temp * b + y1;
return make_pair(x, y);
} // Driver Code int main()
{ // Equation of line is
// ax + by + c = 0
double a = 0.0;
double b = 1.0;
double c = -2;
// Coordinates of point p(x1, y1).
double x1 = 3.0;
double y1 = 3.0;
pair< double , double > foot = findFoot(a, b, c, x1, y1);
cout << foot.first << " " << foot.second;
return 0;
} |
import javafx.util.Pair;
// Java program for implementation of // the above approach class GFG
{ // Function to find foot of perpendicular from // a point in 2 D plane to a Line static Pair<Double, Double> findFoot( double a, double b, double c,
double x1, double y1)
{ double temp = - 1 * (a * x1 + b * y1 + c) / (a * a + b * b);
double x = temp * a + x1;
double y = temp * b + y1;
return new Pair(x, y);
} // Driver Code public static void main(String[] args)
{ // Equation of line is
// ax + by + c = 0
double a = 0.0 ;
double b = 1.0 ;
double c = - 2 ;
// Coordinates of point p(x1, y1).
double x1 = 3.0 ;
double y1 = 3.0 ;
Pair<Double, Double> foot = findFoot(a, b, c, x1, y1);
System.out.println(foot.getKey() + " " + foot.getValue());
}
} // This code contributed by Rajput-Ji |
# Python3 implementation of the approach # Function to find foot of perpendicular # from a point in 2 D plane to a Line def findFoot(a, b, c, x1, y1):
temp = ( - 1 * (a * x1 + b * y1 + c) / /
(a * a + b * b))
x = temp * a + x1
y = temp * b + y1
return (x, y)
# Driver Code if __name__ = = "__main__" :
# Equation of line is
# ax + by + c = 0
a, b, c = 0.0 , 1.0 , - 2
# Coordinates of point p(x1, y1).
x1, y1 = 3.0 , 3.0
foot = findFoot(a, b, c, x1, y1)
print ( int (foot[ 0 ]), int (foot[ 1 ]))
# This code is contributed # by Rituraj Jain |
// C# program for implementation of // the above approach using System;
class GFG
{ // Pair class
public class Pair
{
public double first,second;
public Pair( double a, double b)
{
first = a;
second = b;
}
}
// Function to find foot of perpendicular from // a point in 2 D plane to a Line static Pair findFoot( double a, double b, double c,
double x1, double y1)
{ double temp = -1 * (a * x1 + b * y1 + c) / (a * a + b * b);
double x = temp * a + x1;
double y = temp * b + y1;
return new Pair(x, y);
} // Driver Code public static void Main(String []args)
{ // Equation of line is
// ax + by + c = 0
double a = 0.0;
double b = 1.0;
double c = -2;
// Coordinates of point p(x1, y1).
double x1 = 3.0;
double y1 = 3.0;
Pair foot = findFoot(a, b, c, x1, y1);
Console.WriteLine(foot.first + " " + foot.second);
}
} // This code contributed by Arnab Kundu |
<?php // PHP implementation of the approach // Function to find foot of perpendicular // from a point in 2 D plane to a Line function findFoot( $a , $b , $c , $x1 , $y1 )
{ $temp = floor ((-1 * ( $a * $x1 + $b * $y1 + $c ) /
( $a * $a + $b * $b )));
$x = $temp * $a + $x1 ;
$y = $temp * $b + $y1 ;
return array ( $x , $y );
} // Driver Code // Equation of line is // ax + by + c = 0 $a = 0.0;
$b = 1.0 ;
$c = -2 ;
// Coordinates of point p(x1, y1). $x1 = 3.0 ;
$y1 = 3.0 ;
$foot = findFoot( $a , $b , $c , $x1 , $y1 );
echo floor ( $foot [0]), " " , floor ( $foot [1]);
// This code is contributed by Ryuga ?> |
<script> // JavaScript implementation of the approach
// Function to find foot of perpendicular
// from a point in 2 D plane to a Line
function findFoot(a, b, c, x1, y1) {
var temp = (-1 * (a * x1 + b * y1 + c)) / (a * a + b * b);
var x = temp * a + x1;
var y = temp * b + y1;
return [x, y];
}
// Driver Code
// Equation of line is
// ax + by + c = 0
var a = 0.0;
var b = 1.0;
var c = -2;
// Coordinates of point p(x1, y1).
var x1 = 3.0;
var y1 = 3.0;
var foot = findFoot(a, b, c, x1, y1);
document.write(parseInt(foot[0]) + " " + parseInt(foot[1]));
</script>
|
3 2
Time Complexity: O(1)
Auxiliary Space: O(1)