Find foot of perpendicular from a point in 2 D plane to a Line

0 0 5 minutes read

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++

// 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;
}

Java

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

# 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#

// 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
// 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
?>

Javascript

<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>

Output:

3 2

Time Complexity: O(1)
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!