Find foot of perpendicular from a point in 2 D plane to a Line
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> |
3 2
Time Complexity: O(1)
Auxiliary Space: O(1)
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