查找从2D平面中的点到直线的垂直脚

给定二维平面中的点P和直线方程，任务是找到从P到直线的垂直线的底角。

注意：直线方程的形式为ax + by + c = 0。

例子：

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)

由于线的方程式为ax + x + c = 0的形式。穿过P且垂直于直线的线方程。因此，通过P和Q的线方程变为ay – bx + d = 0 。另外，P穿过穿过P和Q的线，因此我们将P的坐标放在上式中：

ay1 - bx1 + d = 0 
or, d = bx1 - ay1

同样，Q是给定线与穿过P和Q的线的交点。因此，我们可以找到以下解决方案：

ax + by + c = 0
and,
ay - bx + (bx1-ay1) = 0

由于a，b，c，d都是已知的，因此我们可以在此处找到x和y：

以下是上述方法的实现：

C++

// C++ program for implementation of
// the above approach
#include 
using namespace std;
  
// Function to find foot of perpendicular from
// a point in 2 D plane to a Line
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 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 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 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);
    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

输出：

3 2