截面公式(按给定比例划分线的点)

给定两个坐标(x1，y1)和(x2，y2)以及m和n，找到将坐标(x1，y1)和(x2，y2)按比例m：n划分的划分的坐标

例子：

Input : x1 = 1, y1 = 0, x2 = 2 y2 = 5,
        m = 1, n = 1
Output : (1.5, 2.5)
Explanation: co-ordinates (1.5, 2.5)
divides the line in ratio 1 : 1

Input : x1 = 2, y1 = 4, x2 = 4, y2 = 6,
        m = 2, n = 3
Output : (2.8, 4.8)
Explanation: (2.8, 4.8) divides the line
in the ratio 2:3

截面公式告诉我们将给定线段分为两部分的点的坐标，以使它们的长度为m：n

C++

// CPP program to find point that divides
// given line in given ratio.
#include 
using namespace std;
 
// Function to find the section of the line
void section(double x1, double x2, double y1,
              double y2, double m, double n)
{
    // Applying section formula
    double x = ((n * x1) + (m * x2)) /
                            (m + n);
    double y = ((n * y1) + (m * y2)) /
                             (m + n);
 
    // Printing result
    cout << "(" << x << ", ";
    cout << y << ")" << endl;
}
 
// Driver code
int main()
{
    double x1 = 2, x2 = 4, y1 = 4,
           y2 = 6, m = 2, n = 3;
    section(x1, x2, y1, y2, m, n);
    return 0;
}

Java

// Java program to find point that divides
// given line in given ratio.
import java.io.*;
 
class sections {
    static void section(double x1, double x2,
                        double y1, double y2,
                        double m, double n)
    {
        // Applying section formula
        double x = ((n * x1) + (m * x2)) /
                    (m + n);
        double y = ((n * y1) + (m * y2)) /
                    (m + n);
 
 
        // Printing result
        System.out.println("(" + x + ", " + y + ")");
    }
 
    public static void main(String[] args)
    {
        double x1 = 2, x2 = 4, y1 = 4,
               y2 = 6, m = 2, n = 3;
        section(x1, x2, y1, y2, m, n);
    }
}

Python

# Python program to find point that divides
# given line in given ratio.
def section(x1, x2, y1, y2, m, n):
 
    # Applying section formula
    x = (float)((n * x1)+(m * x2))/(m + n)
    y = (float)((n * y1)+(m * y2))/(m + n)
 
    # Printing result
    print (x, y)
 
x1 = 2
x2 = 4
y1 = 4
y2 = 6
m = 2
n = 3
section(x1, x2, y1, y2, m, n)

C#

// C# program to find point that divides
// given line in given ratio.
using System;
 
class GFG {
     
    static void section(double x1, double x2,
                        double y1, double y2,
                          double m, double n)
    {
         
        // Applying section formula
        double x = ((n * x1) + (m * x2)) /
                                    (m + n);
                                     
        double y = ((n * y1) + (m * y2)) /
                                   (m + n);
 
        // Printing result
        Console.WriteLine("(" + x + ", " + y + ")");
    }
 
    // Driver code
    public static void Main()
    {
         
        double x1 = 2, x2 = 4, y1 = 4,
                y2 = 6, m = 2, n = 3;
                 
        section(x1, x2, y1, y2, m, n);
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// JavaScript program to find point that divides
// given line in given ratio
 
    function section(x1, x2, y1, y2, m, n)
    {
        // Applying section formula
        let x = ((n * x1) + (m * x2)) /
                    (m + n);
        let y = ((n * y1) + (m * y2)) /
                    (m + n);
   
   
        // Printing result
        document.write("(" + x + ", " + y + ")");
    }
 
// Driver Code
 
        let x1 = 2, x2 = 4, y1 = 4,
               y2 = 6, m = 2, n = 3;
        section(x1, x2, y1, y2, m, n)
        
       // This code is contributed by avijitmondal1998.

输出：

(2.8, 4.8)

这是如何运作的?

From our diagram, we can see,
PS = x – x1 and RT = x2 – x

We are given,

PR/QR = m/n

Using similarity, we can write
RS/QT = PS/RT = PR/QR

Therefore, we can write
 PS/RR = m/n
 (x - x1) / (x2 - x) = m/n

From above, we get
  x = (mx2 + nx1) / (m + n)

Similarly, we can solve for y.

参考：
http://doubleroot.in/lessons/coordinate-geometry-basics/section-formula/#.WjYXQvbhU8o