查找平行四边形的缺失点

给定三个坐标点A，B和C，找到缺失点D，以使ABCD可以是平行四边形。

例子 ：

Input : A = (1, 0)
        B = (1, 1)
        C = (0, 1)
Output : 0, 0
Explanation:
The three input points form a unit
square with the point (0, 0)

Input : A = (5, 0)
        B = (1, 1)
        C = (2, 5)
Output : 6, 4

如下图所示，可能有多个可能的输出，我们需要打印其中的任何一个。

如果四边形的相对边平行且长度相等，则称其为平行四边形。

由于我们获得了平行四边形的三个点，因此我们可以找到缺失边的斜率及其长度。
该算法可以解释如下

令R为缺失点。现在根据定义，我们有

• PR的长度= QS的长度= L1(对边相等)
• PR的斜率= QS的斜率= M1(对边平行)
• PQ的长度= RS的长度= L2(对边相等)
• PQ的斜率= RS的斜率= M2(相对的边平行)

因此，我们可以找到距P的距离L1处具有斜率M1的点，如下文所述：
在给定斜率的直线上找到给定距离的点。

现在，其中一个要点将满足上述条件，可以轻松检查(使用条件3或4)。
以下是相同的程序。

C++

// C++ program to find missing point of a
// parallelogram
#include 
using namespace std;
  
// struct to represent a co-ordinate point
struct Point {
    float x, y;
    Point()
    {
        x = y = 0;
    }
    Point(float a, float b)
    {
        x = a, y = b;
    }
};
  
// given a source point, slope(m) of line
// passing through it this function calculates
// and return two points at a distance l away
// from the source
pair findPoints(Point source,
                              float m, float l)
{
    Point a, b;
  
    // slope is 0
    if (m == 0) {
        a.x = source.x + l;
        a.y = source.y;
  
        b.x = source.x - l;
        b.y = source.y;
    }
  
    // slope if infinity
    else if (m == std::numeric_limits::max()) {
        a.x = source.x;
        a.y = source.y + l;
  
        b.x = source.x;
        b.y = source.y - l;
    }
  
    // normal case
    else {
        float dx = (l / sqrt(1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx, a.y = source.y + dy;
        b.x = source.x - dx, b.y = source.y - dy;
    }
  
    return pair(a, b);
}
  
// given two points, this function calculates
// the slope of the line/ passing through the
// points
float findSlope(Point p, Point q)
{
    if (p.y == q.y)
        return 0;
    if (p.x == q.x)
        return std::numeric_limits::max();
    return (q.y - p.y) / (q.x - p.x);
}
  
// calculates the distance between two points
float findDistance(Point p, Point q)
{
    return sqrt(pow((q.x - p.x), 2) + pow((q.y - p.y), 2));
}
  
// given three points, it prints a point such
// that a parallelogram is formed
void findMissingPoint(Point a, Point b, Point c)
{
    // calculate points originating from a
    pair d = findPoints(a, findSlope(b, c),
                                      findDistance(b, c));
  
    // now check which of the two points satisfy
    // the conditions
    if (findDistance(d.first, c) == findDistance(a, b))
        cout << d.first.x << ", " << d.first.y << endl;
    else
        cout << d.second.x << ", " << d.second.y << endl;
}
  
// Driver code
int main()
{
    findMissingPoint(Point(1, 0), Point(1, 1), Point(0, 1));
    findMissingPoint(Point(5, 0), Point(1, 1), Point(2, 5));
    return 0;
}

输出 ：

0, 0
6, 4

替代方法：

由于两边相等，AD = BC，AB = CD，我们可以将缺失点(D)的坐标计算为：

AD = BC
(Dx - Ax, Dy - Ay) = (Cx - Bx, Cy - By)
Dx = Ax + Cx - Bx 
Dy = Ay + Cy - By

参考文献： https : //math.stackexchange.com/questions/887095/find-the-4th-vertex-of-the-parallel-ogram

下面是上述方法的实现：

C++

// C++ program to find missing point
// of a parallelogram
#include 
using namespace std;
  
// main method
int main()
{
   int ax = 5, ay = 0; //coordinates of A
   int bx = 1, by = 1; //coordinates of B
   int cx = 2, cy = 5; //coordinates of C
    cout << ax + cx - bx << ", "
         << ay + cy - by;
    return 0;
}

Java

// Java program to 
// find missing point
// of a parallelogram
import java.io.*;
  
class GFG 
{
public static void main (String[] args) 
{
  
    int ax = 5, ay = 0; //coordinates of A
    int bx = 1, by = 1; //coordinates of B
    int cx = 2, cy = 5; //coordinates of C
    System.out.println(ax + (cx - bx) + ", " +
                       ay + (cy - by));
}
}
  
// This code is contributed by m_kit

的Python 3

# Python 3 program to find missing point
# of a parallelogram
  
# Main method
if __name__ == "__main__":
      
    # coordinates of A
      
    ax, ay = 5, 0
      
    # coordinates of B
      
    bx ,by = 1, 1
      
    # coordinates of C
      
    cx ,cy = 2, 5
      
    print(ax + cx - bx , ",", ay + cy - by)
  
# This code is contributed by Smitha

C＃

// C# program to 
// find missing point
// of a parallelogram
using System;
  
class GFG
{
    static public void Main ()
    {
        int ax = 5, ay = 0; //coordinates of A
        int bx = 1, by = 1; //coordinates of B
        int cx = 2, cy = 5; //coordinates of C
        Console.WriteLine(ax + (cx - bx) + ", " +
                             ay + (cy - by));
    }
}
  
// This code is contributed by ajit

输出：

6, 4