在二维平面中找到一个点的镜像

给定二维平面中的一个点 P 和镜面方程，任务是找到由镜面形成的点 Q 的图像。
镜像方程的形式为 ax + by + c =
例子：

Input : P = (1, 0), a = -1, b = 1, c = 0
Output : Q = (0, 1)

Input : P = (3, 3), a = 0, b = 1, c = -2
Output : Q = (3, 1)

解决方案 ：

Let coordinate of P(given point) be (x1, y1)
Let coordinate of Q(image point) be (x2, y2)
Let coordinate of R(point on mirror) be (x3, y3)

由于物象距镜等距，R必是P和Q的中点
由于镜面方程为：ax + by + c = 0。通过 P 和 Q 的线方程垂直于镜面。因此通过 P 和 Q 的线方程变为 ay – bx + d = 0，P 也通过通过 P 和 Q 的线，所以我们把 P 的坐标放在上面的方程中，
a*y1 – b*x1 + d = 0
d = b*x1 – a*y1
同样 R 是镜面和穿过 P 和 Q 的线的交点。所以我们找到了解决方案
ax + by + c = 0
ay -bx + d = 0
由于 a、b、c、d 都是已知的，我们可以在这里找到 x 和 y。由于 R 的坐标现在已知，即 x3，y3 现在已知。
由于 R 是 PQ 的中点，
(x3, y3) = ((x1+x2)/2, (y1+y2)/2)
由于 x1, y1, x3, y3 是已知的，我们得到以下等式 其中 (x, y) 是 Q(image point) 的坐标

我们使用上面的公式找到点 P(x1, y1) 相对于方程 ax + by + c 的镜像的镜像

C++

// C++ code to find mirror image
#include 
using namespace std;
 
// C++ function which finds coordinates
// of mirror image.
// This function return a pair of double
pair mirrorImage(
    double a, double b, double c,
    double x1, double y1)
{
    double temp = -2 * (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 to test above function
int main()
{
    double a = -1.0;
    double b = 1.0;
    double c = 0.0;
    double x1 = 1.0;
    double y1 = 0.0;
 
    pair image = mirrorImage(a, b, c, x1, y1);
    cout << "Image of point (" << x1 << ", " << y1 << ") ";
    cout << "by mirror (" << a << ")x + (" << b
         << ")y + (" << c << ") = 0, is :";
    cout << "(" << image.first << ", " << image.second
         << ")" << endl;
 
    return 0;
}

Java

// Java code to find mirror image
class GFG
{
static class pair
{
    double first, second;
    public pair(double first, double second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// function which finds coordinates
// of mirror image.
// This function return a pair of double
static pair mirrorImage(double a, double b,
                        double c, double x1,
                        double y1)
{
    double temp = -2 * (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)
{
    double a = -1.0;
    double b = 1.0;
    double c = 0.0;
    double x1 = 1.0;
    double y1 = 0.0;
 
    pair image = mirrorImage(a, b, c, x1, y1);
    System.out.print("Image of point (" + x1 +
                                   ", " + y1 + ") ");
    System.out.print("by mirror (" + a +
                          ")x + (" + b +
                          ")y + (" + c + ") = 0, is :");
    System.out.println("(" + image.first +
                       ", " + image.second + ")");
}
}
 
// This code is contributed by 29AjayKumar

Python 3

# Python 3 code to find mirror image
 
# Python function which finds coordinates
# of mirror image.
# This function return a pair of double
def mirrorImage( a, b, c, x1, y1):
    temp = -2 * (a * x1 + b * y1 + c) /(a * a + b * b)
    x = temp * a + x1
    y = temp * b + y1
    return (x, y)
 
# Driver code to test above function
a = -1.0
b = 1.0
c = 0.0
x1 = 1.0
y1 = 0.0
 
x, y = mirrorImage(a, b, c, x1, y1);
print("Image of point (" + str (x1) + ", " + str( y1) + ") ")
print("by mirror (" + str (a) + ")x + (" + str( b) + ")y + (" +str(c) + ") = 0, is :")
print( "(" + str(x) + ", " + str(y) + ")" )
 
# This code is contributed by ApurvaRaj

C#

// C# code to find mirror image
using System;
 
class GFG
{
class pair
{
    public double first, second;
    public pair(double first, double second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// function which finds coordinates
// of mirror image.
// This function return a pair of double
static pair mirrorImage(double a, double b,
                        double c, double x1,
                        double y1)
{
    double temp = -2 * (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)
{
    double a = -1.0;
    double b = 1.0;
    double c = 0.0;
    double x1 = 1.0;
    double y1 = 0.0;
 
    pair image = mirrorImage(a, b, c, x1, y1);
    Console.Write("Image of point (" + x1 +
                                ", " + y1 + ") ");
    Console.Write("by mirror (" + a +
                       ")x + (" + b +
                       ")y + (" + c + ") = 0, is :");
    Console.WriteLine("(" + image.first +
                     ", " + image.second + ")");
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

      // JavaScript code to find mirror image
 
      // JavaScript function which finds coordinates
      // of mirror image.
      // This function return a pair of double
      function mirrorImage(a, b, c, x1, y1) {
        var temp = (-2 * (a * x1 + b * y1 + c)) / (a * a + b * b);
        var x = temp * a + x1;
        var y = temp * b + y1;
        return [x, y];
      }
 
      // Driver code to test above function
      var a = -1.0;
      var b = 1.0;
      var c = 0.0;
      var x1 = 1.0;
      var y1 = 0.0;
 
      var [x, y] = mirrorImage(a, b, c, x1, y1);
      document.write("Image of point (" + x1 + ", " + y1 + ") ");
      document.write(
        "by mirror <br> (" + a + ")x + (" + b + ")y + (" + c + ") = 0, is :"
      );
      document.write("(" + x + ", " + y + ")");
    

输出：

Image of point (1, 0) by mirror 
(-1)x + (1)y + (0) = 0, is :(0, 1)

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