在二维平面中查找点的镜像

给定二维平面中的点P和镜面方程，任务是找到由于镜面而形成的该点Q的图像。
镜像方程式为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)

由于物体和图像与镜子等距，因此R必须是P和Q的中间点

由于镜像的方程式为：ax + by + c =0。通过P和Q的线的方程式垂直于镜像。因此，穿过P和Q的直线方程变为ay – bx + d = 0，而且P穿过穿过P和Q的直线，因此我们将P的坐标放在上面的方程中，
a * y1 – b * x1 + d = 0
d = b * x1 – a * y1

另外，R是通过P和Q的镜像线的交点。因此我们找到了
ax + by + c = 0
ay -bx + d = 0
由于a，b，c，d都是已知的，因此我们可以在这里找到x和y。由于R的坐标现在是已知的，即x3，y3现在是已知的。

由于R是PQ的中点，
(x3，y3)=((x1 + x2)/ 2，(y1 + y2)/ 2)
由于已知x1，y1，x3，y3，我们得到下面的等式，其中(x，y)是Q(像点)的坐标

我们使用上面的公式来找到点P(x1，y1)相对于方程ax + by + c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

输出：

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