In this problem, we are given a point P in a 2-D plane and the points a, b, c of the equation ax + by + c = 0. Our task is to find a mirror image of a point in 2-D plane.
Let’s take an example to understand the problem,
Input
P = (2, 1), a = 1, b = -1, c = 0
Output
(1, 2)
Explanation
The plane looks like,
Solution Approach
To solve the problem, we need to find the equation point P' with coordinates (x', y'). So, we have R, the midpoint where the line form P - P' intersects the mirror line.
The line P-R-P' is perpendicular to the mirror. Hence, the equation of line will be,
ay - by + d = 0
The points are P(x, y) ; P'(x', y') ; R(xm, ym).
The points P and R are known. So, using the equations we will find P’ as,
$$\left(\frac{??'-??}{??}\right)=\left(\frac{??'-??}{??}\right)=\left(\frac{????-????+??}{??^2+x^2}\right)$$
Program to illustrate the working of our solution,
Example
#include <iostream>
using namespace std;
void findMirrorImage( double a, double b, double c, double x, double y){
double points = -2 * (a * x + b * y + c) / (a * a + b * b);
double xm = points * a + x;
double ym = points * b + y;
cout<<"("<<xm<<","<<ym<<")";
}
int main(){
double a = -1.0;
double b = 1.0;
double c = 0.0;
double x = 1.0;
double y = 0.0;
cout<<"Image of point ("<<x<<", "<<y<<") using mirror ("<<a<<")x + ("<<b<<")y + ("<<c<< ") = 0, is :";
findMirrorImage(a, b, c, x, y);
return 0;
}Output
Image of point (1, 0) using mirror (-1)x + (1)y + (0) = 0, is :(0,1)