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I'm having trouble understanding the solution presented below (It's from a textbook). I tried to get something similar, but to no avail. Help me find the way he derived those a,b,x2 and y2 expressions.

Point p is a point which we want to reflect (It's a simple structure with two integers x and y representing coordinates on a 2d plane); (x0,y0) and (x1,y1) are coordinates of endpoints of a line. Expected result is a new point with reflected coordinates.

Point mirror(Point p, int x0, int y0, int x1, int y1) {     double dx,dy,a,b;    long x2,y2;    Point p1; //reflected point to be returned      dx  = (double) (x1 - x0);    dy  = (double (y1 - y0);     a   = (dx * dx - dy * dy) / (dx * dx + dy*dy);    b   = 2 * dx * dy / (dx*dx + dy*dy);     x2  = Math.round(a * (p.x - x0) + b*(p.y - y0) + x0);     y2  = Math.round(b * (p.x - x0) - a*(p.y - y0) + y0);     p1 = Point((int)x2,(int)y2);      return p1;  } 
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    See [here](https://math.stackexchange.com/q/13261/11619) for older discussion (and better quality images).2017-06-12

2 Answers 2

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Here's an explanation. It is easier to follow, if you draw a picture. Let $\vec{u}=(dx,dy)$ be the vector from the point $P_0=(x_0,y_0)$ to the point $P_1=(x_1,y_1)$, i.e. a vector pointing in the direction of the mirror line. Then $\vec{n}=(-dy,dx)$ is perpendicular to it. Let's name the vector from $P_0$ to $P$ $\vec{v}=(p.x-x_0,p.y-y_0)$. The projection of the vector $\vec{v}$ along the normal $\vec{n}$ is then $\vec{p}=\frac{\vec{n}\cdot\vec{v}}{\vec{n}\cdot\vec{n}}\vec{n}=\frac{-(p.x-x_0)dy+(p.y-y_0)dx}{dx^2+dy^2}\vec{n}.$

We compute the mirror image point $P'=(x_2,y_2)$ by comparing the representations of the vectors $\vec{P_0P}=\vec{v}$ and $\vec{P_0P'}=\vec{v'}$ in the (orthogonal) basis $\{\vec{u},\vec{n}\}$. The reflection maps $\vec{u}$ to itself, but the vector perpendicular to mirror is mapped to its negative: $\vec{n}\mapsto -\vec{n}$. Therefore the reflection maps $ \vec{v}\mapsto\vec{v}'=\vec{v}-2\vec{p}. $ The rest amounts to just plugging in the numbers. Write $N=dx^2+dy^2$ for short. We get $ \begin{align} \vec{v}'&=\frac1N\left(N(p.x-x_0)-2dy^2(p.x-x_0)+2dy\,dx(p.y-y_0)\right)\vec{i}\\ &+\frac1N\left(N(p.y-y_0)+2dy\,dx(p.x-x_0)-2dx^2(p.y-y_0)\right)\vec{j}\\ &=\left(a(p.x-x_0)+b(p.y-y_0),b(p.x-x_0)-a(p.y-y_0)\right). \end{align} $ Your formula comes from the fact that the above vector $\vec{v'}$ is the separation vector from $P_0$ to $P'$. Therefore we need to add the coordinates of $P_0$ to the result.

Edit: A crude image here. The points $P,P_0,P_1,P'$ are marked with the dots, and the vectors are $u,n,v,p,v',-2p$. Sorry about the missing arrows on top of those letters - can't do any better :-(

enter image description here

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    @JyrkiLahtonen Ah, thanks, I was just confused about the fact they didn't appear anywhere else, but I see you're just factoring out the common expressions.2017-02-05
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It worked for me in C++. Where p is the point to reflect, and P0(x0, y0) P1(x1, y1) is the mirror line, and p1 is the point reflected. See the graph again

  struct Point{       int x;       int y;   };    Point mirror(Point p, int x0, int y0, int x1, int y1){       double dx,dy,a,b;      long x2,y2;      Point p1; //reflected point to be returned       dx  = (double) (x1 - x0);      dy  = (double) (y1 - y0);       a   = (dx * dx - dy * dy) / (dx * dx + dy*dy);      b   = 2 * dx * dy / (dx*dx + dy*dy);       x2  = round(a * (p.x - x0) + b*(p.y - y0) + x0);      y2  = round(b * (p.x - x0) - a*(p.y - y0) + y0);       p1.x = (int)x2;      p1.y = (int)y2;       return p1;   }