Consider a point $(p,q)$ in $\mathbb{R}^2$. I'm interested in reflecting this point over the line $x+y+10=0$, for example. I know from the formula for the reflection of a point over a line that the reflected point has coordinate $\left(-q-10, -p-10\right)$. However, I'm only interested in reflecting points such that $p+q+10<0$, so all points $(p,q)$ such that $p+q+10\geq 0$ remain $(p,q)$. Is there a way to represent this $\mathbb{R}^2 \to \mathbb{R}^2$ mapping with only absolute values and without cases?
Reflection of point over line with absolute value
functionsgeometry
Related Solutions
For reference:
Let...
$\vec{P} = \langle x-a, y-b \rangle$
$\vec{Q} = \langle c-a, d-b \rangle$
$\theta = \text{ the angle between } \vec{P} \text{ and } \vec{Q}$
First off, let's start with projecting $\vec{P}$ onto $\vec{Q}$. In math...
$\vec{K} = \text{Proj}_{\vec{Q}} \vec{P} = \displaystyle \frac{\vec{P} \cdot \vec{Q}}{||\vec{Q}||^2} \vec{Q}$
This gives us the vector $\vec{K}$ that goes from $A$ to the "intersection" of the two lines $\overline{AB}$ and $\overline{CC'}$. To find the vector from $C$ to that intersection, simply subtract $\vec{K}$ from $\vec{P}$. You can then multiply this vector by two and add to $C$ to get $C'$, or in other words $2(\vec{K}-\vec{P}) + \vec{P}$. This is equivalent to $2\vec{K}-\vec{P}$. Substituting the formula for $\vec{K}$ back in gives:
$\vec{P'} = \displaystyle 2\frac{\vec{P} \cdot \vec{Q}}{||\vec{Q}||^2} \vec{Q} - \vec{P}$
You can now add $\vec{P'}$ to $A$ to get $C'$. That sufficient for your needs?
Here's a simpler idea: represent your new point $C$ in the following parametric form:
$$(1-t)\begin{pmatrix}x_1\\y_1\end{pmatrix}+t\begin{pmatrix}x_2\\y_2\end{pmatrix}$$
where $0\leq t \leq 1$, such that you have the point $\begin{pmatrix}x_1\\y_1\end{pmatrix}$ when $t=0$ and the point $\begin{pmatrix}x_2\\y_2\end{pmatrix}$ when $t=1$. Your task now is to find the value $t^\prime$ such that the length from $t=0$ to $t=t^\prime$ is $c-n$.
Luckily, the distance from $t=0$ to $t=t^\prime$ is easily derived:
$c-n=t^\prime \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
whose derivation I leave as an exercise. Solve for the value of $t^\prime$, substitute into the parametric expression, and you have yourself the desired point.
Best Answer
$$ \pmatrix{p \\ q}\mapsto \pmatrix{ \displaystyle{|p+q+10|+p-q-10\over2} \\ \displaystyle{|p+q+10|-p+q-10\over2}} $$