A simple problem over reflection of point about a line

Question

My question is extremely dull, so please don't bother: I want to have an explicit expression for the reflection of a point $(p,q)$ about a line $y=mx$, in terms of coefficients $p$,$q$ and $m$. But I am unable to get it. (Atleast no textbook or website mentions it to my reach). Please help.

Answer 1

There is a simple derivation using complex numbers.

Let $\alpha$ be the angle the line forms with the positive $x$-semiaxis, so $m=\tan\alpha$; we can choose $-\pi/2<\alpha<\pi/2$.

We can find the point corresponding to $(p,q)$ in three steps:

1. rotate the point around the origin by the angle $-\alpha$;
2. get the reflection of the point about the $x$-axis;
3. rotate the point around the origin by the angle $\alpha$.

If we identify $(p,q)$ with $z=p+iq$ and consider $u=\cos\alpha+i\sin\alpha$, then the three operations above are

1. multiply $z$ by $u^{-1}=\bar{u}$;
2. get the conjugate;
3. multiply by $u$.

The bar denotes conjugation.

Thus the corresponding point is $$ \overline{(z\bar{u})}u=\bar{z}uu=u^2\bar{z} $$ Now it's just a matter of separating the real and imaginary parts. Note that $$ u^2=\cos2\alpha+i\sin2\alpha $$ so we have $$ u^2\bar{z}=(\cos2\alpha+i\sin2\alpha)(p-iq)=(p\cos2\alpha+q\sin2\alpha)+i(p\sin2\alpha-q\cos2\alpha) $$ Taking into account that $$ \cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{1-m^2}{1+m^2},\qquad \sin2\alpha=\frac{2\tan\alpha}{1+\tan^2\alpha}=\frac{2m}{1+m^2} $$ we have $$ (p,q)\mapsto\left(\frac{1-m^2}{1+m^2}p+\frac{2m}{1+m^2}q,\frac{2m}{1+m^2}p-\frac{1-m^2}{1+m^2}q\right) $$ In matrix form, identifying a point with a column vector, the transformation is $$ \begin{bmatrix} p \\ q \end{bmatrix} \mapsto \begin{bmatrix} \dfrac{1-m^2}{1+m^2} & \dfrac{2m}{1+m^2} \\ \dfrac{2m}{1+m^2} & -\dfrac{1-m^2}{1+m^2} \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} \cos2\alpha & \sin2\alpha \\ \sin2\alpha & -\cos2\alpha \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} $$ Note that the trigonometric representation is also valid for any angle, including $\pi/2$.

Answer 2

Let the image of $(p,q)$ be $(a,b)$.

Since the line joining these two points is orthogonal to $y=mx$, its slope $=\frac{b-q}{a-p}=-1/m$. Further, the midpoint of the line segment joining these two points lies on $y=mx\implies \frac{b+q}2=m\cdot\frac{a+p}2$. Solve these two equations to get $(a,b)=(\frac{2mq-p(m^2-1)}{m^2+1},\frac{2mp+q(m^2-1)}{m^2+1})$.