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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.

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 $y=mximplies frac{b+q}2=mcdotfrac{a+p}2$. Solve these two equations to get $(a,b)=(frac{2mq-p(m^2-1)}{m^2+1},frac{2mp+q(m^2-1)}{m^2+1})$.

There is a simple derivation using complex numbers.

Let $alpha$ be the angle the line forms with the positive $x$-semiaxis, so $m=tanalpha$; 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 $u=cosalpha+isinalpha$, 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
$$
overline{(zbar{u})}u=bar{z}uu=u^2bar{z}
$$
Now it's just a matter of separating the real and imaginary parts. Note that
$$
u^2=cos2alpha+isin2alpha
$$
so we have
$$
u^2bar{z}=(cos2alpha+isin2alpha)(p-iq)=(pcos2alpha+qsin2alpha)+i(psin2alpha-qcos2alpha)
$$
Taking into account that
$$
cos2alpha=frac{1-tan^2alpha}{1+tan^2alpha}=frac{1-m^2}{1+m^2},qquad
sin2alpha=frac{2tanalpha}{1+tan^2alpha}=frac{2m}{1+m^2}
$$
we have
$$
(p,q)mapstoleft(frac{1-m^2}{1+m^2}p+frac{2m}{1+m^2}q,frac{2m}{1+m^2}p-frac{1-m^2}{1+m^2}qright)
$$
In matrix form, identifying a point with a column vector, the transformation is
$$
begin{bmatrix} p \ q end{bmatrix}
mapsto
begin{bmatrix}
dfrac{1-m^2}{1+m^2} & dfrac{2m}{1+m^2} \
dfrac{2m}{1+m^2} & -dfrac{1-m^2}{1+m^2}
end{bmatrix}
begin{bmatrix} p \ q end{bmatrix}
=
begin{bmatrix}
cos2alpha & sin2alpha \
sin2alpha & -cos2alpha
end{bmatrix}
begin{bmatrix} p \ q end{bmatrix}
$$
Note that the trigonometric representation is also valid for any angle, including $pi/2$.