Hello, alexgeek!
$\displaystyle \text{Find the reflection of a point }(12, 1)\text{ in the line }y\,=\,7x$
Code:

Q 
o  / y = 7x
*  /
* /
 * M/
 o
 / *
 / *
 / *
 / o (12,1)
/ P
   +            

We want a line perpendicular to $\displaystyle y \,=\,7x.$ .[1]
This line has slope $\displaystyle \text{}\frac{1}{7}$ and contains (12,1).
. . Its equation is: .$\displaystyle y 1 \:=\:\text{}\frac{1}{7}(x12) \quad\Rightarrow\quad y \:=\:\text{}\frac{1}{7}x + \frac{19}{7}$ .[2]
$\displaystyle M$ is the intersection of [1] and [2].
. . $\displaystyle 7x \:=\:\text{}\frac{1}{7}x + \frac{19}{7} \quad\Rightarrow\quad 7x + \frac{1}{7}x \:=\:\frac{19}{7} \quad\Rightarrow\quad \frac{50}{7}x \:=\:\frac{19}{7}$
. . $\displaystyle x \:=\:\frac{19}{50} \quad\Rightarrow\quad y \:=\:\frac{133}{50}$
Point $\displaystyle M$ is $\displaystyle \left(\dfrac{19}{50},\:\dfrac{133}{50}\right)$
Let point $\displaystyle Q$ be $\displaystyle (x,y)$
Note that $\displaystyle M$ is the midpoint of $\displaystyle PQ.$
So we have: .$\displaystyle \begin{Bmatrix}\dfrac{x+12}{2} &=& \dfrac{19}{50} \\ \\[3mm] \dfrac{y+1}{2} &=& \dfrac{133}{50} \end{Bmatrix}$
Solve the two equations: .$\displaystyle x \,=\,\text{}\dfrac{281}{25},\;y \,=\,\dfrac{108}{25}$
Therefore: .$\displaystyle Q\left(\text{}\dfrac{281}{25},\;\dfrac{108}{25}\right) $