Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$ 
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$

Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\tan\theta=-5$
Ref$(\theta)$=$\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}$
$$
\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-25}{1+25}=\frac{-24}{26}=\frac{-12}{13}\\
\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{-10}{26}=\frac{-5}{13}\\
Ref(\theta)\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13}
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}\\
=\frac{1}{13}\begin{bmatrix}-48+65\\-20-156\end{bmatrix}=\frac{1}{13}\begin{bmatrix}17\\-176\end{bmatrix}
$$
Why am I not getting the required solution in Method two using matrix method ?
Thanx @ganeshie8 for the remarks, so in that case how do I find the operator for reflection of a point over the line not passing through the origin ?
 A: As ganeshie8 suggested, your matrix formula is not working because the line does not pass through the origin.
When you translate everything up by $6$ units, the line now passes through the origin and you can continue as follows:
$$\begin{bmatrix}
\dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13}
\end{bmatrix}\begin{bmatrix}4\\-7\end{bmatrix}\\$$
$$=\frac{1}{13} \begin{bmatrix}
-48+35  \\ -20-84
\end{bmatrix}\\$$
$$= (-1, -8)$$
And now translate down by $6$ units to find that the original coordinate is at $(-1,-14)$.
A: The simplest (and shortest) way is to do some affine geometry:
Find first  the projection of the point $A(4,-13)$ onto the line $5x+y+6=0$. As a directing vector for this projection is $\vec n (5,1)$, you have a parametric equation of the line of projection:
$$\overrightarrow{OM}=A+t\mkern 1.5mu\vec n, $$it suffices  to find $t$ so the point $M$ satisfies the equation $5x+y+6=0$. Then, the reflection of $A$ is the  point   $$A'=A+2t\mkern 1.5mu \vec n.$$ 
A: Translating up by $6$ units again, we find that the angle between the $x$-axis and $(4,-7)$ is $-\tan^{-1} \frac{7}{4}$. In addition, the angle between the line and the $x$-axis is $-\tan^{-1} 5$. So the reflected point must have an angle of:
$$\theta =-\tan^{-1} 5  -\tan^{-1} 5 + \tan^{-1} \frac{7}{4}$$
In addition, the distance between the origin and the original point is the same as the distance between the origin and the reflected point (since the reflected angles are equal). This gives:
$$r = \sqrt{4^2 + 7^2}$$
Now, the reflected point is at $(r \cos \theta, r \sin \theta) = (-1,-8)$. Translating down by $6$ units, we again get $(-1, -14)$.
A: Method 3. Translate the origin to the point $(\color{red}{\frac32,-\frac{27}{2}})$ and find the coordinates of the point ${\color{blue}{4\choose -13}}$ in the new system:
$${x'\choose y'}={x\color{red}{-\frac32}\choose y\color{red}{+\frac{27}{2}}}={\color{blue}4-\frac32\choose \color{blue}{-13}+\frac{27}{2}}={\frac52 \choose \frac{1}{2}}$$
Rotate it by $180^\circ$:
$${\begin{pmatrix}-1&0\\ 0&-1\end{pmatrix}}{\frac52\choose \frac12}={-\frac52\choose -\frac12}$$
Now translate it back:
$${x\choose y}={-\frac52+\frac32\choose -\frac12-\frac{27}{2}}={-1\choose -14}.$$
