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 ?
Best Answer
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 $(-1,-14)$.