Let $L$ be a straight line with equation $y = mx+c$. Refecting across $L$ is an affine transformation and as a result, we can express it as a matrix.
We can translate and rotate and make the straight line coincide with the $x-$axis then go back and get the original points.
The translation and rotation matrices are given as follows:
$$
\begin{pmatrix}
1 & 0 & -2\\
0 & 1 & -15\\
0 & 0 & 1
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
\frac{1}{\sqrt{37}} & \frac{6}{\sqrt{37}} & 0\\
-\frac{6}{\sqrt{37}} & \frac{1}{\sqrt{37}} & 0\\
0&0&1
\end{pmatrix}
$$
$\textbf{We have to find the equation of the straight line.}$
From the rotation matrix we have $\sin \theta = -\frac{6}{\sqrt{37}}$ and $\cos \theta = \frac{1}{\sqrt{37}}$ thus slope $m = \tan \theta = -6$.
Using the translation the line should pass through origin thus the constant terms should be zero.
$$y – 15 = -6(x-2) + c$$ Equating the constant terms we have $-15 = 12+c$. Thus $c = -27$.
Hence equation of line is $y = -6x -27$.
Is the solution correct?
Best Answer
It should be $y = 6x + 3$.
If the transformed line is $~Y = 0, $ using rotation matrix, $\sin\theta = - \frac{6}{\sqrt{37}}, \cos\theta = \frac{1}{\sqrt{37}}$.
So, $Y = 0 = x' \sin\theta + y' \cos\theta \implies y' = 6x' ~$ before rotation.
And using translation matrix,
$(y+k) = 6 (x + h) ~ $ before translation, and as $h = -2, k = -15$,
$y - 15 = 6x - 12 \implies y = 6x + 3 ~$ must be the original line.