# Equation of line given translation and rotation that makes the line coincide with $x-$axis.

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?

• If the given translation and rotation matrix takes line $y = mx + c$ to x-axis then the equation of $y = mx + c$ that you obtain seems incorrect. Jan 10, 2022 at 18:35
• @MathLover Is the equation $y = 6x-3$ correct? Jan 11, 2022 at 2:09
• sorry I saw this late. It should be $y = 6x + 3$. Please see my answer. Jan 11, 2022 at 16:07

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.
$$(y+k) = 6 (x + h) ~$$ before translation, and as $$h = -2, k = -15$$,
$$y - 15 = 6x - 12 \implies y = 6x + 3 ~$$ must be the original line.