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

Question

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 $$ \begin{pmatrix} \frac{1}{\sqrt{37}} & \frac{6}{\sqrt{37}} & 0\\ -\frac{6}{\sqrt{37}} & \frac{1}{\sqrt{37}} & 0\\ 0&0&1 \end{pmatrix} $$

$\textbf{We have to find the equation of the straight line.}$

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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 $y = -6x -27$.

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Is the solution correct?

Answer 1

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.