Let $σ_P$ be the rotation about the point $P = (a, b)$ through an $180◦$ angle.
a) Find the equations of this transformation.
For this, using the translation and rotation isometries: $T_{(a,b)} \circ σ_P \circ T_{-(a,b)} = T_{(a,b)}$$ \begin{pmatrix} cos \pi & -sin\pi \\ sin\pi & cos\pi \\ \end{pmatrix} $$ $$ \begin{pmatrix} x-a \\ y-b \\ \end{pmatrix} $$ = $$ \begin{pmatrix} 2a-x \\ y \\ \end{pmatrix} $$ $
Where $T$ represents a translation. I skipped a few steps to save space on here but I believe this is what is being asked. I'm not sure if my thought process is correct on this.
b) What is the image of the line with equation $y = 5x + 7$ under $σ_P$ when $P = (−3, 2)$?
Using what I got in part a, $y = 5x + 7$ is a line passing through points $(0,7),(-1.4,0)$. Applying the rotation $σ_P$ to the points, I output, $(-6,7),(-4.6,0)$ respectively. The line passing through those points would thus be $y=-5x-23$ which is the intended image of the original line. My question is, did I do this rotation correctly despite the fact that I only rotated two points and used their slope to find a line?
Any help would be appreciated on this.
Rotations and translations are rigid motions that carry lines into lines, and two points uniquely determine a line, so your approach to part b is legitimate.
Here’s an alternative: Parametrize the equation of the line, apply the transformation, and then eliminate the parameter, i.e., compute $\sigma_P\left(\begin{smallmatrix}t\\5t+7\end{smallmatrix}\right)$ and then eliminate $t$.