Coordinates of the reflection of a point across a line of specific angle I am going through an exercise to derive the point $P_1$ which is a reflection of of $P_0$ across line $D$.
The following diagram (I couldn't make it accurate for the post, I am sorry) shows the idea:

We know the following:
$ON_0 = ON_1 = x_0$
$N_0P_0 = N_1P_1 = y_0$
I need to find $x_1,y_1$ which are the coordinates of $P_1$ and reflections of $x_0,y_0$.
For $x1$:
$x_1 = OB + BA$
$BA = KP_1$
$KP_1 = N_1P_1\cdot \sin(2a) = y_0 \sin(2a)$
$OB = ON_1 \cdot \cos(2a) = x_0 \cdot \cos(2a)$
Hence: $x_1 = OB + BA = OB + KP_1 = x_0 \cdot \cos(2a) + y_0 \sin(2a)$
That is the $x$ coordinate of $P_1$.
But I am stuck with $y_1$.
I think:
$y_1 = ZN = ON\cdot \sin(2a)$
$ON = ON_1 - KN_1 = x_0 - KN_1$
but after that I am not sure how to proceed.
Additionally I suspect my approach so far is too complicated. I already know the answer and managed to figure out a way to derive it from $x1$ but not for $y1$.
How can I derive $y_1$ i.e. the $y$ coordinate for $P_1$?
Update:
I am looking for an approach using only trigonometry/geometry and not depend on matrices and linear algebra or vector methods/analytic geometry. Only basic trigonometry/geometry like how I tried to derive $x_1$
 A: A) - First solution using linear algebra:
Plainly use the matrix of the symmetry with respect to a line with polar angle $a$:
$$\begin{pmatrix}x_1\\y_1\end{pmatrix}=\underbrace{\begin{pmatrix}\cos(2a)& \ \ \ \sin(2a)\\ \sin(2a)&-\cos(2a)\end{pmatrix}}_{S_a}\begin{pmatrix}x_0\\y_0\end{pmatrix}, \tag{1}$$
Explanation: the first column of this matrix is easily obtained as the image of $\begin{pmatrix}1\\0\end{pmatrix}$ ; the second column is taken as the unique unit vector making with the first vector an orthonormal basis with determinant $-1$  (the determinant $-1$ is characteristic of an operation that reverses shapes' orientations).
Sanity check: $S_a^TS_a=I$: it is indeed an isometry matrix !

B) - Second solution by analytic geometry, using vectors:
$$\vec{OP_1}=\underbrace{\vec{ON_1}}_{= \ n_1\vec{d_1}}+\underbrace{\vec{N_1P_1}}_{= \ n_2\vec{d_2}}\tag{2}$$
where $n_1,n_2$ are their norms and $d_1,d_2$ their unit directing vectors.
What is $n_1$ ? The same as the norm of $\vec{ON_0}$ which is $x_0$ (because a symmetry preserves lengths).
What is $n_2$ ? The same as the norm of $\vec{N_0P_0}$, i.e., $y_0$.
What is $\vec{d_1}$ ? Unit vector $\binom{\cos(2\theta)}{\sin(2\theta)}.$
What is $\vec{d_2}$ ? A unit vector which is orthogonal to the previous one, which means $\sigma \binom{\ \ \sin(2\theta)}{-\cos(2\theta)}$ where $\sigma=\pm 1$. A priori two choices ; just understand that $\sigma=+1$ has to be chosen...
In this way, (2) becomes:
$$\binom{x_1}{y_1}=x_0\binom{\cos(2\theta)}{\sin(2\theta)} 
+y_0\binom{\ \ \sin(2\theta)}{-\cos(2\theta)}\tag{3}$$
which is exactly the same expression as (1).
A: HINT
Are you allowed to use Linear Algebra?
Suppose we have a line passing through the origin along the direction $v = (1,m)$.
Since reflection is a linear transformation, let us denote it by $T:\mathbb{R}^{2}\to\mathbb{R}^{2}$.
Then we have that $T(1,m) = (1,m)$ and $T(-m,1) = (m,-1)$.
If we denote such basis by $\mathcal{B} = \{(1,m), (-m,1)\}$, we conclude that
\begin{align*}
[T]_{\mathcal{B}} =
\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}
\end{align*}
Having said that, let us set that $\mathcal{B}' = \{(1,0),(0,1)\}$. Then it results that
\begin{align*}
[T]_{\mathcal{B}'} = [I]_{\mathcal{B}}^{\mathcal{B}'}[T]_{\mathcal{B}}[I]_{\mathcal{B}'}^{\mathcal{B}}
\end{align*}
Once you obtain the expression for $T$ in the basis $\mathcal{B}'$, you are done.
Can you take it from here?
A: Asking for coordinates but refusing analytical geometry or vectors is somewhat contradictory.
By trigonometry, $OP_0$ has length $l_0=\sqrt{x_0^2+y_0^2}$ and direction $\theta_0=\arctan\dfrac{y_0}{x_0}$. Then $l_1=l_0$, $\theta_1=2a-\theta_0$ and $P_1$ has the coordinates
$$l_0\,(\cos(2a-\theta_0), \sin(2a-\theta_0)).$$
You can expand as
$$l_0\,(\cos(2a)\cos(\theta_0)+\sin(2a)\sin(\theta_0), \sin(2a)\cos(\theta_0)-\cos(2a)\sin(\theta_0))$$
or
$$(x_1,y_1)=(\cos(2a)\,x_0+\sin(2a)\,y_0, \sin(2a)\,x_0-\cos(2a)\,y_0).$$
