Take $x,y$ the global coordinates, $x^\prime,y^\prime$ the local coordinates, $(h,k)$ the global coordinates of the local system's origin, and $\varphi$ the anticlockwise angle from the global system's horizontal ($x$) to the local system's "horizontal" ($x^\prime$).
What is needed, then, to convert from $(x^\prime,y^\prime)$ to $(x,y)$ is to 1. shift the origin; and 2. rotate by $\varphi$ clockwise. This sequence is most easily expressed in matrix-vector notation:
$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\cos\,\varphi&\sin\,\varphi\\-\sin\,\varphi&\cos\,\varphi\end{pmatrix}\cdot\left(\begin{pmatrix}x^\prime\\y^\prime\end{pmatrix}-\begin{pmatrix}h\\k\end{pmatrix}\right)$$
or explicitly,
$$\begin{align*}x&=(x^\prime-h)\cos\,\varphi+(y^\prime-k)\sin\,\varphi\\y&=-(x^\prime-h)\sin\,\varphi+(y^\prime-k)\cos\,\varphi\end{align*}$$
if $(a,b)$ is the cordinates of the known point. then unknown point is $(a+r\cos \theta,b+r\sin \theta)$ where $r$ is the known length and $\theta$ the angle.
Best Answer
There is a simple technique using determinants (see there).
Consider for example points $$\begin{cases}A&=&(x_A,y_A,z_A)&=&(0,2,2)\\ B&=&(x_B,y_B,z_B)&=&(2,0,2)\\ C&=&(x_C,y_C,z_C)&=&(2,2,0)\end{cases}$$
which can be considered as vertices of a cube.
Let $M(1,1,z)$ belonging to plane $ABC$ for which we desire coordinate $z$.
The coplanarity of 4 points is equivalent to the fact that the following determinant is zero:
$$\begin{vmatrix}x_A&y_A&z_A&1\\ x_B&y_B&z_B&1\\ x_C&y_C&z_C&1\\ x_M&y_M&z_M&1 \end{vmatrix}=0 \ \text{giving here} \ \begin{vmatrix}0&1&1&1\\ 1&0&1&1\\ 1&1&0&1\\ 1&1&z&1 \end{vmatrix}=0 $$
giving a first degree equation for $z$. Technicaly: subtract the last column to the two first columns, one gets an easy expansion giving, as expected, the value $z=0$.