How to find a "least squares" line with a known slope? I have gps trackings that I know they fall into a certain pattern - a line with a known angle. How do I find the line that minimizes the distances of the points from it but is in the correct angle?
Unfortunately, I can't post an image for example.
 A: If I well understand, you have a set of $n$ points $(x_1,y_1),(x_2,y_2),...,(x_k,y_k),..., (x_n,y_n)$ and you want to fit a straight line $y=ax+b$ which coefficient $a$ is known.
This is a regression problem of the kind  $Y(x)=b$ where $Y(x)=y(x)-ax$
Hense $$b\simeq \frac{1}{n}\sum_{k=1}^n (y_k-ax_k)$$
In this particular case ($a$ known), the adjusted value of $b$ is the same if we consider the orthogonal distances. 
A: Consider your data as points $(x_i,y_i)$ and you have $n$ data points, then you want to find $\beta_1 ,\beta_2$ such that $y\simeq\beta_1+\beta_2x$.
$$
       X\beta= \begin{bmatrix}
        1 & x_1 \\
        1 & x_2 \\
        1 & x_3  \\
        \vdots & \vdots\\
        1 & x_n\\
        \end{bmatrix}
\begin{bmatrix}
\beta_1\\
\beta_2\\
\end{bmatrix}=\begin{bmatrix}
y_1\\
y_2\\
\vdots\\
y_n
\end{bmatrix}=y
$$
If you have received lots of data from gps your $X$ matrix is tall and a pseudoinverse can be found as:
$$
X^+=(X^TX)^{-1}X^T
$$
And then you can find the solution:
$$
\beta=X^+y
$$
It's MATLAB code is 
beta=X\y;

See here for more information.

If $\beta_2$ is known and you want to find $\beta_1$ such that, $y-\beta_2x\simeq\beta_1$ then
$$
X\beta= \begin{bmatrix}
    1  \\
    1  \\
    1   \\
    \vdots\\
    1 \\
    \end{bmatrix}
\begin{bmatrix}
\beta_1\\
\end{bmatrix}=\begin{bmatrix}
y_1-\beta_2x_1\\
y_2-\beta_2x_2\\
\vdots\\
y_n-\beta_2x_n
\end{bmatrix}
$$
$$
X^+=(X^TX)^{-1}X^T=(\begin{bmatrix}
    1  &
    1  &
    1   &
    \cdots&
    1 
    \end{bmatrix}
\begin{bmatrix}
    1  \\
    1  \\
    1   \\
    \vdots\\
    1 \\
    \end{bmatrix})^{-1}
\begin{bmatrix}
    1  &
    1  &
    1   &
    \cdots&
    1 
    \end{bmatrix}
$$
And the answer will be:
$$
\beta_1=X^+y=\frac{1}{n}\begin{bmatrix}
    1  &
    1  &
    1   &
    \cdots&
    1 
    \end{bmatrix}\begin{bmatrix}
y_1-\beta_2x_1\\
y_2-\beta_2x_2\\
\vdots\\
y_n-\beta_2x_n
\end{bmatrix}$$
$$
\beta_1=\frac{1}{n}\sum_{k=1}^n (y_i-\beta_2x_i)
$$
