Consider two lines defined by:
$$\begin{aligned}y_1 &= m_1 x + b_1\\y_2 &= m_2 x + b_2\end{aligned}$$
where for the sake of argument, the domain of both lines is the same and everything is a real number.
Is there an analytical solution to a new line, $y_3$, that minimizes the error between the original two lines? For example, $m_3 = \frac{1}{2}(m_1+m_2)$ and $b_3 = \frac{1}{2}(b_1+b_2)$. This would work for lines where $m_1 = -m_2$ and $b_1 = -b_2$. However, for two parallel lines averaging the intercepts would be great but the slope would need to be unchanged.
Is there a general expression for the new line that best approximates the original two lines?
There is an analytic solution, but it is much more cumbersome than just averaging slopes and intercepts.
By symmetry for non-parallel lines the smallest error is given by the line that passes through the point of their intersection and bisects the angle between them. Recall that $m=\tan\theta$, where $\theta$ is the angle between the line and the $x$-axis. To get the bisector slope you need to average angles, not slopes, i.e. $m_3=\tan\theta_3:=\tan\frac{\theta_1+\theta_2}{2}$. Expressing $m$ in terms of $m_i=\tan\theta_i$ doesn't look pretty but can be done using trig formulas. Then solving for the intersection point gives you some $(x_*,y_*)$, and the slope-point equation of the bisector line is $y-y_*=m(x-x_*)$. You can see that the new intercept is $b_3=y_*-m_3x_*$, so it will not be any prettier than $m_3$.