Phase difference and path difference relation confusion

Question

If two light waves of same wavelength reaches the point $ P$, then what is the relationship between path difference and phase difference? In many books I have seen that they assume two 1-D equation for each wave. Such as,

\begin{align} y_{1}= a \sin\frac{2π}{\lambda}(ct-x_1) \end{align}

\begin{align} y_{2}= a \sin\frac{2π}{\lambda}(ct-x_2) \end{align}

Now,phase difference, \begin{align} \delta = \frac{2π}{\lambda}(ct-x_1)-\frac{2π}{\lambda}(ct-x_2) \end{align} \begin{align} \Rightarrow \delta = \frac{2π}{\lambda}(x_2 - x_1) \end{align} \begin{align} \Rightarrow \delta = \frac{2π}{\lambda} ∆x \end{align}

But the problem is that both wave sources can be different and the waves may not be parallel or along same axis. So how can we use this formula then?

Answer 1

$\delta$ is the phase-difference $\theta_2 - \theta_1$ at the meeting point, where $\Delta x$ is the path-difference (difference in path lengths) $\ell_2 - \ell_1$ between the two paths from the sources to the meeting point.

The paths need not be parallel. However, the paths must meet at a meeting point. (The tangent vectors to the paths need not be parallel at the meeting point.) At that meeting point, the difference of their phases is calculated.