Why do the $(V,I)$ points for various $R_L$ lie on a line?

This follows from Thevenin's theorem- if the circuit contains linear elements (resistors, voltage sources and current sources), it will behave just like a voltage source in series with the Thevenin-equivalent resistance $R_{\rm Th}$. The behavior of such a circuit will follow Ohm's Law for the Thevenin-equivalent circuit, with $R_L$ in series with $R_{\rm Th}$. For a given $R_L$, the current across the load resistor is $I_L=V_{\rm Th}/(R_{\rm Th}+R_L)$ and the voltage across the load resistor $V_L$ will be $V_L = I_L R_L = R_L(V_L/(R_{\rm Th}+R_L))$. After a bit of algebra, you can show that $I_L = (V_{\rm Th}-V_L)/R_{\rm Th}$, which is exactly an expression of a straight $I$ vs. $V$ load line with slope $-1/R_{\rm Th}$ and y-intercept $I_N = V_{\rm Th}/R_{\rm Th}$.