Derivation of load line:
1. Let's say we know that we will always deal with some circuit form in which a voltage source, a resistor (linear element) will always be in series with another element which can be linear or non-linear. Then it makes sense to try to come up some sort of rules that can be easily applied in the encounters of such problem. To do this, refer to Figure1 above. Where X can be anything linear (such as resistor) and non-linear (diode, BJT...).
2. If we consider the cases of Vx changing from 0 to Vdd, we can then draw the line of I versus Vx, and this line is our load line, with the slope equal to (-1/Rin). Where VRin is the voltage across Rin.
Load line when X block is a resistor:
3. To ensure that this method makes sense, let block X be another resistor, Rx. Compare the value from typical normal derivation where I = Vdd / (Rin + Rx)), compare this result to the interception of two straight lines shown on the Figure 3 above (graphical approach), which can be seen as identical.
Load line when X block is a diode:
4. Now to make things more relevant in real life, let's assume that X block is a diode and super-impose the diode IV curve on the diagram. Although there is no way for voltage across diode to reach Vdd (assuming Vforward of ~0.7V, Vdd of 3.3V or more), but we can still extend the load line till Vdd, creating some sort of asymptotes.
Example usage of load line:
Now, given a diode curve (Figure 4), if we need to pick a Rin value that allowed to circuit to have a bias current of Ibias, we just need to draw a straight line (load line) from x-axis Vdd to the point of diode curve that has the y-axis value of Ibias, then the y-axis intercept current, Io can be used to calculate Rin, with Rin = Vdd/Io.