The first diagram shows the charactaristics of a common emitter amplifer.
Values of Ic have been plotted against values of Vc.
This has been done for four different values of Ib.

The load line  is drawn between two points.
(a) Where the transistor is cut off. There is no Ic. There is no voltage drop across the load resistor and Vc = Vcc. (point A)

(b) When the transistor is fully saturated. Vc = 0.       Ic = Vcc/The load resistor. 

In the diagram, the load line has been constructed for three values of load resistor. (point B for a 3.3k load, point C for a 2.2k load and point D for a 4.9k load)
Let's use the centre one with a 3.3k load resistor.

Q is the quiescent ( no input signal) operating point.

It can be seen that Ib = 40uA, Ic = 1.5 mA and Vc = 5 volts

In the second diagram we are looking at the dynamic conditions (with an input signal).

The base current swings up to 60 uA and down to 20 uA.

This causes Vc to swing down to 2.6V and up to 6.6V.
At the same time Ic swings between 2.2 mA and 1mA.

The current gain is change in Ic/ change in Ib =2.2 mA -1mA/60uA-20uA
              =1.2 mA/ 40uA  = a gain of 30 times.

If an incorrect load line is selected, then Vc will swing one way more than the other, and distortion of the signal will result.

Try the effect of changing Vcc or the load resistor.