What does it mean for a transistor to be biased beyond cutoff?
First, remember that the load line drawing solves a particular set of equations. Where the lines cross gives the the operating point for that combination of power supply, load resistor, and transistor base current. 

Second, it's correct that there is no characteristic curve for the BJT that goes through the region you circled. The reason is conservation of energy. If the BJT operated in that region, it would mean that the BJT was delivering energy to the circuit, rather than taking energy provided by the power supply and turning it into heat. Since a BJT doesn't contain a reserve of energy that can be released in steady-state conditions, it simply can't operate in that region.

There is, however, probably a small region right near the origin where the transistor characteristic curves do pass through quadrant IV of the graph. Consider this circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

This is essentially taking the resistor/power-supply load line and moving it down for the case where Vcc goes to 0. In this case, the base-collector junction will be forward biased and some power from the base bias supply will be delivered to the load resistor. And the load line will give a solution in quadrant IV, but very close to the origin.

If we were talking about a MOSFET instead of a BJT, even this solution would not be possible, since there's no way for current to transfer from the gate side to the drain side of the FET.