The answer just gives you one quadrant to graph the results, so it presumes that the voltmeter V and current meter A are connected to give positive readings.

The graph you give looks right:

- When there is zero current, the voltage measured is ε Volts

- When the variable resistor=0, the voltage will be zero, and the current will be at a maximum = ε/r Amps.

- The equation of the line is V=ε-rI

- But the gradient is "-r", not "r", giving a positive resistance

As a "sanity check", this looks right because, with realistic passive components:

- Resistors are ≥ 0 Ohms

- The battery output power is finite:

- If you put too much load on it, the power output (VxA) drops to zero because V=0

- If you don't put any load on it, the power output (VxA) is zero because I=0

- There is some maximum output power in the middle, where the variable resistor is the same as r

- The idea of having a circuit where the slope of the line is positive means that as you increase the load, the output voltage also increases

- So if you reduced the load to zero, the battery output power would be infinite!

- This is quite unlike ordinary batteries!

PS: This assumes "ideal" measuring instruments: A Voltmeter with infinite impedance, and an Ammeter with zero impedance.