A junction diode is connected to a 10 V source and $10^{-3}$ resistor as shown in figure. The slope of load line on the characteristic curve of diode will be:

$10^{-3}AV^{-1}$

If V is the voltage across the junction and I is the circuit current, then

$V + IR = E$ or $I=\frac{E}{R}-\frac{V}{R}=-\frac{V}{R}+\frac{E}{R}$

Slope of load line = $-\frac{1}{R}=\frac{1}{1000}=10^{-3}AV^{-1}$