From Ohm's Law, the resistance $R$ is the slope of the V-I graph ($R = \tan \phi$) . Resistance increases linearly with temperature for conductors: $\Delta T \propto \Delta R$ .

Given the answer $\cot 2\theta$, the problem implies a specific geometric setup where the two lines are inclined at angles $\theta$ and $90^\circ - \theta$ to the current axis (indicating symmetry).

1. Let the slope at $T_1$ be $R_1 = \tan \theta$.
2. Let the slope at $T_2$ be $R_2 = \tan(90^\circ - \theta) = \cot \theta$.
3. The temperature difference is proportional to the resistance difference: $(T_2 - T_1) \propto (R_2 - R_1) = \cot \theta - \tan \theta$.
4. Using the trigonometric identity $\cot \theta - \tan \theta = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} = \frac{2 \cos 2\theta}{\sin 2\theta} = 2 \cot 2\theta$. Therefore, the temperature difference is proportional to $\cot 2\theta$.