The current I vs voltage V graphs for a given metallic wire at two different temperatures $T_1$ and $T_2$ are shown in the figure. It is concluded that

$$\begin{gathered} \text{ R increases with increasing temp: } \\ \mathrm{V}=\mathrm{IR} \\ \text{ Slope of graph }=\frac{\mathrm{I}}{\mathrm{V}}=\frac{1}{\mathrm{R}};\text{ Slope of }{\mathrm{T}}_1\text{ is more i.e }\frac{1}{{\mathrm{R}}_1}\text{ is more, hence }{\mathrm{R}}_1 \\ \text{ is less. This concludes that }{\mathrm{T}}_1\text{ will be less than }{\mathrm{T}}_2\text{ as }{\mathrm{R}}_1\text{ is less than }{\mathrm{R}}_2\text{ . } \end{gathered}$$