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

 

R increasing with increasing temperature $V=IR$  
Slope of graph $=\frac{I}{V}=\frac{1}{R};$ Slope of $T_1$ is more i.e. $\frac{1}{R_1}$ is more, hence $R_1$ is less. This concludes that $T_1$ will be less than $T_2$ as $R_1$ is less than $R_2.$