Intergrated rate law for zero order reactions is $\left[{\mathrm{C}}_{\mathrm{t}}\right]=\left[{\mathrm{C}}_{\mathrm{o}}\right]-\mathrm{k}\mathrm{t}$ or $\left[{\mathrm{C}}_{\mathrm{t}}\right]=-\mathrm{k}\mathrm{t}+\left[{\mathrm{C}}_{\mathrm{o}}\right]$ or $\left[\mathrm{R}\right]=-\mathrm{k}\mathrm{t}+\left[{\mathrm{C}}_{\mathrm{o}}\right]$ ( ${\mathrm{C}}_{\mathrm{t}}=\mathrm{R}$ given)
On comparing with $\mathrm{y}=\text{mx}+\mathrm{C}$
slope $\mathrm{m}=-\mathrm{k}$ for graph between (R) and t with intercept ${\mathrm{C}}_{\mathrm{o}}$
first graph is of zero order reactions.
Integrated rate law for first order reaction is
$\mathcal{l}\mathrm{n}\left[\mathrm{R}\right]=-\mathrm{k}\mathrm{t}+\mathcal{l}\mathrm{n}\left({\mathrm{C}}_{\mathrm{o}}\right)$
On comparing with $\mathrm{y}=\text{mx}+\mathrm{C}$
Slope $\mathrm{m}=-\mathrm{k}$ for graph between 1n (R) and t with intercept 1n $\left({\mathrm{C}}_{\mathrm{o}}\right)$
Second graph is of first order