The general expression for the half-life of ${\mathrm{n}}^{\mathrm{th}}$ order reaction is as follows:

${\mathrm{t}}_{1/2}=\mathrm{k}\text{'}·{\mathrm{a}}^{1-\mathrm{n}}$

$\left(\text{Here,} \mathrm{k}\text{'} \text{is not rate constant}\right)$

Taking log on both sides,

$\log \left({\mathrm{t}}_{1/2}\right)=\log {\mathrm{k}}^{\text{'}}+(1-\mathrm{n}) \log \mathrm{a}$

The given graph is a straight line with slope $45°$, thus the equation of the line will be:

$\log \left({\mathrm{t}}_{1/2}\right)=\log \mathrm{a} + \mathrm{C}$

On comparing both the equations, $1-\mathrm{n}=1\Rightarrow \mathrm{n}=0$

Thus, the given graph represents a zero-order reaction.