The circumcentre of the triangle formed by $(0, 0), (2, -1)$ and $(-1, 3)$ is $(5/2,5/2).$ Then the orthocentre is

Let A=(0, 0), B=(2, -1), C=(-1, 3) are vertices of ${∆}^{\mathrm{le}} \mathrm{ABC}$

Given Circum centre $\left(\mathrm{S}\right)=\left(\frac{5}{2}, \frac{5}{2}\right)$

Centriod $\left(\mathrm{G}\right)=\left(\frac{1}{3}, \frac{2}{3}\right)$

Since ‘G’ divides $\overline{HS}$  in the ratio 2:1 then 3G=2S+H

$$\begin{gathered} \Rightarrow \mathrm{H}=3\mathrm{G}-2\mathrm{S} \\ =3\left(\frac{1}{3}, \frac{2}{3}\right)-2\left(\frac{5}{2}, \frac{5}{2}\right) \\ =(1-5, 2-5) \\ =(-4, -3) \end{gathered}$$