Vertices of triangle are \$(0,1), (2,-1)\$ and \$(-1,3)\$.

Let \$A(0,1), B(2, -1)\$ and \$C(-1,3)\$.

Let AD and BE be the altitudes.

Slope of line joining points \$B(2, -1)\$ and \$C(-1, 3)\$ is

\begin{align} \require{cancel} \text{Slpoe of BC } &= \frac{3 + 1}{-1 - 2} \\ &= \frac{4}{-3} \\ &= - \frac{4}{3} \\ \end{align}

Since AD is perpendicular to BC.

The slope of the altitude AD = \$ \frac{3}{4} \$

Equation of altitude AD is

\begin{align} \require{cancel} && y - 1 &= \frac{3}{4}(x - 0) \\ \Rightarrow && 4(y - 1) &= 3(x - 0) \\ \Rightarrow && 4y - 4 &= 3x \\ \Rightarrow && 3x - 4y &= -4 \tag{1} \\ \end{align}

Slope of line joining points \$(-1, 3)\$ and \$(0, 1)\$.

\begin{align} \require{cancel} \text{Slpoe of AC } &= \frac{3 - 1}{-1 - 0} \\ &= \frac{2}{-1} \\ &= -2 \\ \end{align}

BE is perpendicular to AC.

The slope of the altitude BE \$= \frac{1}{2} \$

Equation of BE is

\begin{align} \require{cancel} y + 1 &= \frac{1}{2}(x – 2) \\ 2y + 2 &= x - 2 \\ x - 2y &= 4 \tag{2} \\ \end{align}

Solving equation (1) and (2) \$ 3x - 4y = -4 \$ and \$ x - 2y = 4 \$,

(1) \$\times\$ 1
(2) \$\times\$ 3

\begin{align} 3x – 4y = -4& \\ \underline{+ 3x + 6y = 12}& \\ 2y = - 16& \\ \\ y = -\frac{16}{2} \\ y = - 8 \\ \end{align}

Substitute \$y = -8\$ in (2).

\begin{align} \require{cancel} x - 2(-8) &= 4 \\ x + 16 &= 4 \\ x &= 4 - 16 \\ x &= - 12 \\ \end{align}

Answer

The orthocentre is \$(-12, -8)\$.