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

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

Let AD and BE be the altitudes.

Slope of line joining points \$B(2, -1)\$ and \$C(0, 0)\$ is

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

Since AD is perpendicular to BC.

The slope of the altitude AD = \$ 2\$.

Equation of altitude AD is

\begin{align} \require{cancel} && y - 3 &= 2(x + 1) \\ \Rightarrow && y - 3 &= 2x + 4 \\ \Rightarrow && 2x - y &= -5 \tag{1}\\ \end{align}

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

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

BE is perpendicular to AC.

The slope of the altitude BE \$= \frac{1}{3} \$.

Equation of BE is

\begin{align} \require{cancel} y + 1 &= \frac{1}{3}(x – 2) \\ 3y – 3 &= x - 2 \\ x - 3y &= 5 \tag{2} \\ \end{align}

Solving equation (1) and (2) \$2x - y = -5 \$ and \$x - 3y = 5 \$,

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

\begin{align} 2x – y = -5& \\ \underline{+ 2x – 6y = 10}& \\ 5y = - 15& \\ \\ y = -\frac{15}{5} \\ y = - 3 \\ \end{align}

Substitute \$y = -3\$ in (1).

\begin{align} \require{cancel} 2x - (-3) &= -5 \\ 2x + 3 &= -5 \\ 2x &= - 8 \\ x &= - 4 \end{align}

Answer

The orthocentre is \$(-4, -3)\$.