vertices of triangle are \$(3, 1), (0, 4)\$ and \$(–3, 1)\$.

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

Let AD and BE be the altitudes.

Slope of line joining points B \$(0, 4)\$ and \$(-3, 1)\$

\begin{align} \require{cancel} \text{Slpoe of BC } &= \frac{1 - 4}{-3 - 0} \\ &= \frac{-3}{-3} \\ &= 1 \\ \end{align}

AD is perpendicular to BC.
⇒ The slope of the altitude AD = –1.

If the slope of altitude AD = -1, then the equation of altitude AD is

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

Slope of line joining points A \$(3,1)\$ and C \$(-3 ,1)\$

\begin{align} \require{cancel} \text{Slpoe of AC } &= \frac{1 - 1}{-3 -3} \\ &= \frac{0}{6} \\ &= 0 \\ \end{align}

Therefore slope of the altitude BE \$= \frac{-1}{0} = \infty \$, since BE is perpendicular to AC.

Equation of BE is

\begin{align} \require{cancel} y – 4 &= \frac{-1}{0} (x – 0) \\ 0 &= -x + 0 \\ x &= 0 \tag{2} \\ \end{align}

Solving equation (1) and (2) \$x + y -4 = 0\$ and \$x = 0\$, we get

\begin{align} \require{cancel} y - 4 = 0 \\ y = 4 \\ \end{align}

Answer

The ortho centre is \$(0, 4)\$.