Vertices of triangle are \$(1, 2), (2, 3)\$ and \$(4, 3)\$.

Let \$A(1,2), B(2, 3)\$ and \$C(4,3)\$.

Let AD and BE be the altitudes.

Slope of line joining points \$B(2, 3)\$ and \$C(4,3)\$ is

\begin{align} \require{cancel} \text{Slpoe of BC } &= \frac{3 - 3}{4 - 2} \\ &= \frac{0}{2} \\ &= 0 \\ \end{align}

Since AD is perpendicular to BC.

The slope of the altitude AD = \$ - \frac{1}{0} = \infty\$

Equation of altitude AD is

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

Slope of line joining points \$(1, 2)\$ and \$(4, 3)\$.

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

BE is perpendicular to AC.

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

Equation of BE is

\begin{align} \require{cancel} y – 3 &= -3(x – 2) \\ y – 3 &= -3x + 6 \\ 3x + y &= 9 \tag{2} \\ \end{align}

Solving equation (1) and (2) \$x = 1\$ and \$3x + y = 9\$,

substitute \$x = 1\$ in(2),we get

\begin{align} \require{cancel} 3(1) + y &= 9 \\ 3 + y &= 9 \\ y &= 6 \end{align}

Answer

The orthocentre is \$(1, 6)\$.