Let the coordinates are $\mathrm{A}(1,3), \mathrm{B}(-3,5), \mathrm{C}(5,-1)$
Slope of $\mathrm{AC}=\frac{-1-3}{5-1}=\frac{-4}{4}=-1=m_1$
Slope of $\mathrm{BC}=\frac{-1-5}{5+3}=\frac{-6}{8}=\frac{-3}{4}=m_2$
Slope $\mathrm{BE}=\left(m^{\prime}\right)$ is $m_1 m^{\prime}=-1$
$m^{\prime}=\frac{-1}{m_1}=\frac{-1}{-1}=1$
Slope of AD $\left(m^{\prime \prime}\right)$ is $m_2 m^{\prime \prime}=-1$
$$
m^{\prime \prime}=\frac{-1}{m_2}=\frac{-1}{\frac{-3}{4}}=\frac{4}{3}
$$
Equation of line BE is $(y-5)=m^{\prime}(x-(-3))$
$$
\begin{aligned}
& \Rightarrow(y-5)=1(x+3) \\
& \Rightarrow(y-5)=x+3 \\
& \Rightarrow x-y=-8
\end{aligned}
$$
Equation of line $\mathrm{AD}$ is $(y-3)=\frac{4}{3}(x-1)$
$$
\begin{aligned}
& \Rightarrow 3 y-9=4 x-4 \\
& \Rightarrow 4 x-3 y=-5
\end{aligned}
$$
Multiply eq. (i) by 3 .
$$
3 x-3 y=-24
$$
Subtract eq. (ii) from (iii)
$$
\begin{gathered}
3 x-3 y=-24 \\
4 x-3 y=-5 \\
(-)(+) \quad(+) \\
\hline-x=-19 \\
x=19
\end{gathered}
$$
From (i),
$$
\begin{aligned}
& x-y=-8 \\
& 19+8=y \\
& y=27
\end{aligned}
$$
Therefore, orthocentre $\mathrm{H}$ is $(19,27)$.