The equation of line passing through the points $(3,0)$ and $\left(1, \frac{4}{3}\right)$ is
$$
\begin{array}{rlrl}
y-0 & =\frac{\frac{4}{3}-0}{1-3}(x-3) \\
\Rightarrow & y =\frac{-2}{3}(x-3) \\
\Rightarrow & 3 y =-2 x+6 \\
\Rightarrow & 2 x+3 y =6
\end{array}
$$
Any line perpendicular to the line Eq. (i) and passing through $(8,1)$ is
$$
\begin{aligned}
y-1 & =\frac{-1}{\left(\frac{-2}{3}\right)}(x-8) \\
\Rightarrow \quad y-1 & =\frac{3}{2}(x-8) \Rightarrow 2 y-2=3 x-24 \\
\Rightarrow \quad 3 x-2 y & =22
\end{aligned}
$$

Intersecting point of line Eqs. (i) and (ii) is $(6,-2)$. The line $2 x+3 y=6$ intersect $Y$-axis at the point $(0,2)$ and line $3 x-2 y=22$ intersect the $Y$-axis at the point $(0,-11)$.
$\therefore$ The area of required triangle
$$
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
6 & -2 & 1 \\
0 & 2 & 1 \\
0 & -11 & 1
\end{array}\right| \\
& =\frac{1}{2} \cdot 6(2+11)=39 \mathrm{sq} \text { unit }
\end{aligned}
$$