A straight line $\frac{x}{a}-\frac{y}{b}=1$ passes thought $(8,6)$.
Then, $\frac{8}{a}-\frac{6}{b}=1...(i)$
Area of triangle formed by axis and line $=12 \mathrm{sq}$ unit



$$
\begin{aligned}
& \text { Area of triangle }=\frac{1}{2} \times \text { base } \times \text { height } \\
& \qquad \begin{aligned}
12 & =\frac{1}{2}|(a) \times(-b)| \Rightarrow 24=|-a b| \\
|a b|=24, a b & = \pm 24, b=\frac{ \pm 24}{a}
\end{aligned}
\end{aligned}
$$
Now, in Eq. (i) put $b=24 / a$
$\begin{array}{rlrl} & & \frac{8}{a}-\frac{6}{b} & =1 \\ \Rightarrow & & \frac{8}{a}-\frac{6}{\frac{24}{a}} & =1 \Rightarrow \frac{8}{a}-\frac{a}{4}=1 \\ \Rightarrow & & \frac{32-a^2}{4 a} & =1 \Rightarrow 32-a^2=4 a \\ \Rightarrow & a^2+4 a-32 & =0 \Rightarrow(a+8)(a-4)=0 \\ & a=-8 \mid a & =4 \\ & b=-3 \mid b & =6\end{array}$
$\therefore$ Equation of lines are
\begin{array}{c|l}
\frac{x}{-8}-\frac{y}{-3}=1 & \frac{x}{4}-\frac{y}{6}=1 \\
-3 x+8 y=24 & \frac{3 x-2 y}{12}=1 \\
3 x-8 y+24=0 & \begin{array}{l}3 x-2 y=12 \\
3 x-2 y-12=0\end{array}
\end{array}