Straight line $x+3 y-4=0, x+y-4=0$ and $3 x+y-4=0$


$$
\begin{array}{ll}
\text { So, } \quad & x+3 y-4=0 \\
& 3 x+y-4=0 \\
\Rightarrow \quad & 3 x+9 y-12=0
\end{array}
$$
$\begin{aligned} & \begin{array}{l}3 x+y-4=0 \\ -\quad-+\end{array} \\ & 8 y=8 \\ & y=1 \text { Now, } x=1\end{aligned}$


(ii) Now, for $B$ point
$$
\begin{aligned}
& x+3 y-4=0 \\
& x+y-4=0 \\
& \frac{-\quad+}{2 y=0} \\
& \Rightarrow \quad y=0, x=4 \\
&
\end{aligned}
$$
(iii) for $C$ point
$$
\begin{gathered}
3 x+y-4=0 \\
x+y-4=0 \\
-\quad-\quad+ \\
\hline 2 x=0 \\
x=0 \Rightarrow y=4
\end{gathered}
$$


Now,
$$
\begin{aligned}
& A B=\sqrt{9+1}=\sqrt{10} \\
& B C=\sqrt{32}=4 \sqrt{2} \\
& C A=\sqrt{10} \\
& \because \quad A B=C A=\sqrt{10} \\
&
\end{aligned}
$$
$\therefore A B C$ an isosceles triangle.