Given that, $x+3 y-9=0, \quad 4 x+b y-2=0$ $2 x-y-4=0$ are concurrent.
$\begin{aligned}
& \therefore \quad\left|\begin{array}{ccc}
1 & 3 & -9 \\
4 & b & -2 \\
2 & -1 & -4
\end{array}\right|=0 \\
& 1(-4 b-2)-3(-16+4)-9(-4-2 b)=0 \\
& -4 b-2+36+36+18 b=0 \\
& 14 b=-70 \quad \Rightarrow b=-5
\end{aligned}$
From equations $x+3 y-9=0$ and $2 x-y-4=0$
$\Rightarrow \quad x=3 \text { and } y=2$
$\therefore \quad$ Concurrency point is $(3,2)$
Equation of line passing through $(-5,0)$ and $(3,2)$ is
$\begin{aligned}
& y-0=\frac{2-0}{3+5}(x+5) \\
& y=\frac{1}{4}(x+5) \quad \Rightarrow 4 y=x+5 \\
& \Rightarrow x-4 y+5=0
\end{aligned}$