The equation of the passing through the point of intersection of the line 2x + y -4 =0, x – 3y + 5 =0 and lying at a distance of $\sqrt{5}units$ from the origin, is

:  Given line 2x + y -4 =0—(1),   x – 3y + 5 =0—(2)

$\begin{array}{l}Solving\left(1\right)\&\left(2\right)\\ x y 1\\ \begin{array}{cccc}1& -4& 2& 1\\ -3& 5& 1& -3\end{array}\\ \Rightarrow \frac{x}{5-12}=\frac{y}{-4-10}=\frac{1}{-6-1}\\ \Rightarrow \frac{x}{-7}=\frac{y}{-14}=\frac{1}{-7}\\ x=\frac{-7}{-7}\Rightarrow 1,y=\frac{-14}{-7}\Rightarrow 2\end{array}$

(x,  y) = (1,  2) and  let m be the slope Equation of line

Y – 2 = m ( x – 1)—(3)

mx – y + 2 – m = 0

$\begin{array}{l}Given\perp rdis\tan cefrom\left(0,0\right)to\left(3\right)=\sqrt{5}\\ \Rightarrow \frac{\left|2-m\right|}{\sqrt{m^2+1}}=\sqrt{5}\\ \Rightarrow 4+m^2+4m+1=0\\ \Rightarrow {\left(2m+1\right)}^2=0\Rightarrow 2m+1=0\Rightarrow m=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ From\left(2\right)y-2=\frac{-1}{2}\left(x-1\right)\\ \Rightarrow 2y-4=-x+1\\ \Rightarrow x+2y-5=0\end{array}$