If a straight line is passing through the point of intersection of the lines 3x – 4y + 1 = 0, 5x + y – 1 = 0 and making equal non – zero intercepts on the coordinate axes, then the area (in sq. units) of the triangle formed by this line with the coordinate axes, is

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

Solving  (1) & (2)

$\begin{array}{l}x y 1\\ \begin{array}{cccc}-4& 1& 3& -4\\ 1& -1& 5& 1\end{array}\\ \Rightarrow \frac{x}{4-1}=\frac{y}{5+3}=\frac{1}{3+20}\\ \Rightarrow \frac{x}{3}=\frac{y}{8}=\frac{1}{23}\\ \therefore \left(x,y\right)=\left(\frac{3}{23},\frac{8}{23}\right)\end{array}$

$\begin{array}{l}Letequationoflineis\frac{x}{a}+\frac{y}{b}=1\\ Givena=b\\ \Rightarrow \frac{x}{a}+\frac{y}{b}=1\\ \Rightarrow x+y=a--\left(3\right)\\ Ifpassesthrough\left(\frac{3}{23},\frac{8}{23}\right)\\ \Rightarrow \frac{3}{23}+\frac{8}{23}=a\Rightarrow a=\frac{11}{23}\\ From\left(3\right)x+y-\frac{11}{23}=0\\ Areawithcoordinateaxis=\frac{c^2}{2\left|ab\right|}=\frac{{\left(\frac{11}{23}\right)}^2}{2\left|1\times 1\right|}\Rightarrow \frac{\text{121}}{2\times 529}=\frac{121}{1058}\end{array}$