The equation of the line passing through the point of intersection of the lines \(3 x-y=5\) and \(x+3 y=1\) and making equal intercepts on the axes is

Required equation passes through point of intersection of lines \(3 x-y=5\) and \(x+3 y=1\)
\(\therefore \quad\) Point of intersection \(=\left(\frac{16}{10}, \frac{-2}{10}\right)\)
Equation of line in double intercept form is \(\frac{x}{a}+\frac{y}{b}=1\)
But \(a=b\)
...[Given]
So, equation of line is \(x+y=\mathrm{a}\)
Since line passes through \(\left(\frac{16}{10}, \frac{-2}{10}\right)\)
\(\begin{aligned}
\therefore \quad & \frac{16}{10}+\frac{-2}{10}=\mathrm{a} \\
& \Rightarrow \frac{14}{10}=\mathrm{a} \\
& \Rightarrow \mathrm{a}=\frac{7}{5}
\end{aligned}\)
\(\therefore \quad\) The required equation of line is \(x+y=\frac{7}{5}\)
\(\Rightarrow 5 x+5 y-7=0\)