The equation of the line passing through the point of intersection of the lines $3 x-y=5$ and $x+3 y=1$ and…
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
$5 x+5 y-7=0$
$5 x-5 y-7=0$
$2 x+y-7=0$
$x-y+7=0$
Solution
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 $x+y=\frac{7}{5}$
$\Rightarrow 5 x+5 y-7=0$