The joint equation of a pair of lines passing through the origin and making an angle of \(\frac{\pi}{4}\) with the line \(3 x+2 y-8=0\) is

The slope of line \(3 x+2 y-8=0\) is
\(\mathrm{m}_1=\frac{-3}{2}\)
Let \(\mathrm{m}\) be the slope of one of the lines making an angle \(\frac{\pi}{4}\) with \(3 x+2 y-8=0\)
\(\begin{aligned}
\therefore \quad \tan \frac{\pi}{4} & =\left|\frac{\mathrm{m}-\mathrm{m}_1}{1+\mathrm{mm}_1}\right| \\
\Rightarrow 1 & =\left|\frac{\mathrm{m}-\left(\frac{-3}{2}\right)}{1+\mathrm{m}\left(\frac{-3}{2}\right)}\right| \\
\Rightarrow 1 & =\left|\frac{2 \mathrm{~m}+3}{2-3 \mathrm{~m}}\right|
\end{aligned}\)
Squaring on both sides, we get
\(\begin{aligned}
& (2-3 m)^2=(2 m+3)^2 \\
& \Rightarrow 5 m^2-24 m-5=0
\end{aligned}\)
This is the auxiliary equation of two lines and their joint equation is obtained by putting
\(\mathrm{m}=\frac{y}{x}\)
\(\therefore \quad\) The joint equation of the lines is
\(\begin{gathered}
5\left(\frac{y}{x}\right)^2-24\left(\frac{y}{x}\right)-5=0 \\
\text { i.e., } 5 x^2+24 x y-5 y^2=0
\end{gathered}\)