The joint equation of a pair of lines passing through the origin and making an angle of $\frac{\pi}{4}$ with…
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
$5 x^2+24 x y-5 y^2=0$
$5 x^2-24 x y+5 y^2=0$
$5 x^2-24 x y-5 y^2=0$
$5 x^2+24 x y+5 y^2=0$
Solution
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
$\begin{gathered}
5\left(\frac{y}{x}\right)^2-24\left(\frac{y}{x}\right)-5=0 \\
\text { i.e., } 5 x^2+24 x y-5 y^2=0
\end{gathered}$