The joint equation of two lines through the origin, each of which making an angle of
\(30^{\circ}\) with line \(x+y=0\) is

Given equation of line is \(x+y=0\), having slope \(=-1\) \(\ldots(1)\)
Required line makes angle of \(30^{\circ}\) with given line (1)
\(\tan 30^{\circ}=\left|\frac{m+1}{1-m}\right| \Rightarrow \frac{1}{\sqrt{3}}=\left|\frac{m+1}{1-m}\right|\)
On squaring both sides, we get
\(\frac{1}{3}=\frac{(m+1)^{2}}{(1-m)^{2}} \)
\( (1-m)^{2}=3(m+1)^{2} \Rightarrow 1-2 m+m^{2}=3(m^{2}+\)\(2 m+1) \)
\( 3 m^{2}+6 m+3-1+2 m-m^{2}=0 \Rightarrow 2 m^{2}+8 m+2=0 \)
\( m^{2}+4 m+1=0\)
\(\begin{aligned}
\text { Put } \mathrm{m}=& \frac{\mathrm{y}}{\mathrm{x}} \\
\therefore & \frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}+\frac{4 \mathrm{y}}{\mathrm{x}}+1=0 \Rightarrow \mathrm{y}^{2}+4 \mathrm{xy}+\mathrm{x}^{2}=0
\end{aligned}\)