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The joint equation of two lines through the origin, each of which making an angle of
$30^{\circ}$ with line $x+y=0$ is
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$30^{\circ}$ with line $x+y=0$ is
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Verified Answer
The correct answer is:
$x^{2}+4 x y+y^{2}=0$
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
$\begin{array}{l}
\frac{1}{3}=\frac{(m+1)^{2}}{(1-m)^{2}} \\
(1-m)^{2}=3(m+1)^{2} \Rightarrow 1-2 m+m^{2}=3\left(m^{2}+2 m+1\right) \\
3 m^{2}+6 m+3-1+2 m-m^{2}=0 \Rightarrow 2 m^{2}+8 m+2=0 \\
m^{2}+4 m+1=0
\end{array}$
$\begin{aligned}
\text { Put } \mathrm{m}=& \frac{\mathrm{y}}{\mathrm{x}} \\
\therefore & \frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}+\frac{4 \mathrm{y}}{\mathrm{x}}+1=0 \Rightarrow \mathrm{y}^{2}+4 \mathrm{xy}+\mathrm{x}^{2}=0
\end{aligned}$
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
$\begin{array}{l}
\frac{1}{3}=\frac{(m+1)^{2}}{(1-m)^{2}} \\
(1-m)^{2}=3(m+1)^{2} \Rightarrow 1-2 m+m^{2}=3\left(m^{2}+2 m+1\right) \\
3 m^{2}+6 m+3-1+2 m-m^{2}=0 \Rightarrow 2 m^{2}+8 m+2=0 \\
m^{2}+4 m+1=0
\end{array}$
$\begin{aligned}
\text { Put } \mathrm{m}=& \frac{\mathrm{y}}{\mathrm{x}} \\
\therefore & \frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}+\frac{4 \mathrm{y}}{\mathrm{x}}+1=0 \Rightarrow \mathrm{y}^{2}+4 \mathrm{xy}+\mathrm{x}^{2}=0
\end{aligned}$
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