The joint equation of pair of lines through the origin and making an angle of $\frac{\pi}{6}$ with the line…

The joint equation of pair of lines through the origin and making an angle of $\frac{\pi}{6}$ with the line $3 x+y-6=0$ is
  1. $13 x^2+12 x y+3 y^2=0$
  2. $13 x^2-12 x y+3 y^2=0$
  3. $13 x^2+12 x y-3 y^2=0$
  4. $13 x^2-12 x y-3 y^2=0$

Solution

The slope of the line $3 x+y-6=0$ is $m_1=-3$. Let $m$ be the slope of one of the lines making an angle $\frac{\pi}{6}$ with $3 x+y-6=0$. $\begin{aligned} & \therefore \quad \tan \frac{\pi}{6}=\left|\frac{\mathrm{m}-\mathrm{m}_1}{1+\mathrm{m}_1}\right| \\ & \quad \Rightarrow \frac{1}{\sqrt{3}}=\left|\frac{\mathrm{m}-(-3)}{1+\mathrm{m}(-3)}\right| \\ & \quad \Rightarrow \frac{1}{\sqrt{3}}=\left|\frac{\mathrm{m}+3}{1-3 \mathrm{~m}}\right| \end{aligned}$
Squaring on both sides, we get $\begin{aligned} & (1-3 m)^2=3(m+3)^2 \\ & \Rightarrow 6 m^2-24 m-26=0 \\ & \Rightarrow 3 m^2-12 m-13=0 \end{aligned}$
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 $\begin{aligned} & 3\left(\frac{y}{x}\right)^2-12\left(\frac{y}{x}\right)-13=0 \\ & \Rightarrow 13 x^2+12 x y-3 y^2=0 \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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