For integer $k$, if the area of the triangle formed by the pair of lines $S=3 x^2-2 k x y+y^2=0$ with the line $L=2 x-y-6=0$ is 36 sq. units, then for the angle $\theta$ between the lines $S=0, \sin \theta=$

• A

  $\frac{1}{2}$
• B

  $\frac{\sqrt{3}}{2}$
• C

  $\frac{1}{\sqrt{3}}$
• Correct answer
  D

  $\frac{1}{\sqrt{5}}$

Solution

Equation of given pair of straight lines $S=3 x^2-2 k x y+y^2=0$ and the line $L=2 x-y-6=0$ On solving, we get $3 x^2-2 k x(2 x-6)+(2 x-6)^2=0$ $\Rightarrow \quad(7-4 k) x^2+12(k-2) x+36=0$ Let the intersection points are $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ $\therefore$ The area of $\triangle A B C$ is $\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 0 \\x_1 & y_1 & 1 \\x_2 & y_2 & 1\end{array}\right||=36 \Rightarrow| \frac{y_1}{x_1}-\frac{y_2}{x_2} \mid=\frac{72}{\left|x_2 x_1\right|}$
$\because \quad m_1+m_2=2 k, m_1 m_2=3$ and $\frac{y_1}{x_1}=m_1$ $\frac{y_2}{x_2}=m_2$ and $\left|x_1 x_2\right|=\frac{36}{|7-4 k|}$ So, $\quad\left|m_1-m_2\right|=\frac{72}{36}|7-4 k|$ $\Rightarrow \quad\left(m_1+m_2\right)^2-4 m_1 m_2=4(7-4 k)^2$ (on squaring both sides) $\Rightarrow \quad 4 k^2-12=4\left(49+16 k^2-56 k\right)$ $\Rightarrow 15 k^2-56 k+52=0 \Rightarrow 15 k^2-30 k-26 k+52=0$ $\Rightarrow 15 k(k-2)-26(k-2)=0$ $\Rightarrow \quad k=2 \quad(\because k$ is an integer $)$ $\therefore \quad \tan \theta=\frac{2 \sqrt{k^2-3}}{3+1}=\frac{1}{2} \Rightarrow \sin \theta=\frac{1}{\sqrt{5}}$

Hence, the correct answer is option: 'D'.