Question

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

• A. $\frac{1}{2}$
• B. $◯\frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

Answer: (D)

Equation of given pair of straight lines $S=3x^2-2kxy+y^2=0$ and the line
$L=2x-y-6=0$
On solving, we get
$3x^2-2kx(2x-6)+(2x-6{)}^2=0$
$\Rightarrow (7-4k)x^2+12(k-2)x+36=0$
Let the intersection points are $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$
$\therefore$ The area of $△ABC$ is
$\frac{1}{2}\left|\begin{array}{ccc}0& 0& 0\\ x_1& y_1& 1\\ x_2& y_2& 1<br/>\end{array}\right|=36$
$\Rightarrow \left|\frac{y_1}{x_1}-\frac{y_2}{x_2}\right|=\frac{72}{|x_2{x}_1|}$

$\because m_1+m_2=2k,m_1{m}_2=3$ and $\frac{y_1}{x_1}=m_1$
$\frac{y_2}{x_2}=m_2$ and $\left|x_1{x}_2\right|=\frac{36}{|7-4k|}$
So, $\left|m_1-m_2\right|=\frac{72}{36}|7-4k|$
$\Rightarrow {\left(m_1+m_2\right)}^2-4m_1{m}_2=4(7-4k{)}^2$
(on squaring both sides)
$\Rightarrow 4k^2-12=4\left(49+16k^2-56k\right)$
$\Rightarrow 15k^2-56k+52=0$
$\Rightarrow 15k^2-30k-26k+52=0$
$\Rightarrow 15k(k-2)-26(k-2)=0$
$\Rightarrow k=2(\because k$ is an integer $)$
$\therefore \tan \theta =\frac{2\sqrt{k^2-3}}{3+1}=\frac{1}{2}$
$\Rightarrow \sin \theta =\frac{1}{\sqrt{5}}$