Question

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. $\bigcirc \frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

Answer: (D)

Equation of given pair of straight lines $S=3 x^{2}-2 k x y +y^{2}=0$ and the line
$L=2 x-y-6=0$
On solving, we get
$3 x^{2}-2 k x(2 x-6)+(2 x-6)^{2}=0$
$\Rightarrow (7-4 k) x^{2}+12(k-2) x+36=0$
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
$\frac{1}{2}\begin{vmatrix} 0 & 0 & 0 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1
\end{vmatrix}=36$
$\Rightarrow \left|\frac{y_{1}}{x_{1}}-\frac{y_{2}}{x_{2}}\right|=\frac{72}{\left|x_{2} x_{1}\right|}$

$\because m_{1}+m_{2}=2 k, m_{1} m_{2}=3$ 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, $\left |m_{1}-m_{2}\right|=\frac{72}{36}|7-4 k|$
$\Rightarrow \left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}=4(7-4 k)^{2}$
(on squaring both sides)
$\Rightarrow 4 k^{2}-12=4\left(49+16 k^{2}-56 k\right)$
$\Rightarrow 15 k^{2}-56 k+52=0$
$\Rightarrow 15 k^{2}-30 k-26 k+52=0$
$\Rightarrow 15 k(k-2)-26(k-2)=0$
$\Rightarrow k=2\,\, (\because k$ is an integer $)$
$\therefore \tan \theta=\frac{2 \sqrt{k^{2}-3}}{3+1}=\frac{1}{2}$
$\Rightarrow \sin \theta=\frac{1}{\sqrt{5}}$