For an integer k,if the area of the triangle | vedclass

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(D) The pair of lines is given by $S = 3x^2 - 2kxy + y^2 = 0$. The line is $L = 2x - y - 6 = 0$,which implies $y = 2x - 6$.Substituting $y$ in $S=0$,w

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

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(D) The pair of lines is given by $S = 3x^2 - 2kxy + y^2 = 0$. The line is $L = 2x - y - 6 = 0$,which implies $y = 2x - 6$.Substituting $y$ in $S=0$,we get $3x^2 - 2kx(2x - 6) + (2x - 6)^2 = 0$.This simplifies to $(3 - 4k + 4)x^2 + (12k - 24)x + 36 = 0$,or $(7 - 4k)x^2 + 12(k - 2)x + 36 = 0$.Let the intersection points be $A(x_1, y_1)$ and $B(x_2, y_2)$. The origin $C$ is $(0, 0)$.The area of $	riangle ABC = \frac{1}{2} |x_1y_2 - x_2y_1| = 36$,which implies $|x_1x_2| |m_1 - m_2| = 72$.From the quadratic equation,$x_1x_2 = \frac{36}{7 - 4k}$.Also,for $S = ax^2 + 2hxy + by^2 = 0$,$|m_1 - m_2| = \frac{2\sqrt{h^2 - ab}}{|b|} = \frac{2\sqrt{k^2 - 3}}{1} = 2\sqrt{k^2 - 3}$.Substituting these into the area formula: $|\frac{36}{7 - 4k}| \cdot 2\sqrt{k^2 - 3} = 72 \Rightarrow |\sqrt{k^2 - 3}| = |7 - 4k|$.Squaring both sides: $k^2 - 3 = (7 - 4k)^2 = 49 + 16k^2 - 56k$.$15k^2 - 56k + 52 = 0 \Rightarrow (k - 2)(15k - 26) = 0$.Since $k$ is an integer,$k = 2$.For $k = 2$,$	an 	heta = \frac{2\sqrt{k^2 - 3}}{a + b} = \frac{2\sqrt{4 - 3}}{3 + 1} = \frac{2}{4} = \frac{1}{2}$.Since $	an 	heta = \frac{1}{2}$,$\sin 	heta = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}}$.

For an 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 $36$ sq. units,then for the angle $\theta$ between the lines $S = 0$,$\sin \theta =$

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