If the slope of one of the lines represented | vedclass

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(A) The equation of the pair of lines is $2x^2 + 3xy + ky^2 = 0$. Dividing by $x^2$,we get $k(\frac{y}{x})^2 + 3(\frac{y}{x}) + 2 = 0$. Let $m = \frac

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$\frac{\pi}{2}$

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(A) The equation of the pair of lines is $2x^2 + 3xy + ky^2 = 0$. Dividing by $x^2$,we get $k(\frac{y}{x})^2 + 3(\frac{y}{x}) + 2 = 0$. Let $m = \frac{y}{x}$ be the slope of the lines. Then $km^2 + 3m + 2 = 0$. Given that one slope $m_1 = 2$,substituting this into the equation: $k(2)^2 + 3(2) + 2 = 0 \implies 4k + 6 + 2 = 0 \implies 4k = -8 \implies k = -2$. The equation becomes $2x^2 + 3xy - 2y^2 = 0$. Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a = 2$,$2h = 3 \implies h = \frac{3}{2}$,and $b = -2$. The angle $	heta$ between the lines is given by $	an 	heta = |\frac{2\sqrt{h^2 - ab}}{a + b}|$. Here $a + b = 2 + (-2) = 0$. Since the sum of coefficients of $x^2$ and $y^2$ is zero,the lines are perpendicular. Therefore,$	heta = \frac{\pi}{2}$.

If the slope of one of the lines represented by $2x^2 + 3xy + ky^2 = 0$ is $2$,then the angle between the pair of lines is

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