The joint equation of the pair of lines passi | vedclass

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Question

(A) Let the slope of the required lines be $m$. The slope of the given line $3x + y - 6 = 0$ is $m_1 = -3$.The angle between the lines is $	heta = \f

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$2x^2 + 3xy - 2y^2 = 0$

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(A) Let the slope of the required lines be $m$. The slope of the given line $3x + y - 6 = 0$ is $m_1 = -3$.The angle between the lines is $	heta = \frac{\pi}{4}$. Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m m_1} ight|$, we get:$	an \frac{\pi}{4} = \left| \frac{m - (-3)}{1 + m(-3)} ight| \implies 1 = \left| \frac{m + 3}{1 - 3m} ight|$.This gives two cases: $1 = \frac{m + 3}{1 - 3m}$ or $1 = -\frac{m + 3}{1 - 3m}$.Case $1$: $1 - 3m = m + 3 \implies 4m = -2 \implies m = -\frac{1}{2}$.Case $2$: $1 - 3m = -(m + 3) \implies 1 - 3m = -m - 3 \implies 2m = 4 \implies m = 2$.The equations of the lines are $y = -\frac{1}{2}x$ and $y = 2x$, which are $x + 2y = 0$ and $2x - y = 0$.The joint equation is $(x + 2y)(2x - y) = 0$.$2x^2 - xy + 4xy - 2y^2 = 0 \implies 2x^2 + 3xy - 2y^2 = 0$.

The joint equation of the pair of lines passing through the origin and making an angle of $\frac{\pi}{4}$ with the line $3x + y - 6 = 0$ is

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