The joint equation of the pair of lines passi | vedclass

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(B) Let $L_{1}$ and $L_{2}$ be the required lines passing through the origin $(0,0)$.Since the triangle formed by the lines and the line $y=4$ is equi

Vedclass · Accepted Answer

$3x^{2}-y^{2}=0$

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(B) Let $L_{1}$ and $L_{2}$ be the required lines passing through the origin $(0,0)$.Since the triangle formed by the lines and the line $y=4$ is equilateral,the angle made by each line with the positive $x$-axis can be determined.The line $y=4$ is horizontal. The lines pass through the origin and make an equilateral triangle with $y=4$,meaning the angle at the origin is $60^{\circ}$.Thus,the lines make angles of $90^{\circ} - 30^{\circ} = 60^{\circ}$ and $90^{\circ} + 30^{\circ} = 120^{\circ}$ with the positive $x$-axis.The slopes of the lines are $m_{1} = 	an(60^{\circ}) = \sqrt{3}$ and $m_{2} = 	an(120^{\circ}) = -\sqrt{3}$.The equations of the lines are $y = \sqrt{3}x$ and $y = -\sqrt{3}x$.Rearranging,we get $y - \sqrt{3}x = 0$ and $y + \sqrt{3}x = 0$.The joint equation is $(y - \sqrt{3}x)(y + \sqrt{3}x) = 0$.This simplifies to $y^{2} - 3x^{2} = 0$,which is equivalent to $3x^{2} - y^{2} = 0$.

The joint equation of the pair of lines passing through the origin and forming an equilateral triangle with the line $y=4$ is

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