The joint equation of the pair of lines passi | vedclass

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(A) The lines pass through the origin and form an equilateral triangle with the line $y=5$. Since the triangle is equilateral,the angle each line make

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The lines pass through the origin and form an equilateral triangle with the line $y=5$. Since the triangle is equilateral,the angle each line makes with the $y$-axis is $30^{\circ}$.Thus,the angles the lines make with the positive $x$-axis are $90^{\circ}-30^{\circ}=60^{\circ}$ and $90^{\circ}+30^{\circ}=120^{\circ}$.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 $(y - \sqrt{3}x)(y + \sqrt{3}x) = 0$.$y^2 - 3x^2 = 0$,which can be written as $3x^2 - y^2 = 0$.

The joint equation of the pair of lines passing through the origin and forming an equilateral triangle with the line $y=5$ is

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