The joint equation of the pair of lines passi | vedclass

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Question

(C) The lines pass through the origin and form an equilateral triangle with the line $x=3$. Since the triangle is equilateral,the angle at the origin

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The lines pass through the origin and form an equilateral triangle with the line $x=3$. Since the triangle is equilateral,the angle at the origin $O$ is $60^{\circ}$. Given the symmetry about the $x$-axis,the lines make angles of $30^{\circ}$ and $-30^{\circ}$ with the $x$-axis. The slopes of the lines are $m_1 = 	an(30^{\circ}) = \frac{1}{\sqrt{3}}$ and $m_2 = 	an(-30^{\circ}) = -\frac{1}{\sqrt{3}}$. The equations of the lines are $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$,which can be written as $x - \sqrt{3}y = 0$ and $x + \sqrt{3}y = 0$. The joint equation is $(x - \sqrt{3}y)(x + \sqrt{3}y) = 0$,which simplifies to $x^2 - 3y^2 = 0$.

The joint equation of the pair of lines passing through the origin and forming an equilateral triangle with the line $x=3$ is

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