The joint equation of the lines passing throu | vedclass

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Question

(A) The first quadrant is divided into three equal parts by two lines passing through the origin. Since the total angle of the first quadrant is $90^{

Vedclass · Accepted Answer

$\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0$

Vedclass · Answer

(A) The first quadrant is divided into three equal parts by two lines passing through the origin. Since the total angle of the first quadrant is $90^{\circ}$,each line makes an angle of $30^{\circ}$ and $60^{\circ}$ with the positive $x$-axis.The slopes of these lines are $m_1 = 	an(30^{\circ}) = \frac{1}{\sqrt{3}}$ and $m_2 = 	an(60^{\circ}) = \sqrt{3}$.The equations of the lines are $y = \frac{1}{\sqrt{3}}x$ and $y = \sqrt{3}x$,which can be written as $x - \sqrt{3}y = 0$ and $\sqrt{3}x - y = 0$.The joint equation is $(x - \sqrt{3}y)(\sqrt{3}x - y) = 0$.Expanding this,we get $\sqrt{3}x^2 - xy - 3xy + \sqrt{3}y^2 = 0$.$\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$.

The joint equation of the lines passing through the origin and trisecting the first quadrant is . . . . . .

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