What type of triangle is formed by the line x | vedclass

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(C) The given pair of lines is $x^2 - 3y^2 = 0$,which can be written as $x^2 = 3y^2$,or $x = \pm \sqrt{3}y$. This represents two lines passing through

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Equilateral

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(C) The given pair of lines is $x^2 - 3y^2 = 0$,which can be written as $x^2 = 3y^2$,or $x = \pm \sqrt{3}y$. This represents two lines passing through the origin: $L_1: x - \sqrt{3}y = 0$ and $L_2: x + \sqrt{3}y = 0$. The third line is $x = a$. To find the vertices of the triangle,we find the intersection points: $1$. Intersection of $L_1$ and $x = a$: $a = \sqrt{3}y \implies y = \frac{a}{\sqrt{3}}$. Point $P = (a, \frac{a}{\sqrt{3}})$. $2$. Intersection of $L_2$ and $x = a$: $a = -\sqrt{3}y \implies y = -\frac{a}{\sqrt{3}}$. Point $Q = (a, -\frac{a}{\sqrt{3}})$. $3$. Intersection of $L_1$ and $L_2$: The origin $O = (0, 0)$. The lengths of the sides are: $PQ = \sqrt{(a-a)^2 + (\frac{a}{\sqrt{3}} - (-\frac{a}{\sqrt{3}}))^2} = \sqrt{0 + (\frac{2a}{\sqrt{3}})^2} = \frac{2|a|}{\sqrt{3}}$. $OP = \sqrt{a^2 + (\frac{a}{\sqrt{3}})^2} = \sqrt{a^2 + \frac{a^2}{3}} = \sqrt{\frac{4a^2}{3}} = \frac{2|a|}{\sqrt{3}}$. $OQ = \sqrt{a^2 + (-\frac{a}{\sqrt{3}})^2} = \sqrt{a^2 + \frac{a^2}{3}} = \frac{2|a|}{\sqrt{3}}$. Since $OP = OQ = PQ = \frac{2|a|}{\sqrt{3}}$,the triangle is equilateral.

What type of triangle is formed by the line $x = a$ and the pair of lines $x^2 - 3y^2 = 0$?

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