A straight line passing through the point (2, | vedclass

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(C) The given lines are $L_1: \sqrt{3}x + y = 0$ and $L_2: \sqrt{3}x - y = 0$.Slope of $L_1$ is $m_1 = -\sqrt{3}$,so the angle with the $x$-axis is $\

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$y - 2 = 0$

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(C) The given lines are $L_1: \sqrt{3}x + y = 0$ and $L_2: \sqrt{3}x - y = 0$.Slope of $L_1$ is $m_1 = -\sqrt{3}$,so the angle with the $x$-axis is $	heta_1 = 120^\circ$.Slope of $L_2$ is $m_2 = \sqrt{3}$,so the angle with the $x$-axis is $	heta_2 = 60^\circ$.The angle between the lines is $120^\circ - 60^\circ = 60^\circ$.For $	riangle OAB$ to be equilateral,the line $AB$ must make an angle of $60^\circ$ with both lines $L_1$ and $L_2$.Since the angle between $L_1$ and $L_2$ is $60^\circ$ and they are symmetric about the $x$-axis,the line $AB$ must be parallel to the $x$-axis to form an equilateral triangle with the origin.$A$ line parallel to the $x$-axis has the form $y = k$.Since it passes through $(2, 2)$,the equation is $y = 2$,or $y - 2 = 0$.

$A$ straight line passing through the point $(2, 2)$ intersects the lines $\sqrt{3}x + y = 0$ and $\sqrt{3}x - y = 0$ at points $A$ and $B$ respectively. Find the equation of the line $AB$ such that the triangle $OAB$ is an equilateral triangle,where $O$ is the origin.

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