If P(-1, 0),Q(0, 0),and R(3, 3sqrt{3}) are th | vedclass

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(B) The coordinates of the points are $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$.First,find the slopes of lines $QP$ and $QR$.The slope of $QP$ $(m_1)

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$\sqrt{3}x + y = 0$

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(B) The coordinates of the points are $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$.First,find the slopes of lines $QP$ and $QR$.The slope of $QP$ $(m_1)$ is $\frac{0 - 0}{0 - (-1)} = 0$.The slope of $QR$ $(m_2)$ is $\frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$.The angle $	heta_1$ of line $QP$ with the positive $x$-axis is $0^\circ$.The angle $	heta_2$ of line $QR$ with the positive $x$-axis is $	an^{-1}(\sqrt{3}) = 60^\circ$.The angle bisector of $\angle PQR$ will make an angle $	heta = \frac{0^\circ + 60^\circ}{2} = 30^\circ$ with the positive $x$-axis.The slope of the bisector is $m = 	an(30^\circ) = \frac{1}{\sqrt{3}}$.Since the bisector passes through the origin $Q(0, 0)$,its equation is $y - 0 = \frac{1}{\sqrt{3}}(x - 0)$,which simplifies to $\sqrt{3}y = x$ or $x - \sqrt{3}y = 0$.However,checking the geometry,the angle $\angle PQR$ is between the negative $x$-axis (line $QP$) and the line $y = \sqrt{3}x$ (line $QR$). The bisector of this angle is $y = 	an(120^\circ)x = -\sqrt{3}x$,which is $\sqrt{3}x + y = 0$.

If $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$ are three points,then the equation of the bisector of the $\angle PQR$ is

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