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

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(C) The slope of $QP$ is $0$ (since $P$ and $Q$ lie on the $x$-axis,specifically the negative $x$-axis).The slope of $QR$ is $\frac{3\sqrt{3} - 0}{3 -

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

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(C) The slope of $QP$ is $0$ (since $P$ and $Q$ lie on the $x$-axis,specifically the negative $x$-axis).The slope of $QR$ is $\frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$.This means the angle made by $QR$ with the positive $x$-axis is $60^\circ$.Since $QP$ lies along the negative $x$-axis,the angle $\angle PQR$ is $180^\circ - 60^\circ = 120^\circ$.The angle bisector of $\angle PQR$ will make an angle of $120^\circ - 60^\circ = 60^\circ$ with $QP$ (the negative $x$-axis).Thus,the angle made by the bisector with the positive $x$-axis is $180^\circ - 60^\circ = 120^\circ$.The equation of a line passing through the origin $(0, 0)$ with slope $m = 	an(120^\circ) = -\sqrt{3}$ is $y = -\sqrt{3}x$,which simplifies to $\sqrt{3}x + y = 0$.

Let $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$ be three points. Then the equation of the bisector of the angle $\angle PQR$ is

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