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

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(C) The line $QP$ lies along the negative $x$-axis,so its angle with the positive $x$-axis is $\pi$. The line $QR$ passes through $(0, 0)$ and $(3, 3\

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

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(C) The line $QP$ lies along the negative $x$-axis,so its angle with the positive $x$-axis is $\pi$. The line $QR$ passes through $(0, 0)$ and $(3, 3\sqrt{3})$,so its slope is $m = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$. The angle $	heta$ that $QR$ makes with the positive $x$-axis is $	an 	heta = \sqrt{3}$,which means $	heta = \frac{\pi}{3}$.The angle $\angle PQR$ is the angle between the line $QP$ (angle $\pi$) and the line $QR$ (angle $\frac{\pi}{3}$).The measure of $\angle PQR = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.The bisector of $\angle PQR$ will make an angle $\alpha$ with the positive $x$-axis,where $\alpha = \frac{\pi + \frac{\pi}{3}}{2} = \frac{4\pi/3}{2} = \frac{2\pi}{3}$.The slope of the bisector is $	an(\frac{2\pi}{3}) = -\sqrt{3}$.Since the bisector passes through the origin $Q(0, 0)$,its equation is $y - 0 = -\sqrt{3}(x - 0)$,which simplifies to $y = -\sqrt{3}x$,or $\sqrt{3}x + y = 0$.

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

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