Let P equiv (-1, 0),Q equiv (0, 0),and R = (3 | vedclass

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(D) Given points are $P = (-1, 0)$,$Q = (0, 0)$,and $R = (3, 3\sqrt{3})$.Line $QP$ lies along the negative $x$-axis,so its angle of inclination is $\p

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

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(D) Given points are $P = (-1, 0)$,$Q = (0, 0)$,and $R = (3, 3\sqrt{3})$.Line $QP$ lies along the negative $x$-axis,so its angle of inclination is $\pi$ radians.Line $QR$ passes through $(0, 0)$ and $(3, 3\sqrt{3})$. Its slope is $m = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$.The angle of inclination $\phi$ of line $QR$ satisfies $	an \phi = \sqrt{3}$,so $\phi = \frac{\pi}{3}$.The angle $\angle PQR$ is the angle between the negative $x$-axis and the line $QR$,which is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.The angle bisector of $\angle PQR$ makes an angle $\alpha$ with the positive $x$-axis. Since the bisector divides $\angle PQR$ into two equal parts of $\frac{\pi}{3}$ each,the angle of the bisector with the positive $x$-axis is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.The slope of the bisector is $m = 	an(\frac{2\pi}{3}) = -\sqrt{3}$.The equation of the line passing through $(0, 0)$ with slope $-\sqrt{3}$ is $y = -\sqrt{3}x$,which simplifies to $\sqrt{3}x + y = 0$.

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

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