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

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(A) The coordinates of the 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 with the positiv

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

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(A) The coordinates of the 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 with the positive $x$-axis is $180^\circ$ (or $\pi$ 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 $	heta$ that $QR$ makes with the positive $x$-axis is given by $	an 	heta = \sqrt{3}$,so $	heta = 60^\circ$ (or $\frac{\pi}{3}$ radians).The angle $\angle PQR$ is the angle between the line $QP$ $(180^\circ)$ and $QR$ $(60^\circ)$,which is $180^\circ - 60^\circ = 120^\circ$.The angle bisector of $\angle PQR$ will make an angle of $60^\circ + \frac{120^\circ}{2} = 120^\circ$ (or $\frac{2\pi}{3}$ radians) with the positive $x$-axis.The slope of the bisector is $	an(120^\circ) = -\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. Then the equation of the bisector of the $\angle PQR$ is

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