Let three points be P(-1, 0),Q(0, 0),and R(3, | vedclass

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Question

(A) The coordinates are $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$.Line $QP$ lies on the $x$-axis,so its angle with the positive $x$-axis is $	heta_1

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

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(A) The coordinates are $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$.Line $QP$ lies on the $x$-axis,so its angle with the positive $x$-axis is $	heta_1 = 180^{\circ}$.Line $QR$ passes through $(0, 0)$ and $(3, 3\sqrt{3})$. Its slope is $m = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$. Thus,the angle with the positive $x$-axis is $	heta_2 = 60^{\circ}$.The angle $\angle PQR$ is the angle between these two lines,which is $180^{\circ} - 60^{\circ} = 120^{\circ}$.The angle bisector divides this angle into two $60^{\circ}$ parts. The angle of the bisector with the positive $x$-axis is $180^{\circ} - 60^{\circ} = 120^{\circ}$.The slope of the bisector is $	an(120^{\circ}) = -\sqrt{3}$.Since it passes through the origin $Q(0, 0)$,the equation is $y = -\sqrt{3}x$,which simplifies to $\sqrt{3}x + y = 0$.

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

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