Let P equiv (-5, 0),Q equiv (0, 0),and R equi | vedclass

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(D) The coordinates are $P(-5, 0)$,$Q(0, 0)$,and $R(2, 2\sqrt{3})$.$QP$ lies along the negative $X$-axis,so the angle it makes with the positive $X$-a

Vedclass · Accepted Answer

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

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(D) The coordinates are $P(-5, 0)$,$Q(0, 0)$,and $R(2, 2\sqrt{3})$.$QP$ lies along the negative $X$-axis,so the angle it makes with the positive $X$-axis is $180^{\circ}$.The slope of $QR$ is $m = \frac{2\sqrt{3} - 0}{2 - 0} = \sqrt{3}$.Since $	an 	heta = \sqrt{3}$,the angle $	heta$ that $QR$ makes with the positive $X$-axis is $60^{\circ}$.The angle $\angle PQR$ is the angle between $QP$ and $QR$,which is $180^{\circ} - 60^{\circ} = 120^{\circ}$.The angle bisector of $\angle PQR$ divides this $120^{\circ}$ angle into two $60^{\circ}$ angles.Since $QP$ is at $180^{\circ}$,the bisector makes an angle of $180^{\circ} - 60^{\circ} = 120^{\circ}$ with the positive $X$-axis.The slope of the bisector is $	an 120^{\circ} = -\sqrt{3}$.The equation of the line passing through the origin $(0, 0)$ with slope $-\sqrt{3}$ is $y = -\sqrt{3}x$,which simplifies to $\sqrt{3}x + y = 0$.

Let $P \equiv (-5, 0)$,$Q \equiv (0, 0)$,and $R \equiv (2, 2\sqrt{3})$ be three points. Then the equation of the bisector of the angle $\angle PQR$ is

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