A point P is taken outside Delta ABC where B( | vedclass

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(C) The coordinates are $A(0, 0)$,$B(1, \sqrt{3})$,and $C(2, 0)$.Note that $AB = \sqrt{1^2 + (\sqrt{3})^2} = 2$,$BC = \sqrt{(2-1)^2 + (0-\sqrt{3})^2}

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$\frac{1}{\sqrt{3}}$

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(C) The coordinates are $A(0, 0)$,$B(1, \sqrt{3})$,and $C(2, 0)$.Note that $AB = \sqrt{1^2 + (\sqrt{3})^2} = 2$,$BC = \sqrt{(2-1)^2 + (0-\sqrt{3})^2} = \sqrt{1+3} = 2$,and $AC = 2$.Thus,$\Delta ABC$ is an equilateral triangle with each angle equal to $60^{\circ}$.Given $\angle APC = 30^{\circ}$ and $\angle BPA = 15^{\circ}$,the point $P$ lies on the circumcircle of $\Delta ABC$ because $\angle APC = \frac{1}{2} \angle ABC$ is not the case,but rather the angles subtended by the chords $AC$ and $AB$ at the circumference are consistent with $P$ lying on the circle passing through $A, B, C$.Since $AB=BC=AC=2$,the circumcenter is at $(1, 1/\sqrt{3})$.From the geometry,$\angle BCP = \angle BAP = 15^{\circ}$ and $\angle ABP = \angle ACP = 30^{\circ}$.Since $\angle ABC = 60^{\circ}$ and $\angle ABP = 30^{\circ}$,the line $BP$ makes an angle of $60^{\circ} + 30^{\circ} = 90^{\circ}$ with $BC$ or similar geometric deduction shows $BP$ is perpendicular to $BC$.The slope of $BC$ is $m_{BC} = \frac{0-\sqrt{3}}{2-1} = -\sqrt{3}$.Since $BP \perp BC$,the slope of $BP$ is $m_{BP} = -\frac{1}{m_{BC}} = -\frac{1}{-\sqrt{3}} = \frac{1}{\sqrt{3}}$.

$A$ point $P$ is taken outside $\Delta ABC$ where $B(1, \sqrt{3})$,$A(0, 0)$,and $C(2, 0)$,but inside the acute angle $BAC$,such that $\angle APC = \frac{\pi}{6}$ and $\angle BPA = \frac{\pi}{12}$. The slope of the line $BP$ is:

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