SOLUTION: Let R=(a,b) be the image of rotating point P=(7,2) clockwise by 150^\circ degrees around Q=(12,-5). What is b?

Question 1210144: Let R=(a,b) be the image of rotating point P=(7,2) clockwise by 150^\circ degrees around Q=(12,-5). What is b?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $P = (7, 2)$ and $Q = (12, -5)$. We want to rotate $P$ clockwise by $150^\circ$ around $Q$ to obtain point $R = (a, b)$.
First, let's find the vector $\vec{QP} = P - Q = (7-12, 2-(-5)) = (-5, 7)$.
We want to rotate this vector clockwise by $150^\circ$. This is equivalent to rotating counterclockwise by $-150^\circ$.
We can represent the vector $\vec{QP}$ as a complex number: $z = -5 + 7i$.
To rotate $z$ counterclockwise by $-150^\circ$, we multiply it by $e^{-i5\pi/6} = \cos(-150^\circ) + i\sin(-150^\circ) = \cos(210^\circ) + i\sin(210^\circ) = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$.
The rotated vector is:
$$z' = z \cdot e^{-i5\pi/6} = (-5 + 7i)\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = \frac{5\sqrt{3}}{2} + \frac{5}{2}i - \frac{7i\sqrt{3}}{2} + \frac{7}{2} = \left(\frac{5\sqrt{3}}{2} + \frac{7}{2}\right) + i\left(\frac{5}{2} - \frac{7\sqrt{3}}{2}\right)$$
This corresponds to the vector $\vec{QR} = \left(\frac{5\sqrt{3}}{2} + \frac{7}{2}, \frac{5}{2} - \frac{7\sqrt{3}}{2}\right)$.
So, $R = Q + \vec{QR} = \left(12 + \frac{5\sqrt{3}}{2} + \frac{7}{2}, -5 + \frac{5}{2} - \frac{7\sqrt{3}}{2}\right) = \left(\frac{31}{2} + \frac{5\sqrt{3}}{2}, -\frac{5}{2} - \frac{7\sqrt{3}}{2}\right)$.
We want to find $b$, which is the $y$-coordinate of $R$.
$$b = -\frac{5}{2} - \frac{7\sqrt{3}}{2} = \frac{-5 - 7\sqrt{3}}{2}$$
Final Answer: The final answer is $\boxed{\frac{-5-7\sqrt{3}}{2}}$

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