Question 1210143: Let R be the image of rotating point P=(4,0) counterclockwise by 60^\circ degrees around Q=(12,-7). What is PR?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $P = (4, 0)$ and $Q = (12, -7)$.
We want to rotate $P$ counterclockwise by $60^\circ$ around $Q$ to obtain point $R$.
We want to find the distance $PR$.
First, let's find the vector $\vec{QP} = P - Q = (4-12, 0-(-7)) = (-8, 7)$.
Let's rotate this vector counterclockwise by $60^\circ$.
We can represent the vector $\vec{QP}$ as a complex number: $z = -8 + 7i$.
To rotate $z$ counterclockwise by $60^\circ$, we multiply it by $e^{i\pi/3} = \cos(60^\circ) + i\sin(60^\circ) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.
The rotated vector is:
$$z' = z \cdot e^{i\pi/3} = (-8 + 7i)\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -4 - 4i\sqrt{3} + \frac{7i}{2} - \frac{7\sqrt{3}}{2} = \left(-4 - \frac{7\sqrt{3}}{2}\right) + i\left(\frac{7}{2} - 4\sqrt{3}\right)$$
This corresponds to the vector $\vec{QR} = \left(-4 - \frac{7\sqrt{3}}{2}, \frac{7}{2} - 4\sqrt{3}\right)$.
So, $R = Q + \vec{QR} = \left(12 - 4 - \frac{7\sqrt{3}}{2}, -7 + \frac{7}{2} - 4\sqrt{3}\right) = \left(8 - \frac{7\sqrt{3}}{2}, -\frac{7}{2} - 4\sqrt{3}\right)$.
We want to find $PR$.
Let's first find $PQ = \sqrt{(12-4)^2 + (-7-0)^2} = \sqrt{8^2 + 7^2} = \sqrt{64+49} = \sqrt{113}$.
Since rotation preserves distances, $QR = PQ = \sqrt{113}$.
We want $PR$, not $QR$.
Since we are rotating $P$ around $Q$, the distance $QP$ is the same as $QR$.
We want to find the distance between $P$ and $R$.
We know $QP = QR = \sqrt{113}$.
Let $PR = d$.
We have a triangle $PQR$ with $QP = QR = \sqrt{113}$ and $\angle PQR = 60^\circ$.
Since $QP = QR$, $\triangle PQR$ is an isosceles triangle.
Since $\angle PQR = 60^\circ$, the other two angles are equal, and their sum is $180^\circ - 60^\circ = 120^\circ$.
Thus, $\angle QPR = \angle QRP = 60^\circ$.
Therefore, $\triangle PQR$ is an equilateral triangle, so $PR = QP = QR = \sqrt{113}$.
Final Answer: The final answer is $\boxed{\sqrt{113}}$
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