document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210143>Question 1210143</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let R be the image of rotating point P=(4,0) counterclockwise by 60^\circ degrees around Q=(12,-7). What is PR? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851365 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851365\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $P = (4, 0)$ and $Q = (12, -7)$.<BR>\n" );
document.write( "We want to rotate $P$ counterclockwise by $60^\circ$ around $Q$ to obtain point $R$.<BR>\n" );
document.write( "We want to find the distance $PR$.\r<BR>\n" );
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document.write( "First, let's find the vector $\vec{QP} = P - Q = (4-12, 0-(-7)) = (-8, 7)$.<BR>\n" );
document.write( "Let's rotate this vector counterclockwise by $60^\circ$.\r<BR>\n" );
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document.write( "We can represent the vector $\vec{QP}$ as a complex number: $z = -8 + 7i$.<BR>\n" );
document.write( "To rotate $z$ counterclockwise by $60^\circ$, we multiply it by $e^{i\pi/3} = \cos(60^\circ) + i\sin(60^\circ) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.<BR>\n" );
document.write( "The rotated vector is:<BR>\n" );
document.write( "$$z' = z \cdot e^{i\pi/3} = (-8 + 7i)\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -4 - 4i\sqrt{3} + \frac{7i}{2} - \frac{7\sqrt{3}}{2} = \left(-4 - \frac{7\sqrt{3}}{2}\right) + i\left(\frac{7}{2} - 4\sqrt{3}\right)$$<BR>\n" );
document.write( "This corresponds to the vector $\vec{QR} = \left(-4 - \frac{7\sqrt{3}}{2}, \frac{7}{2} - 4\sqrt{3}\right)$.<BR>\n" );
document.write( "So, $R = Q + \vec{QR} = \left(12 - 4 - \frac{7\sqrt{3}}{2}, -7 + \frac{7}{2} - 4\sqrt{3}\right) = \left(8 - \frac{7\sqrt{3}}{2}, -\frac{7}{2} - 4\sqrt{3}\right)$.\r<BR>\n" );
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document.write( "We want to find $PR$.<BR>\n" );
document.write( "Let's first find $PQ = \sqrt{(12-4)^2 + (-7-0)^2} = \sqrt{8^2 + 7^2} = \sqrt{64+49} = \sqrt{113}$.<BR>\n" );
document.write( "Since rotation preserves distances, $QR = PQ = \sqrt{113}$.<BR>\n" );
document.write( "We want $PR$, not $QR$.\r<BR>\n" );
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document.write( "Since we are rotating $P$ around $Q$, the distance $QP$ is the same as $QR$.<BR>\n" );
document.write( "We want to find the distance between $P$ and $R$.<BR>\n" );
document.write( "We know $QP = QR = \sqrt{113}$.<BR>\n" );
document.write( "Let $PR = d$.<BR>\n" );
document.write( "We have a triangle $PQR$ with $QP = QR = \sqrt{113}$ and $\angle PQR = 60^\circ$.<BR>\n" );
document.write( "Since $QP = QR$, $\triangle PQR$ is an isosceles triangle.<BR>\n" );
document.write( "Since $\angle PQR = 60^\circ$, the other two angles are equal, and their sum is $180^\circ - 60^\circ = 120^\circ$.<BR>\n" );
document.write( "Thus, $\angle QPR = \angle QRP = 60^\circ$.<BR>\n" );
document.write( "Therefore, $\triangle PQR$ is an equilateral triangle, so $PR = QP = QR = \sqrt{113}$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{\sqrt{113}}$<BR>\n" );
document.write( " </SPAN>\n" );
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