document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210142>Question 1210142</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Point B is the reflection of point A=(2,-1) over the line y=1. Point C is the reflection of point B over the line x + 2y = a. If  C=(-5,8), what is a? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851364 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=851364\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $A = (2, -1)$.<BR>\n" );
document.write( "Since $B$ is the reflection of $A$ over the line $y=1$, the $x$-coordinate of $B$ is the same as the $x$-coordinate of $A$, and the $y$-coordinate of $B$ is such that the midpoint of $AB$ lies on the line $y=1$.<BR>\n" );
document.write( "Let $B = (x_B, y_B)$.<BR>\n" );
document.write( "The midpoint of $AB$ is $\left(\frac{2+x_B}{2}, \frac{-1+y_B}{2}\right)$.<BR>\n" );
document.write( "Since this midpoint lies on $y=1$, we have $\frac{-1+y_B}{2} = 1$, so $-1+y_B = 2$, and $y_B = 3$.<BR>\n" );
document.write( "Thus, $B = (2, 3)$.\r<BR>\n" );
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document.write( "Now, $C = (-5, 8)$ is the reflection of $B = (2, 3)$ over the line $x + 2y = a$.<BR>\n" );
document.write( "Let the midpoint of $BC$ be $M = \left(\frac{2+(-5)}{2}, \frac{3+8}{2}\right) = \left(-\frac{3}{2}, \frac{11}{2}\right)$.<BR>\n" );
document.write( "Since $M$ lies on the line $x + 2y = a$, we have<BR>\n" );
document.write( "$$-\frac{3}{2} + 2\left(\frac{11}{2}\right) = a$$<BR>\n" );
document.write( "$$-\frac{3}{2} + 11 = a$$<BR>\n" );
document.write( "$$-\frac{3}{2} + \frac{22}{2} = a$$<BR>\n" );
document.write( "$$a = \frac{19}{2}$$\r<BR>\n" );
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document.write( "The line $BC$ is perpendicular to the line $x + 2y = a$.<BR>\n" );
document.write( "The slope of $x + 2y = a$ is $-\frac{1}{2}$.<BR>\n" );
document.write( "The slope of $BC$ is $\frac{8-3}{-5-2} = \frac{5}{-7} = -\frac{5}{7}$.<BR>\n" );
document.write( "Since the slopes are not negative reciprocals, our midpoint calculation is sufficient.\r<BR>\n" );
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document.write( "Therefore, $a = \frac{19}{2}$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{\frac{19}{2}}$<BR>\n" );
document.write( " </SPAN>\n" );
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