document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210146>Question 1210146</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the equation of the reflection of the line y = 2x + 1 across the line y = 5x - 3? Give your answer in slope-intercept form.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851368 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=851368\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $L_1$ be the line $y = 2x + 1$, and let $L_2$ be the line $y = 5x - 3$.<BR>\n" );
document.write( "We want to find the equation of the reflection of $L_1$ across $L_2$.\r<BR>\n" );
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document.write( "Let $P(x_1, y_1)$ be a point on $L_1$, so $y_1 = 2x_1 + 1$.<BR>\n" );
document.write( "Let $P'(x_2, y_2)$ be the reflection of $P$ across $L_2$.<BR>\n" );
document.write( "The midpoint of $PP'$ lies on $L_2$, and $PP'$ is perpendicular to $L_2$.\r<BR>\n" );
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document.write( "The midpoint of $PP'$ is $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.<BR>\n" );
document.write( "Since $M$ lies on $L_2$, we have<BR>\n" );
document.write( "$$\frac{y_1+y_2}{2} = 5\left(\frac{x_1+x_2}{2}\right) - 3$$<BR>\n" );
document.write( "$$y_1+y_2 = 5(x_1+x_2) - 6$$<BR>\n" );
document.write( "$$y_1+y_2 = 5x_1+5x_2 - 6$$<BR>\n" );
document.write( "Substituting $y_1 = 2x_1 + 1$, we get<BR>\n" );
document.write( "$$2x_1+1+y_2 = 5x_1+5x_2 - 6$$<BR>\n" );
document.write( "$$y_2 = 3x_1+5x_2 - 7 \quad (*)$$\r<BR>\n" );
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document.write( "Since $PP'$ is perpendicular to $L_2$, the slope of $PP'$ is the negative reciprocal of the slope of $L_2$.<BR>\n" );
document.write( "The slope of $L_2$ is 5, so the slope of $PP'$ is $-\frac{1}{5}$.<BR>\n" );
document.write( "Thus,<BR>\n" );
document.write( "$$\frac{y_2-y_1}{x_2-x_1} = -\frac{1}{5}$$<BR>\n" );
document.write( "$$5(y_2-y_1) = -(x_2-x_1)$$<BR>\n" );
document.write( "$$5y_2 - 5y_1 = -x_2 + x_1$$<BR>\n" );
document.write( "Substituting $y_1 = 2x_1 + 1$, we get<BR>\n" );
document.write( "$$5y_2 - 5(2x_1+1) = -x_2 + x_1$$<BR>\n" );
document.write( "$$5y_2 - 10x_1 - 5 = -x_2 + x_1$$<BR>\n" );
document.write( "$$5y_2 = 11x_1 - x_2 + 5 \quad (**)$$\r<BR>\n" );
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document.write( "From (*), we have $3x_1 = y_2 - 5x_2 + 7$.<BR>\n" );
document.write( "Substituting this into (**), we get<BR>\n" );
document.write( "$$5y_2 = \frac{11}{3}(y_2 - 5x_2 + 7) - x_2 + 5$$<BR>\n" );
document.write( "$$15y_2 = 11y_2 - 55x_2 + 77 - 3x_2 + 15$$<BR>\n" );
document.write( "$$4y_2 = -58x_2 + 92$$<BR>\n" );
document.write( "$$y_2 = -\frac{58}{4}x_2 + \frac{92}{4}$$<BR>\n" );
document.write( "$$y_2 = -\frac{29}{2}x_2 + 23$$\r<BR>\n" );
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document.write( "Thus, the equation of the reflected line is $y = -\frac{29}{2}x + 23$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{y = -\frac{29}{2} x + 23}$<BR>\n" );
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