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 March 31st, 2018, 06:18 AM #1 Senior Member   Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry [ASK] Line of a Mirror The point B(3, -1) is reflected by the line g and results in B'(5, 7). The equation of line g is .... A. 4y + x - 15 = 0 B. 4y + x - 9 = 0 C. 4y + x + 15 = 0 D. 4y - x - 15 = 0 E. 4y - x - 9 = 0 Since I didn't know how to approach the problem in a formal, textbook way, I tried to get... creative. The point of reflection must be exactly in the middle of (3, -1) and (5, 7), that is, (4, 3). Since the mirror must be a line perpendicular to BB' (which has the slope 4) and going through (4, 3), the slope of the mirror is $\displaystyle -\frac{1}{4}$ and I substituted it in the $\displaystyle y-y_1=m(x-x_1)$ equation. This is what I got: $\displaystyle y-3=-\frac{1}{4}(x-4)$ 4(y - 3) = -(x - 4) 4y - 12 = -x + 4 4y + x - 12 - 4 = 0 4y + x - 16 = 0 which is not in any of the options, but really close to the option A. Can we just assume that the option A was a typo? Or did I make a mistake somewhere? Last edited by Monox D. I-Fly; March 31st, 2018 at 06:21 AM. Tags line, mirror Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ganesh Ujwal Physics 4 January 11th, 2015 05:26 AM aditya ranjan Physics 7 January 5th, 2015 05:06 PM rakmo Algebra 1 November 26th, 2013 04:44 AM layneinchains Linear Algebra 1 July 16th, 2013 04:36 PM aedji Algebra 2 May 24th, 2010 02:49 PM

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