My Math Forum [ASK] Line of a Mirror

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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.

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