Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2018-03-31 02:18:31

Monox D. I-Fly
Member
From: Indonesia
Registered: 2015-12-02
Posts: 2,000

[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

and I substituted it in the
equation. This is what I got:

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 (2018-03-31 02:20:46)


Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.

Offline

#2 2018-03-31 05:56:51

Bob
Administrator
Registered: 2010-06-20
Posts: 10,136

Re: [ASK] Line of a Mirror

I think your method is good and I get 16 too.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

Board footer

Powered by FluxBB