Reflecting B(3,-1) on Line g: 4y + x - 15 = 0

[Monoxdifly]

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 $$-\frac{1}{4}$$ and I substituted it in the $$y-y_1=m(x-x_1)$$ equation. This is what I got:
$$y-3=-\frac{1}{4}(x-4)$$
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?

 

[MarkFL]

What I would do is observe that the line must pass through the midpoint of the two given points, and be perpendicular to the line through the two given points.The midpoint is:

$$\left(\frac{3+5}{2},\frac{-1+7}{2}\right)=(4,3)$$

The slope is:

$$m=-\frac{\Delta x}{\Delta y}=-\frac{5-3}{7+1}=-\frac{1}{4}$$

Thus, our line is:

$$y-3=-\frac{1}{4}(x-4)$$

Or:

$$4y-12=-x+4$$

Or:

$$4y+x-16=0$$

I agree with your answer. :)

 

[Monoxdifly]

Isn't that basically what I did?

 

[MarkFL]

Isn't that basically what I did?

Yes...it's just easier for me to work the problem and then compare results.