How Do Reflections Along Different Lines Affect Point Coordinates?

  • Thread starter Thread starter blafu
  • Start date Start date
  • Tags Tags
    Lines Reflection
AI Thread Summary
Reflections of points along the lines y = x+c and y = -x+c yield specific transformations for point coordinates. For y = x+c, a point (a, b) becomes (b-c, a+c), while for y = -x+c, it transforms to (-b+c, -a+c). The discussion confirms the correctness of these transformations by checking that points on the lines remain unchanged and that the lines connecting original and reflected points are perpendicular to the reflection lines. Follow-up questions explore the effects of reflections along lines with different slopes (m) and whether generalizations can be made regarding the slopes of lines formed by original and reflected points. The conversation emphasizes the importance of understanding reflections in coordinate geometry.
blafu
Messages
2
Reaction score
0
We were asked to make generalisations for the following:

a.) if a point is reflected along y = x+c
b.) if a point is reflected along y = -x+c

i came up w/ the ff.:
a.) if a point is reflected along y = x+c, then the general point (a, b) will become (b-c, a+c)

b.) if a point is reflected along y = -x+c, then the general point (a, b) will become (-b+c, -a+c)

let me knw if these are correct... if not then pls guide me and tell me the correct generalisations thank u!
 
Physics news on Phys.org
One easy way to check: if points are reflected around a line, then points on the line are not changed. (0, c) is on the line. If "(a, b) will become (b- c, a+ c)" then (0, c) will become (0, c). That looks good. More generally, (x, x+ c) will become ((x+c)-c, x+ c)= (x, c).

If y= -x+ c then (x, -x+ c), using (-b+c, -a+c) becomes (-(-x+c)+ c, -x+ c)= (x, -x+c) again.

To complete the check, to see if the line from (a, b) to (b-c, a+ c) is always perpendicular to y= x+ c. The slope of the line from (a, b) to (b-c, a+ c) is (a+c- b)/(b-c-a)= -1. The slope of y= x+ c is 1. Yes they are perpendicular.
 
Thank you very much...just a few followup questions (I hope it is ok to ask in this thread as well):

1) what would happen to point (a, b) if it is reflected along y = mx+c, where m is not 1 or -1 (that is, a line which slope is not -1 or 1)?
2) can "when a point A is reflected along line y = -mx+c, the line made by points A and A' have slope of 1" be made as a generalisation too?
3) can "when a point B is reflected along line y = mx+c, the line made by points B and B' have slope of -1" be made as a generalisation too?

for 2 and 3, provided that m is not 0 or undefined..., and that A and A', and B and B' are not invariant points.
 
Last edited:
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top