reflection about y=x+2

I think this should be a relatively basic concept to grasp, but for some reason im struggling to understand it.

I can sort of explain why a point (x,y) transforms to (y-2,x+2) but my solutions feels a lot more like guessing and hoping than a concrete formula.

If anyone could help out, it would be greatly appreciated.

Reply 1

mosaurlodon

The way I do it is...

so for a reflection of y=x, its a rule that (x,y) transforms to (y,x) - (dont actually know why)
so for y=x+2 I imagine it as the mirror moving up by 2, so the transformed y-coordinate is 2 higher -> (__, x+2)
but you can also imagine it as the mirror moving to the left by 2 (x=y-2), so the transformed x-coordinate is 2 to the left -> (y-2,___)
so combined gives you (x,y) -> (y-2,x+2)

but to me this method is a lot more guessing and hoping than coming up with a formula.

I also dont get why you only move each transformed coord by 2 rather than 4 - why doesnt the mirror double the distance?
Take a reflection about y=1 for eg, and for (x,y), then the transformed coordinate is (x, 2-y), so here the distance is doubled. using a 2 rather than a 1.

(edited 11 months ago)

Reply 2

tonyiptony

Generally speaking, it's difficult to find a formula for reflecting along any line. IIRC, the formula for reflecting along y=2x is already not worth remembering. It's a classic vector question though, which is related to finding the projection of the position vector v onto the reflecting line r (so something like

(r\cdot v)/(v\cdot v)

would show up).

In this particular case, however, one way to think about this is it's actually three transformations combined into one:
(1) translate downward by 2; then
(2) reflect along x=y; then
(3) translate upward back by 2,

of which each of them is easy.

(edited 11 months ago)

Reply 3

mosaurlodon

I like your breakdown of the reflection of y=x+2, but im slightly confused about your vectors.

why doesnt v dot r become 0, as the line thats perpendicular to r is what youre looking for (that joins the original point and the transformed point) which should make the dot product 0?

edit: is it because the vector (v-r_g) dot r = 0
where r_g is a general point on line r?

(edited 11 months ago)

Reply 4

mqb2766

Original post by mosaurlodon

The way I do it is...
so for a reflection of y=x, its a rule that (x,y) transforms to (y,x) - (dont actually know why)
so for y=x+2 I imagine it as the mirror moving up by 2, so the transformed y-coordinate is 2 higher -> (__, x+2)
but you can also imagine it as the mirror moving to the left by 2 (x=y-2), so the transformed x-coordinate is 2 to the left -> (y-2,___)
so combined gives you (x,y) -> (y-2,x+2)

but to me this method is a lot more guessing and hoping than coming up with a formula.
I also dont get why you only move each transformed coord by 2 rather than 4 - why doesnt the mirror double the distance?
Take a reflection about y=1 for eg, and for (x,y), then the transformed coordinate is (x, 2-y), so here the distance is doubled. using a 2 rather than a 1.

For lines with gradients of +/-1, its easy to get the reflections by projecting a point (a,b) vertically onto the line then horizontally the same distance. This works for the line y=x, so the point (a,b) projected vertically would become (a,a) and the vertical distance is a-b. Then subtracting this (a-b) off the x coordinate gives the reflected point as (b,a).

So for y=x+2, the point (a,b) projects vertically onto the line (a,a+2) and the distance moved is a+2-b. So project the same distance horizontally, the reflected point becomes (b-2,a+2).

This works when the grad of the lines / normal is +/-1, otherwise you have to be a bit more rigorous with the vector algebra / dot product and as tony says, its a well known result, so subtract twice the distance from the line along the normal and that gives the reflection. A similar way but less dot productey
https://undergroundmathematics.org/geometry-of-equations/r6230/solution
Edit - note a problem solving way to do this question would be to get the midpoint for each of the options and simply check which one lay on the line.

Similarly, thinking about the equation of a line as
nhat.x = c
where nhat is the unit hormal vector and x = (x,y) and c is the distance of the origin from the line means its a bit easier to think about reflections and then generalises to the plane representation.

(edited 11 months ago)

Reply 5

mosaurlodon

ok, thank you for that, I think it should stick in my brain a little better.

I was also looking at another way to do reflections involving matrices.
so for a reflection in like y=mx you know its just a linear transfomation so
(I dont know how to type this notation)
[a b]_ [x]
[c d] * [y]
so for eg, for a reflection in y = 2x, (1,2) -> (1,2) and the invariant line, y=-1/2x so any points on the line goes back to the line like (2,-1) -> (-2,1)
then
[a b] _[1] = [1]
[c d] * [2] _[2]
and
[a b] _ [2] = [-2]
[c d] * [-1] _ [1]
then its surprisingly simple to solve for ABCD and then you get (x,y) -> (-3/5 x+ 4/5 y, 4/5 x + 3/5 y)

another thing, I thought it would be a lot harder to reflect an actual graph, so say reflect y = x^2 from y= x+2, but if you just know (x,y) -> (y-2,x+2) then you literally just replace the y=x^2 with (x+2) = (y-2)^2 and thats it.

(edited 11 months ago)

Reply 6

DFranklin

Original post by mosaurlodon

I think this should be a relatively basic concept to grasp, but for some reason im struggling to understand it.
I can sort of explain why a point (x,y) transforms to (y-2,x+2) but my solutions feels a lot more like guessing and hoping than a concrete formula.
If anyone could help out, it would be greatly appreciated.

A very mechanical way of solving this is:

First, work out where (0, 0) maps to, call it a.
Then work out where (1, 0) and (0, 1) map to; call these b and c.

Then (x, y) maps to a + x(b-a) + y(c-a).

This works for any affine transformation.