Math Help (solved by Luthienjish)

urgent ;c

cos and sin won't work as it's not a circle. It's a triangle.

bump, i suck at math :3

Unfortunately can't help myself, but I'd recommend you to also post it at http://math.stackexchange.com . So far they never failed to deliver an answer to my questions.

Oh algebra... my mortal enemy

Everytime I see math bump.

View attached image.

Not really my cup of tea but this looks promising... https://en.m.wikipedia.org/wiki/Householder_transformation

maybe you're using a scientific calculator and it's on RAD setting not on DEG?

I teach math, in other language unfortunately. Though I'm afraid I cannot explain well in English, I'll try.

(a, b) -> (X, Y), reflected by y=4/3x
First, {Y-b}/{X-a} = -3/4 (for they need to be crossed by right angle)
Second, {Y+b}/2 = 4/3[{X+a}/2] (for their middle point needs to be on the said line)
Solve two equations.

X = {-7a+24b}/25, Y = {24a+7b}/25
The matrix
[ -7/25 24/25]
[24/25 7/25]

(Edit) In terms of theta... we use them only for rotation, not reflect. But you may know theta by using calculator.

Bump hoping someone else can come along and solve for OP.

I do not know the answer by heart... But, hey, we can figure it out; you want a linear transformation which takes:

i) v1=(cosθ,sinθ) into v1=(cosθ,sinθ) -- that will keep the line y=mx in place, immovable.
ii) v2=(-sinθ,cosθ) into -v2=(sinθ,-cosθ) -- that will take (a vector which is orthogonal to the line) to its negative.

And that is it -- if you know how the transformation operates on a basis, you know the transformation itself! Just take a matrix with the "outputs" in the columns:

A=[cosθ   sinθ]
  [sinθ  -cosθ]

And make another one with the "inputs" in the columns:

B=[cosθ  -sinθ]
  [sinθ   cosθ]

The matrix you want is A*B^(-1), that is:

M=[cosθ   sinθ]  *  [ cosθ sinθ] = [cos2θ   sin2θ]
  [sinθ  -cosθ]     [-sinθ cosθ]   [sin2θ  -cos2θ]

just as you found online... Why doesn't it work for you? Taking M*P should give you P'.

Let us test this: when y=4/3 x, we have cosθ=3/5 and sinθ=4/5.
So cos2θ=(3/5)^2-(4/5)^2=-7/25 and sin2θ=2(3/5)(4/5)=24/25, and M is:

M=[-7/25  24/25]
  [24/25  7/25]

just as Luthienjish found around here. :D

So, according to this, the symmetric of P=[2;1] would be P'=MP, you can finish the calculation.

Cheers!

Luthienjish's response gives the right intuition and the easiest way of solving the problem, OP.

But note that the matrix you found online was actually correct: reflection across a line can be factored into the two linear transformations, as Luthienjish explained. First, you reflect the plane across the 45 degree line; and then you rotate the plane counterclockwise by twice the difference of the angles, from the x-axis, of the line of reflection and the 45 degree line until the transformed line of reflection coincides with the 45 degree line.

The first transformation interchanges the unit vectors (1,0) and (0,1), and has matrix ((0,1), (1,0)) like balloonmial said. A counterclockwise rotation by φ sends (1,0) to (cos φ, sin φ) and (0,1) to (-sin φ, cos φ), which implies that the second transformation, rotation by φ=2(θ-π/4), has matrix ((cos(2θ-π/2), -sin(2θ-π/2)), (sin(2θ-π/2), cos(2θ-π/2))=((sin 2θ, cos2θ), (-cos2θ, sin2θ)). Thus the final transformation has matrix ((cos 2θ, sin 2θ), (sin2θ, -cos2θ)).

Expressing the slope m=arctan(θ) of the line of reflection as a ratio of numbers y and x (the height and base, respectively, of the triangle between the x-axis and the line of reflection) and setting h=sqrt(x^2+y^2) (the hypotenuse of this triangle), we determine that the final transformation has matrix (1/h)^2 ((x^2-y^2), 2xy), (2xy, y^2-x^2)). In other words, the point (a,b) maps to (a cosθ + b sin2θ, a sin2θ - b cos2θ) = (1/h)^2 (a(y^2-x^2) + 2bxy, b(x^2 - y^2) + 2 axy).

In particular, when m=4/3, the transformation has matrix (1/25)((-7,24), (24, 7)) and (a,b) maps to (1/25) (-7 a + 24 b, 24 a + 7 b)). In your original example, the point (2,1) is reflected across the line y=4/3x to (2/5, 11/5).

All matrices above are written as lists of column vectors, i.e. ((a,b), (c,d)) stands for
a c
b d
with the appropriate brackets.

I'm glad the Mathematicians of SG worked it out :D