# rotation on a line

• May 9th 2011, 10:07 AM
eleahy
rotation on a line
I was given a question in an exam where it asked to get the reflection of a line through the points (6,2) and (4,1).
I know you have to translate it to the origin then rotate it to lie on the origin and then reflect it but I dont know how to get the translation vector, and also unsure of how to get the angle of rotation to the axis before i reflect it( i think it could be tan inverse of the slope??)
• May 9th 2011, 02:04 PM
Deveno
the easiest way (conceptually, at least), is to translate one of the points to the origin. you can pick either one. suppose you chose to translate (6,2) to the origin. we are thus moving the point (6,2) to (0,0): that is, we are sending (x,y) --> (x-6,y-2). this transformation sends (4,1) to (-2,-1).

so now, we have translating the original problem of reflecting through the line that goes from (6,2) to (4,1), to one of reflecting through the line that goes from (0,0) to (-2,-1).

this is the line y = x/2 (it has slope -1/(-2) = 1/2). in terms of the angle θ this line makes with the x-axis we have:

tanθ = 1/2
cosθ = 1/(√5/2) = 2/√5
sinθ = (1/2)/(√5/2) = 1/√5

so to rotate this line to the x-axis, we have to rotate it -θ, which is multiplying (x,y) (as a column vector) by the matrix:

[ 2/√5 1/√5]
[-1/√5 2/√5], which sends (x,y) to ((2/√5)x + (1/√5)y, -(1/√5)x + (2/√5)y).

now, our line of reflection is just the x-axis, and this just sends (x,y) ---> (x,-y).

finally, we "rotate back", and then "translate back".

so let's put this all together: let's call the translation T, the rotation (by θ) R, and the horizontal reflection H.

the complete transformation is T^-1(R(H(R^-1(T(x,y))))).

T(x,y) = (x-6,y-2). R^-1(T(x,y)) = ((2/√5)(x-6) + (1/√5)(y-2), -(1/√5)(x-6) + (2/√5)(y-2)).

H(R^-1(T(x,y))) = ((2/√5)(x-6) + (1/√5)(y-2), (1/√5)(x-6) - (2/√5)(y-2)).

R(H(R^-1(T(x,y)))) =

[(2/√5)((2/√5)(x-6) + (1/√5)(y-2)) - (1/√5)((1/√5)(x-6) - (2/√5)(y-2)), (1/√5)((2/√5)(x-6) + (1/√5)(y-2)) + (2/√5)((1/√5)(x-6) - (2/√5)(y-2))]

= ((4/5)(x-6) + (2/5)(y-2) - (1/5)(x-6) + (2/5)(y-2), (2/5)(x-6) + (1/5)(y-2) + (2/5)(x-6) - (4/5)(y-2))

= ((3/5)(x-6) + (4/5)(y-2), (4/5)(x-6) - (3/5)(y-2))

finally, we find T^-1(R(H(R^-1(T(x,y))))) = ((3/5)(x-6) + (4/5)(y-2) + 6, (4/5)(x-6) - (3/5)(y-2) + 2)

= (3x/5 + 4y/5 + 4/5, 4x/5 - 3y/5 - 8/5) (you may want to check my arithmetic).
• May 9th 2011, 02:16 PM
eleahy
thank you sooo much really appreciate all the work you put into that and actually understand it all for once :):) thanks again