This question involves geometric transformations represented by a 2 x 2 matrix m

Question

This question involves geometric transformations represented by a 2 x 2 matrix m, where a point is mapped to by the formula Reflection in the x-axis is represented by the matrix Reflection in the x-axis is represented by the matrix my=0= . Where is the point mapped to, after reflection in the y-axis? Consider the line through the origin, with slope tan(theta), namely y = x tan(theta). Where is the point mapped to, after reflection in this line y = x tan(theta)? Where is the point mapped to, after reflection in this line y = x tan(theta)? Combine your results from parts (c) and (d) to write down the matrix representing the reflection of any point in the line y = x tan(theta). Express your result in (e) as a combination of a reflection in the x-axis and a rotation.

Explanation / Answer

a) (2 -3)T    (you get this either by multiplying or noting that a reflection on the y-axis leaves the x-coordinate unchanged and swiches the y-coordinate to the negative of the value.

b) (-2 3)T is reflection in the y-axis

c) (1 0) is relected about y = x tan(theta). The slope is cot(theta), or we move from (1 0) to

(1 - c tan(theta), c) on the line y = tan(theta) x

c = tan(theta)(1 - c tan(theta)) = tan(theta) - c tan^2(theta)

c + c tan^2(theta) = tan(theta)

The point on the line is (sin(theta)cos(theta)/tan(theta), sin(theta) cos(theta)) =

(cos^2(theta), sin(theta), cos(theta))

Reflecting, we get

(2 cos^2(theta) - 1, 2 sin(theta) cos(theta)) or

(cos(2 theta), sin(2 theta))

This is equivalent to rotating the point by 2 theta

d) Next, we map (0 1)T through the line

We get (c, 1 - c cot(theta)), where the point is on the line y = x tan(theta), so

1 - c cot(theta) = c tan(theta)

c(tan(theta)+cot(theta) = 1

c = 1/(tan(theta)+cot(theta))

c = 1/(sin^2(theta)+cos^2(theta))/sin(thetacos(theta)) = 1/(1/(sin(theta)cos(theta))=

sin(theta)cos(theta)

Thus, the point is

c(tan(theta)+cost(theta))

c sec^2(theta) = tan(theta)

c = tan(theta)cos^2(theta)

c = sin(theta)cos(theta)

The point on the line is (sin(theta)cos(theta), sin(theta)cos(theta)tan(theta)) =

(sin(theta)cos(theta), sin^2(theta))

Then, the reflected point is (2 sin(theta)cos(theta), 2 sin^2(theta) - 1) =

(sin(2 theta), - cos(2 theta))

This is equivalent to rotating by - pi + 2 theta

e) (x0 cos(2 theta) + y0 sin(2 theta), x0 sin(2 theta) - y0 cos(2 theta)))

f) A reflection of (x0 y0)T in the x-axis changes it to (x0 -y0)T

Thus, as R =

cos(mu)       -sin(mu)

sin(mu)      cos(mu)

We just let mu = 2 theta, to get the desired relection + translation of

cos(2 theta) -sin(2 theta)    *         (x0 -y0)T    =

sin(2 theta) cos(2 theta)

(x0 cos(2 theta) + y0 sin(2 theta), x0 sin(2 theta) - y0 cos(2 theta)))

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