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# Determine the determine angle of rotation if a triangle is reflected in the mirror lines y=2x+4and y=-1/2x-5?

Update:

sorry 4 the typographical error

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• Anonymous
1 decade ago
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I'm not sure if I understood the question correctly, but I've spent a few hours on it anyway.

I was thinking of somehow finding the two reflection transformation matrices and transforming some points and seeing where they ended up.

Then I realized that it didn't matter what the shape was so a line would do rather than a triangle.

Then I realized that we are not interested in where the points of the object ended up, but rather how much the object had rotated about its own center.

This can be solved without matrices.

If we consider the line x = 0, we can sketch it on a graph over the top of the y axis. This will serve as our triangle.

On the same graph we also plot the line y = 2x + 4.

Now we can see where the x = 0 line is reflected by the line y = 2x + 4.

From the gradient of y = 2x + 4 we know the angle this line makes with the x axis.

m = 2

tan a = 2

a = arctan 2

The angle the line y=2x+4 makes with the x axis is arctan 2.

From this we can find the angle the line y=2x+4 makes with our "triangle" line x=0.

The sum of the angles enclosed by our two lines and the x axis equals π.

π/2 + a + b = π

b = π – π/2 – a

b = π/2 – a

b is half of the angle of rotation due to the reflection about the first line so the full angle of rotation is twice that.

c = 2b

We follow the same procedure for the rotation due to the reflection in the second line.

We can find the angle that our once rotated line makes with the x axis.

π/2 + c + d = π

d = π - π/2 - c

d = π/2 - c

(I hope you are sketching all this.)

From the gradient of y = (-1/2)x - 5 we know the angle this line makes with the x axis.

m = -1/2

tan e = -1/2

e = arctan(-1/2)

Since we only care about the magnitude of e we fiddle the math a bit.

e = |arctan(-1/2)|

e = arctan(1/2)

The angle the line y = (-1/2)x - 5 makes with the x axis is arctan(1/2).

Now that we know the angle that our once rotated line makes with the x axis (d) and the angle the line y = (-1/2)x - 5 makes with the x axis (e), we can find the angle between these two lines.

f = d + e

Since f is the angle our once rotated “triangle” line makes with our second line of reflection, the total rotation angle for the second reflection is twice that.

g = 2 f

Now we have the two angles of rotation due to the two reflections, we only have to add them together to get the total angle of rotation.

r = c + g

It was a lot easier with just a couple of scribble pictures - harder to describe in words.

Substituting to get the total rotation r.

r = c + g

r = 2b + 2f

r = 2(π/2 – a) + 2 (d + e)

r = π – 2a + 2d + 2e

r = π – 2 arctan 2 + 2(π/2 - c) + 2 arctan(1/2)

r = π – 2 arctan 2 + π - 2c + 2 arctan(1/2)

r = 2π – 2 arctan 2 – 2(2b) + 2 arctan(1/2)

r = 2π – 2 arctan 2 – 4b + 2 arctan(1/2)

r = 2π – 2 arctan 2 – 4(π/2 – a) + 2 arctan(1/2)

r = – 2 arctan 2 + 4a + 2 arctan(1/2)

r = – 2 arctan 2 + 4 arctan 2 + 2 arctan(1/2)

r = 2 arctan 2 + 2 arctan(1/2)

r = 2 (arctan 2 + arctan(1/2))

r = 2 (π/2)

r = π

After having done all this. Couldn’t it be solved easier by treating all the lines as if they pass through the origin. You could perhaps use transformation matrices for reflections about lines through the origin. These are easy to construct and multiply. Hmmmm.

See "Reflection across a line of given slope" at "http://planetmath.org/encyclopedia/DerivationOf2DR... .

Can also be done geometrically without the matrices (just angles). Like the method above but all the lines through the origin. Remember only the gradients are important.

Even better. The lines y = 2x and y = (-1/2)x are at right angles to each other. Graphically we can see that two reflections about two lines at right angles to each other will give a resultant rotation of π.

Now prove it.

Forgive me if I have made an error.

Thanks for the practice.

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