Transformations and Invariant Points

Reflection in this axis keeps the x-coordinate unchanged.

1. The point is P=(3,2).
2. Reflection in the x-axis changes the sign of the y-coordinate, giving (3,-2).

Add the horizontal and vertical components separately.

1. Start at P=(-2,1).
2. The image is (-2+4,1-3)=(2,-2).

A quarter-turn anticlockwise takes rightwards to upwards.

1. The point is (2,-3).
2. A 90\degree anticlockwise turn sends (x,y) to (-y,x), so the image is (3,2).

Corresponding points lie at equal perpendicular distances on opposite sides of the mirror line.

1. The vertices at x=-3 and x=5 have midpoint x=1.
2. The other pair at x=-1 and x=3 has the same midpoint.
3. The mirror line is x=1.

An invariant point stays in exactly the same position.

1. Points on the mirror line remain fixed.
2. Only A=(-3,3) lies on y=-x.

The centre is the midpoint of a point and its half-turn image.

1. Here P=(3,2) and C=(1,-1).
2. The vector from C to P is (2,3); the image has vector (-2,-3).
3. The image is (1-2,-1-3)=(-1,-4).

Consider the distance of a fixed point from the enlargement centre.

1. The centre does not move because its distance from itself is zero.
2. Every other point has its nonzero distance from C multiplied by 3.
3. The only invariant point is C=(-2,1).

The same nonzero movement is applied to every point.

1. Every point moves by the same nonzero vector.
2. No point in the plane remains at its original position.

Check which original vertices lie on the mirror line.

1. The vertices A and B lie on the x-axis, so they stay fixed.
2. C moves below the axis. There are 2 invariant vertices.

A shape covering its original position does not mean that each vertex stays fixed.

1. Each vertex moves to the position previously occupied by a different vertex.
2. The centre is fixed, but it is not a vertex. There are 0 invariant vertices.

Two reflections in parallel lines give a translation across twice the gap.

1. Follow one point through both reflections. Take (0,0).
2. Reflect in L_1:x=-1: the origin is 1 unit right of the line, so it goes to 1 unit left of it, (-2,0).
3. Reflect in L_2:x=2: (-2,0) is 4 units left of the line, so it goes to 4 units right of it, (6,0).
4. The point has moved 6 units right, which is twice the 3-unit gap from L_1 to L_2. Every point moves the same way, so the single transformation is the translation \binom60.
5. \binom30 is the common slip of using the gap between the lines only once.

Track both signs of a general point through the two reflections.

1. The first reflection sends (x,y) to (x,-y).
2. The second sends this to (-x,-y), which is a 180\degree rotation about the origin.

Apply the transformations in the stated order.

1. Start at P=(-2,3).
2. Reflection in y=x swaps coordinates, giving (3,-2).
3. Reflecting the intermediate point (3,-2) in the x-axis gives the final point (3,2).

Undo the translation first, then undo the reflection.

1. The final point is Q=(-2,4).
2. Undoing the translation gives (-2+1,4-3)=(-1,1).
3. Reflection in x=2 sends x=-1 to x=5, so P=(5,1).

Work out where a general point (x,y) ends up, then ask when it ends up where it started.

1. A 180\degree rotation about the origin sends (x,y) to (-x,-y).
2. The translation then subtracts 4 from the x-coordinate and adds 6 to the y-coordinate, giving (-x-4,\,-y+6).
3. An invariant point ends up where it started, so -x-4=x and -y+6=y.
4. 2x=-4 gives x=-2, and 2y=6 gives y=3.
5. Check: (-2,3) rotates to (2,-3), then translates to (2-4,\,-3+6)=(-2,3). The invariant point is (-2,3).

Trace the origin and use the fact that these mirror lines are parallel.

1. Two reflections in parallel lines give a translation.
2. The origin lies on L_1, so the first reflection leaves it at (0,0).
3. Its reflection in y=x+2 is (-2,2), with midpoint (-1,1) on the mirror line.
4. The translation is therefore \binom{-2}{2}.

For a fixed point, the final x-coordinate must equal the original one.

1. The reflection sends (x,y) to (-x,y), then the translation gives (6-x,y).
2. A point is fixed when 6-x=x, so x=3.
3. Every point on x=3 is invariant.

Follow one point through both rotations and see how far it moves.

1. Follow one point through both rotations, using vectors from each centre. Take P=(0,3), with C=(1,1) and D=(3,1).
2. From C to P is \binom{-1}{2}. A 90\degree clockwise turn sends \binom{a}{b} to \binom{b}{-a}, giving \binom{2}{1}. So P moves to (1+2,\,1+1)=(3,2).
3. From D to (3,2) is \binom{0}{1}. A 90\degree anticlockwise turn sends \binom{a}{b} to \binom{-b}{a}, giving \binom{-1}{0}. So the point moves to (3-1,\,1+0)=(2,1).
4. Overall P moved from (0,3) to (2,1): 2 right and 2 down.
5. The clockwise and anticlockwise quarter-turns cancel, so the shape ends up facing the same way: the single transformation is the translation \binom{2}{-2}. (Following C gives the same: (1,1) stays put, then goes to (3,-1).)

Find where the two mirror lines cross, and follow the origin through both reflections.

1. The mirror lines cross where x=-x+4, so x=2 and y=2: at (2,2).
2. (2,2) lies on both mirror lines, so neither reflection moves it. It is invariant.
3. Follow the origin: it lies on y=x, so the first reflection leaves it at (0,0). Reflecting in y=-x+4 sends it along y=x to (4,4), with (2,2) halfway.
4. So the combined transformation is a rotation of 180\degree about (2,2) (reflections in two perpendicular lines always give a half-turn about the crossing point). Every other point moves, so (2,2) is the only invariant point.

Work out the combined effect before checking the original vertices.

1. The rotation sends (x,y) to (-x,-y).
2. Reflection in the y-axis then gives (x,-y), equivalent to reflection in the x-axis.
3. Only original vertices on y=0 remain fixed: A and D.