Math 8 · 6.NS.C.6B · 8.G.A.1

Reflections

A figure on graph paper is flipped over a line. Where exactly does each corner land, and if you see only the two figures, how do you find the line?

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When you reach for it

A reflection flips a figure over a line, the mirror or line of reflection, the way a shape drawn in wet ink prints onto the other half of a folded page. On plain paper you can fold and trace. On the coordinate plane you can do better: name exactly where every point lands, as a pair of numbers, with no folding and no guessing.

Reach for this move in two situations. The first is going forwards: a figure and a mirror are given, and you want the image. When the mirror is one of the axes, a vertical or horizontal line, or one of the two diagonals y = x and y = -x, the image comes from a one-line rule on the coordinates. The second is going backwards: a figure and its image are given, and you want the mirror, or want to know whether there is one at all. Both rest on the single fact that defines a reflection, which comes from the perpendicular bisector.

The procedure

What a reflection does to one point

Fix a line \ell as the mirror. The reflection in \ell sends a point P to its image P' like this.

If P is on \ell, it stays where it is: P' = P.
If P is not on \ell, then P' is the point that makes \ell the perpendicular bisector of \overline{PP'}. So the segment \overline{PP'} crosses the mirror at a right angle, and the mirror cuts it at its midpoint.

Put another way: drop the perpendicular from P to the mirror, and carry on past the mirror by the same distance again. That is the whole definition, and every rule below is this definition worked out for one particular mirror. The precise treatment, and the proof that a reflection keeps every length and every angle, is in rigid motions defined precisely.

Over the x-axis

Take the point P(x, y). The line through P perpendicular to the x-axis is vertical, so P' is straight above or below P, and its first coordinate is still x. P is at height y, a distance |y| from the axis on one side. P' must be the same distance away on the other side, at height -y. So

(x, y) \mapsto (x, -y)

Check it against the definition. The midpoint of \overline{PP'} is \displaystyle \left(x, \frac{y + (-y)}{2}\right) = (x, 0), which is on the axis, and the segment is vertical, so it meets the horizontal axis at a right angle. A point already on the axis has y = 0, and the rule leaves it alone, since -0 = 0, exactly as the definition requires.

Over the y-axis

The same argument with the roles of the coordinates swapped. The perpendicular to the y-axis is horizontal, so the height y is kept, and the point crosses to the other side at the same distance:

(x, y) \mapsto (-x, y)

In both cases the coordinate that changes sign is the one that measures distance from the mirror. Over the x-axis that distance is the height, so y changes; over the y-axis it is the distance left or right, so x changes.

Over a vertical or horizontal line

Let the mirror be the vertical line x = a. The perpendicular is again horizontal, so y is kept. For the first coordinate, the mirror is at the midpoint of \overline{PP'}, so a is the average of x and the new coordinate x':

\frac{x + x'}{2} = a \qquad\Longrightarrow\qquad x' = 2a - x

The same answer comes from distances. P is x - a to the right of the mirror (a negative amount means it is to the left), and P' is the same amount to the left, at a - (x - a) = 2a - x. For a horizontal mirror y = b the argument runs the other way round. So

\begin{aligned} x = a&: \quad (x, y) \mapsto (2a - x,\ y) \\ y = b&: \quad (x, y) \mapsto (x,\ 2b - y) \end{aligned}

The axes are the cases a = 0 and b = 0, where 2a - x becomes -x and 2b - y becomes -y.

Over the line y = x

This mirror runs at 45^\circ through the origin, and the rule for it is a swap:

(x, y) \mapsto (y, x)

Here the rule is easier to check than to discover, so check it against the definition. Take P(x, y) and Q(y, x). The midpoint of \overline{PQ} is

\left(\frac{x + y}{2},\ \frac{x + y}{2}\right)

and its two coordinates are equal, so it lies on the mirror y = x. If x \neq y, the Slope of \overline{PQ} is

\frac{x - y}{y - x} = -1

and the mirror has slope 1. The product of the slopes is -1, so the segment is perpendicular to the mirror, because perpendicular lines have slopes multiplying to -1. Both conditions of the definition hold, so Q is the image of P. If x = y, then P is on the mirror, and the swap leaves it alone, as it should.

Over the line y = -x

The other diagonal also has a short rule: swap the coordinates and change both signs.

(x, y) \mapsto (-y, -x)

The check is the same. The midpoint of \overline{PQ}, for Q(-y, -x), is \displaystyle \left(\frac{x - y}{2},\ \frac{y - x}{2}\right), whose second coordinate is minus its first, so it lies on y = -x. When x + y \neq 0 the slope of \overline{PQ} is \displaystyle \frac{-x - y}{-y - x} = 1, and the mirror has slope -1, so they are perpendicular.

All six rules side by side:

Mirror	(x, y) goes to
the x-axis	(x, -y)
the y-axis	(-x, y)
the vertical line x = a	(2a - x,\ y)
the horizontal line y = b	(x,\ 2b - y)
the line y = x	(y, x)
the line y = -x	(-y, -x)

To reflect a whole polygon, reflect its vertices and join the images in the same order. A reflection carries segments to segments of the same length, so the edges come along with their endpoints.

Finding the line of reflection

Now go backwards. A point P and its image P' are given, with P' \neq P. By the definition, the mirror is the perpendicular bisector of \overline{PP'}, and a segment has exactly one perpendicular bisector. So one pair pins the mirror down: it passes through the midpoint of \overline{PP'} and is perpendicular to it.

Two quick cases come up often. If P and P' have the same y, the segment is horizontal and the mirror is the vertical line x = (the average of the two x values). If they have the same x, the mirror is the horizontal line y = (the average of the two y values).

For a whole figure, find the mirror from one pair of matching vertices and test it on the others. Every pair must give the same line, because one mirror does all the work. If two pairs give different lines, the figure was not reflected at all, whatever it looks like.

Reflections turn a figure over

A reflection reverses the direction you travel round a figure. If the vertices A, B, C run counterclockwise, their images A', B', C' run clockwise, just as a right hand in a mirror looks like a left hand. A translation or a rotation keeps the direction. So when the matching vertices are known, this gives a second test: an image that runs round the same way as the original cannot have come from a single reflection.

Worked examples

Example 1

Reflecting a triangle over each axis

Triangle ABC has vertices A(1, 2), B(5, 1) and C(4, 5). Reflect it over the x-axis, calling the image A_1B_1C_1, and over the y-axis, calling the image A_2B_2C_2.

Solution

Over the x-axis, each point keeps its x and its y changes sign, by the rule (x, y) \mapsto (x, -y):

A_1(1, -2), \qquad B_1(5, -1), \qquad C_1(4, -5)

Over the y-axis, each point keeps its y and its x changes sign:

A_2(-1, 2), \qquad B_2(-5, 1), \qquad C_2(-4, 5)

Figure 1The triangle of Example 1 and its two mirror images. Each dashed segment crosses its axis at a right angle and is cut in half by it.

Check one vertex against the definition. The midpoint of \overline{AA_1} is (1, 0), on the x-axis, and \overline{AA_1} is vertical, so the axis is its perpendicular bisector. Check one length with the distance formula: \displaystyle AB = \sqrt{4^2 + 1^2} = \sqrt{17} and \displaystyle A_1B_1 = \sqrt{4^2 + 1^2} = \sqrt{17}, since the change in x is still 4 and the change in y has only changed sign.

Notice too that A \to B \to C runs counterclockwise round its triangle, while A_1 \to B_1 \to C_1 runs clockwise. The image is the same triangle, turned over.

Example 2

A segment that crosses the mirror

The segment \overline{PQ} runs from P(-3, 4) to Q(2, -1). Reflect it over the x-axis, and find where the segment and its image each cross the axis.

Solution

Reflect the endpoints: P'(-3, -4) and Q'(2, 1). The image is the segment \overline{P'Q'}.

The original has slope \displaystyle \frac{-1 - 4}{2 - (-3)} = -1, so it lies on y = -x + 1, which meets the axis where 0 = -x + 1, at (1, 0). The image has slope \displaystyle \frac{1 - (-4)}{2 - (-3)} = 1, so it lies on y = x - 1, which meets the axis at (1, 0) too.

They cross the axis at the same point, and that is no accident. The point (1, 0) is on the mirror, so the reflection leaves it where it is; it is on \overline{PQ}, so its image is on \overline{P'Q'}; and its image is itself. The slope changed sign because the reflection changes the sign of every rise and keeps every run.

Example 3

A vertical mirror and a horizontal mirror

Triangle ABC has vertices A(-2, 1), B(0, 4) and C(1, -1).

(a) Reflect it in the line x = 3. \qquad (b) Reflect it in the line y = -1.

Solution

(a) The mirror is vertical, so each point moves horizontally and keeps its y. The new x is 2 \cdot 3 - x = 6 - x:

A'(8, 1), \qquad B'(6, 4), \qquad C'(5, -1)

Figure 2The reflection in x = 3 from Example 3. Each point keeps its height and ends up as far right of the mirror as it was left of it.

Check A by distances: it is 3 - (-2) = 5 units left of the mirror, and A' is 5 units right of it, at 3 + 5 = 8.

(b) The mirror is horizontal, so each point moves vertically and keeps its x. The new y is 2(-1) - y = -2 - y:

\begin{aligned} A(-2, 1) &\mapsto (-2, -3) \\ B(0, 4) &\mapsto (0, -6) \\ C(1, -1) &\mapsto (1, -1) \end{aligned}

The last one is C itself. C has y = -1, so it lies on the mirror, and the rule, which knows nothing about that, gives -2 - (-1) = -1 and leaves it in place, exactly as the definition says it must.

Example 4

Finding a vertical or horizontal mirror

(a) A reflection sends triangle A(-4, 1), B(-1, 2), C(-2, 5) to A'(6, 1), B'(3, 2), C'(4, 5), in that order. Find the line of reflection and the rule.

(b) A reflection sends P(-3, 8) to P'(-3, -2). Find the line of reflection.

Solution

(a) A and A' have the same y, so \overline{AA'} is horizontal and the mirror, being perpendicular to it, is vertical. It passes through the midpoint, whose x is \displaystyle \frac{-4 + 6}{2} = 1. Test the other pairs: \displaystyle \frac{-1 + 3}{2} = 1 and \displaystyle \frac{-2 + 4}{2} = 1, and both pairs share a y as well. All three agree, so the mirror is

x = 1

and the rule is (x, y) \mapsto (2 - x,\ y), from 2a - x with a = 1.

(b) P and P' have the same x, so the segment is vertical and the mirror is horizontal, through the midpoint’s height \displaystyle \frac{8 + (-2)}{2} = 3. The mirror is y = 3, and P is 5 above it while P' is 5 below it.

Example 5

Over the two diagonals

Triangle DEF has vertices D(2, 1), E(5, 2) and F(4, 0).

(a) Reflect it in the line y = x. \qquad (b) Reflect it in the line y = -x.

Solution

(a) The rule is the swap (x, y) \mapsto (y, x):

D'(1, 2), \qquad E'(2, 5), \qquad F'(0, 4)

Figure 3The reflection in y = x from Example 5. Every dashed segment has slope -1, and the mirror cuts each one in half.

Check E. The midpoint of \overline{EE'} is \displaystyle \left(\frac{7}{2}, \frac{7}{2}\right), on the mirror, and the slope of \overline{EE'} is \displaystyle \frac{5 - 2}{2 - 5} = -1, perpendicular to the mirror’s slope 1. Notice that F is on the x-axis and F' is on the y-axis. The mirror y = x swaps the two axes.

(b) The rule is (x, y) \mapsto (-y, -x): swap, then change both signs. For D(2, 1) that gives (-1, -2), for E(5, 2) it gives (-2, -5), and for F(4, 0) it gives (0, -4).

Example 6

Finding a slanted mirror

A reflection sends P(-2, 3), Q(0, 5), R(1, 2) to P'(4, -3), Q'(6, -1), R'(3, 0), in that order. Find the line of reflection. Is it y = x?

Solution

Use the first pair. The midpoint of \overline{PP'} is \displaystyle \left(\frac{-2 + 4}{2}, \frac{3 + (-3)}{2}\right) = (1, 0). The slope of \overline{PP'} is \displaystyle \frac{-3 - 3}{4 - (-2)} = -1, so the mirror has slope 1. The line of slope 1 through (1, 0) is

y = x - 1

Now test the other pairs, since one mirror has to serve all three. The midpoint of \overline{QQ'} is (3, 2), and 2 = 3 - 1; the slope of \overline{QQ'} is \displaystyle \frac{-1 - 5}{6 - 0} = -1. The midpoint of \overline{RR'} is (2, 1), and 1 = 2 - 1; the slope of \overline{RR'} is \displaystyle \frac{0 - 2}{3 - 1} = -1. Both pass.

Figure 4Example 6. The three midpoints all lie on one line, perpendicular to all three dashed segments, and that line is the mirror.

It is not y = x: that mirror would swap P(-2, 3) to (3, -2), not (4, -3). The mirror is the line y = x moved down 1, and the rule shows how. Slide everything up 1, which carries the mirror onto y = x; swap; then slide back down 1:

\begin{aligned} (x, y) &\mapsto (x,\ y + 1) \\ &\mapsto (y + 1,\ x) \\ &\mapsto (y + 1,\ x - 1) \end{aligned}

That rule sends P(-2, 3) to (4, -3), Q(0, 5) to (6, -1) and R(1, 2) to (3, 0), as given.

Example 7

When there is no mirror

A triangle A(3, 1), B(5, 4), C(1, 4) is moved to A'(-3, -1), B'(-5, -4), C'(-1, -4), in that order. Is this a reflection?

Solution

If it were, the mirror would be the perpendicular bisector of \overline{AA'}. Its midpoint is (0, 0) and its slope is \displaystyle \frac{-1 - 1}{-3 - 3} = \frac{1}{3}, so the only candidate is the line of slope -3 through the origin, y = -3x.

Test it on B. The midpoint of \overline{BB'} is (0, 0), which is on y = -3x, but the slope of \overline{BB'} is \displaystyle \frac{-4 - 4}{-5 - 5} = \frac{4}{5}, not \displaystyle \frac{1}{3}, so \overline{BB'} does not meet y = -3x at a right angle. The only possible mirror fails, so this is not a reflection.

Every point went to (-x, -y), the half turn about the origin, which is a rotation. The direction test agrees: A \to B \to C runs counterclockwise, and so does A' \to B' \to C'. A reflection would have reversed it.

Example 8

The image of a line

Find an equation of the image of the line y = -2x + 6 under the reflection in y = x.

Solution

A point (a, b) is on the image exactly when the point it came from, (b, a), is on the original line, that is, when a = -2b + 6. So the image is the line x = -2y + 6: the original equation with x and y swapped. Solve for y:

2y = 6 - x \qquad\Longrightarrow\qquad y = -\frac{1}{2}x + 3

Check with two points. The original meets the axes at (3, 0) and (0, 6). Their images (0, 3) and (6, 0) both satisfy \displaystyle y = -\frac{1}{2}x + 3. The original also crosses the mirror where x = -2x + 6, at (2, 2), and so does the image: -1 + 3 = 2.

Swapping x and y in an equation is exactly how an inverse function is found, so this picture is the graph of an inverse: the inverse of f(x) = -2x + 6 is \displaystyle -\frac{1}{2}x + 3, and its graph is the mirror image of the graph of f in y = x.

Example 9

The shortest path by way of a line

Points A(1, 3) and B(7, 5) lie above the x-axis. A path runs from A to some point P on the axis, then straight on to B. Where should P be to make AP + PB as short as possible, and how short is it?

Solution

Reflect A over the axis to A'(1, -3). For any P on the axis, P is on the mirror, so it is its own image, and the reflection carries \overline{AP} onto \overline{A'P}. Reflections keep lengths, so AP = A'P and

AP + PB = A'P + PB

Figure 5Example 9. The path through P is as long as the straight segment from A' to B, and no path through any other point of the axis is shorter.

The right side is a path from A' to B with a bend at P. By the triangle inequality it is at least the straight distance A'B, and equals it exactly when P lies on the segment \overline{A'B}. So the best P is where \overline{A'B} crosses the axis. Its slope is \displaystyle \frac{5 - (-3)}{7 - 1} = \frac{4}{3}, so it lies on \displaystyle y + 3 = \frac{4}{3}(x - 1). At y = 0, \displaystyle x - 1 = \frac{9}{4}, so the best point is

P\left(\frac{13}{4},\ 0\right)

and the shortest total is

AP + PB = A'B = \sqrt{6^2 + 8^2} = 10

Check directly. \displaystyle AP = \sqrt{\left(\frac{9}{4}\right)^2 + 3^2} = \frac{15}{4} and \displaystyle PB = \sqrt{\left(\frac{15}{4}\right)^2 + 5^2} = \frac{25}{4}, which add to 10.

Boundaries

Each rule belongs to one mirror. The swap (x, y) \mapsto (y, x) is the reflection in y = x and in no other line; Example 6 shows the line y = x - 1 needs a different rule. For a mirror that is not one of the six in the table, go back to the definition: the image is found from the perpendicular through the point and the midpoint, as in rigid motions defined precisely.

Points on the mirror stay put, and no others do. A point off the mirror is a positive distance from it, and its image is the same distance on the other side, so it moves.

One pair finds the mirror only if the point moved. If P' \neq P, the mirror is the one perpendicular bisector of \overline{PP'}. If P' = P, the pair says only that the mirror passes through P, and infinitely many lines do.

A figure needs every pair to agree. One pair always produces a candidate line. The figure has been reflected only if every other matching pair has that same line as its perpendicular bisector, as Example 7 shows. The direction test needs the matching vertices too: which corner went where.

The diagonal rules need the same scale on both axes. The algebra of (x, y) \mapsto (y, x) holds on any graph, but if one unit across is drawn longer than one unit up, the line y = x is no longer drawn at 45^\circ and the picture stops looking like a mirror image. The geometry of perpendiculars and equal distances assumes the plane is drawn true.

A mirror is a line. Flipping through a point, (x, y) \mapsto (-x, -y), is a half turn, not a reflection, as Example 7 found.

Where this goes next

A reflection is the most basic rigid motion: two reflections in a row make a translation when the mirrors are parallel and a rotation when they cross, and every rigid motion of the plane can be built from at most three. Doing one reflection after another is composing transformations, and the order can matter. A figure that a reflection carries onto itself has line symmetry, and the mirror is its line of symmetry.

Because reflections keep lengths and angles, two figures are congruent when slides, turns and flips carry one onto the other, and what rigid motions preserve says exactly what survives.

On graphs, the same rules act on whole curves. Replacing y by -y flips a graph over the x-axis, which is reflecting a graph across the axes; a graph that a flip over the y-axis leaves unchanged has symmetry about the vertical axis; and swapping x and y gives the inverse function, whose graph is the mirror image in y = x.

Exercises

Over the axes

Reflect P(3, -7) over the x-axis.
Reflect Q(-6, 2) over the y-axis.
Triangle JKL has vertices J(2, 0), K(-3, 4) and L(5, -2). Reflect it over the y-axis.
Which points are their own image under the reflection over the x-axis?

Over vertical and horizontal lines

Reflect (6, 1) in the line x = 2.
Reflect (-3, 5) in the line y = 2.
Reflect (4, -2) in the line x = -1.
A rectangle has vertices (-1, 2), (3, 2), (3, 4) and (-1, 4). Reflect it in the line y = -2.
Write the rule for the reflection in the line x = 5, and say why it is right.

Over the diagonals

Reflect (7, -2) in the line y = x.
Reflect (7, -2) in the line y = -x.
A triangle has vertices (0, 3), (3, 6) and (-1, 4). Reflect it in the line y = x.
Reflect (-4, 4) in the line y = -x, and explain the answer.

Finding the line of reflection

A reflection sends A(-5, 2) to A'(3, 2). Find the line of reflection.
A reflection sends B(4, 6) to B'(4, -2). Find the line of reflection.
A reflection sends C(-1, 6) to C'(6, -1). Find the line of reflection.
A reflection sends D(0, -3) to D'(4, 1). Find the line of reflection.
A triangle A(1, 1), B(2, 4), C(4, 2) is moved to A'(-1, 1), B'(-2, 4), C'(-4, -2), in that order. Is this a reflection?

Problems

A straight river runs along the x-axis, with distances in kilometers. A walker goes from a house at A(2, 4) to the river and then to a barn at B(10, 2). At which point of the river should she stop to make the walk as short as possible, and how long is that walk?
Find an equation of the image of the line y = 3x - 6 under the reflection over the y-axis.
Find an equation of the image of the line y = 2x + 4 under the reflection in y = x.

Reasoning

Show that the reflection over the x-axis carries the vertical line x = c onto itself, and find which of its points stay put.
Show that the reflection in the line x = a keeps the distance between any two points.
Show that the line y = -x is the perpendicular bisector of the segment from (x, y) to (-y, -x) whenever x + y \neq 0, and say what happens when x + y = 0.
Explain why one point P and its image P' with P' \neq P are enough to find the mirror, but a point that is its own image is not.

Answers

(3, 7). The x stays, and the y changes sign.
(6, 2). The y stays, and the x changes sign.
J'(-2, 0), K'(3, 4), L'(-5, -2), by (x, y) \mapsto (-x, y).
Exactly the points of the x-axis, those of the form (x, 0). The image (x, -y) equals (x, y) only when -y = y, that is, y = 0.
(-2, 1). The new x is 2 \cdot 2 - 6 = -2, and y stays.
(-3, -1). The new y is 2 \cdot 2 - 5 = -1, and x stays.
(-6, -2). The new x is 2(-1) - 4 = -6.
(-1, -6), (3, -6), (3, -8), (-1, -8). The new y is 2(-2) - y = -4 - y, so a height of 2 goes to -6 and a height of 4 goes to -8.
(x, y) \mapsto (10 - x,\ y). The perpendicular to a vertical mirror is horizontal, so y stays, and the mirror is at the midpoint, so \displaystyle \frac{x + x'}{2} = 5 and x' = 10 - x.
(-2, 7), by the swap.
(2, -7). Swap and change both signs: (-(-2),\ -7).
(3, 0), (6, 3), (4, -1), by swapping each pair.
(-4, 4), the point itself. The rule gives (-4, -(-4)) = (-4, 4). The point is on the mirror, since 4 = -(-4), and points on the mirror stay put.
x = -1. The two points share a y, so the mirror is vertical, at \displaystyle x = \frac{-5 + 3}{2} = -1.
y = 2. The two points share an x, so the mirror is horizontal, at \displaystyle y = \frac{6 + (-2)}{2} = 2.
y = x. The midpoint is \displaystyle \left(\frac{5}{2}, \frac{5}{2}\right) and the slope of \overline{CC'} is \displaystyle \frac{-1 - 6}{6 - (-1)} = -1, so the mirror has slope 1 through the midpoint. The coordinates were swapped, as the rule for y = x says.
y = -x + 1. The midpoint is (2, -1) and the slope of \overline{DD'} is \displaystyle \frac{1 - (-3)}{4 - 0} = 1, so the mirror has slope -1: y + 1 = -(x - 2).
No. \overline{AA'} and \overline{BB'} both have perpendicular bisector x = 0, but \overline{CC'} has midpoint (0, 0) and slope \displaystyle \frac{1}{2}, so its perpendicular bisector is y = -2x. No single line serves all three. The lengths say the same: \displaystyle AC = \sqrt{10} but \displaystyle A'C' = \sqrt{18}, and a reflection keeps every length.
At \displaystyle \left(\frac{22}{3}, 0\right), and the walk is 10 km. Reflect A to A'(2, -4). The segment from A' to B has slope \displaystyle \frac{6}{8} = \frac{3}{4} and meets the axis where \displaystyle x - 2 = \frac{16}{3}. The walk equals \displaystyle A'B = \sqrt{8^2 + 6^2} = 10.
y = -3x - 6. A point (x, y) is on the image when (-x, y) is on the original, so y = 3(-x) - 6.
\displaystyle y = \frac{1}{2}x - 2. Swap x and y: x = 2y + 4, so \displaystyle y = \frac{x - 4}{2}.
Only (c, 0) stays put. A point (c, y) of the line goes to (c, -y), whose first coordinate is still c, so every image is on the line. Every point (c, y) is also the image of (c, -y), which is on the line, so the image is the whole line. A point stays put only when -y = y, that is y = 0, where the line crosses the mirror. The line lands on itself, but flipped end for end about (c, 0).
The points (x_1, y_1) and (x_2, y_2) go to (2a - x_1, y_1) and (2a - x_2, y_2). The change in x becomes (2a - x_2) - (2a - x_1) = -(x_2 - x_1) and the change in y is unchanged. Squaring removes the sign, so the distance formula gives the same distance.
The midpoint is \displaystyle \left(\frac{x - y}{2}, \frac{y - x}{2}\right), whose second coordinate is minus its first, so it is on y = -x. The slope of the segment is \displaystyle \frac{-x - y}{-y - x} = 1, and 1 \cdot (-1) = -1, so the segment is perpendicular to the mirror. When x + y = 0, the point is on the mirror, and (-y, -x) is (x, y) itself, so there is no segment and the point stays put.
If P' \neq P, the mirror must be the perpendicular bisector of \overline{PP'}, and a segment has only one, so the line is fixed. If P' = P, the definition says only that P lies on the mirror, and every line through P passes that test.