A shape and its image
A transformation moves every point of a shape to a new position. The shape you start with is called the object, and the shape it becomes is called the image. If the object is named A, the image is named A', read "A prime".
A reflection is the transformation a mirror makes. It needs one straight line, the mirror line, and it carries every point of the object across that line. The image is the same size and shape as the object, so the two are congruent, but it faces the opposite way, as your reflection in a mirror does.
The gold line is the mirror line, here the y-axis. The triangle A has corners (1, 1), (4, 1) and (4, 3); its image A' has corners (−1, 1), (−4, 1) and (−4, 3).
Straight across, the same distance beyond
To reflect a point, go from it straight to the mirror line, meeting the line at a right angle, and carry on the same distance on the other side. That is where its image is.
For a reflection in the x-axis, the corner (4, 3) is 3 squares above the mirror line. Its image is 3 squares below it, at (4, −3). The point has moved 3 + 3 = 6 squares in all, straight down. A point that is on the mirror line is 0 from it, so it stays where it is.
So the segment from a point to its image crosses the mirror line at a right angle, and the mirror line cuts it into two equal halves. That makes the mirror line the perpendicular bisector of the segment joining any point to its image, the line built in Constructing a Perpendicular Bisector.
Reflected in the x-axis, each corner lands as far below the line as it was above it: (1, 1) to (1, −1), (4, 1) to (4, −1), and (4, 3) to (4, −3).
The axes as coordinate rules
The y-axis is vertical, so going straight across it means moving horizontally. The y-coordinate does not change. The distance from the y-axis is the size of the x-coordinate, and the image is that far on the other side, so the x-coordinate changes sign. (4, 3) lands on (−4, 3), and in general .
The x-axis is horizontal, so the move is vertical. The x-coordinate stays and the y-coordinate changes sign: . One reflection in an axis changes exactly one sign, the sign of the coordinate that measures distance from that axis.
The line y = x swaps the coordinates
The line y = x passes through (0, 0), (1, 1) and (2, 2), at 45° to each axis. Its reflection rule is : the two coordinates trade places. The two facts, a right angle and two equal halves, show why.
Take the point (4, 1) and the point (1, 4). The segment joining them has gradient . The line y = x has gradient 1, and 1 × (−1) = −1, so the segment is perpendicular to the mirror line. Its midpoint is , which is on y = x because its two coordinates are equal. So y = x is the perpendicular bisector of the segment, and (1, 4) is the image of (4, 1).
The same working holds for any point (a, b) and the point (b, a): the gradient between them is , and the midpoint has both coordinates equal to . A point already on the line, such as (3, 3), is its own image.
Reflected in y = x, the corners (2, 0), (4, 1) and (3, 3) land on (0, 2), (1, 4) and (3, 3). The corner (3, 3) is on the mirror line, so it does not move.
The line y = −x
The line y = −x passes through (0, 0), (1, −1) and (−1, 1). Its rule is : the coordinates trade places and both change sign. For (4, 1), the image is (−1, −4).
Check it with the same two facts. The segment from (4, 1) to (−1, −4) has gradient , and the mirror has gradient −1; 1 × (−1) = −1, so they are perpendicular. The midpoint is , whose y-coordinate is minus its x-coordinate, so it is on y = −x.
The gold line is the mirror y = −x. The white line passes through the point (4, 1) and its image (−1, −4), and it crosses the mirror at a right angle, at (1.5, −1.5), halfway between them.
The four reflection rules, each applied to the point (4, 1).
Finding the mirror line
When a point and its image are given, the mirror line is the perpendicular bisector of the segment between them. Take P(1, 1) and its image P'(5, 3).
The midpoint of PP′ is , and the mirror line passes through it. The gradient of PP′ is , so the mirror line, perpendicular to it, has gradient −2, because . Through (3, 2) with gradient −2: y − 2 = −2(x − 3), which is y = −2x + 8.
Check with a point on the mirror line, such as (4, 0). Its distance from P is , and its distance from P' is . Every point of the mirror line is equally far from a point and its image. For a whole shape, one pair of matching points finds the line, and a second pair checks it.
The white line passes through P(1, 1) and its image P'(5, 3). The gold line, y = −2x + 8, crosses it at a right angle at the midpoint (3, 2): it is the mirror line.
The usual mistakes
Changing the wrong sign. The x-axis mirror moves points up and down, so it changes the sign of y: (3, 2) lands on (3, −2), not (−3, 2).
Changing both signs. (3, 2) to (−3, −2) is a half turn about the origin, not a reflection. A reflection in an axis changes one sign.
Changing a sign in y = x. The mirror y = x swaps the coordinates and changes no sign: (3, 2) lands on (2, 3).
Moving the wrong distance, or in the wrong direction. The image is as far beyond the mirror as the object is in front of it, not on the line and not twice as far away. And the move is at a right angle to the mirror: across a slanted mirror such as y = x, it is not along a grid line.
Worked example: A Strip of Paper Folded Off Center
Question A strip of paper is 12 cm long. Its left end is folded over to the right, so that the end lands 2 cm from the right end of the strip. (a) How far from the left end is the fold? (b) A hole is punched through both layers, 1 cm from the fold. When the strip is opened out, how far from the left end is each hole?
1.The left end lands 2 cm from the right end, so it lands 12 − 2 = 10 cm from where it started.
The left end lands 2 cm short of the right end: 12 − 2 = 10 cm from where it started. 2.The end is the same distance from the fold before and after folding, so the fold is halfway along those 10 cm.
The end is as far from the fold before folding as after, so the fold is halfway along the 10 cm. 3.(a) Half of 10 cm is 10 ÷ 2 = 5 cm, so the fold is 5 cm from the left end.
(a) The fold is 10 ÷ 2 = 5 cm from the left end. 4.The hole is 1 cm from the fold. When the strip is opened, one hole is 1 cm to the right of the fold and its twin is 1 cm to the left of it.
Opened, the hole 1 cm right of the fold has a twin 1 cm left of it. 5.(b) The holes are 5 + 1 = 6 cm and 5 − 1 = 4 cm from the left end. Check: the part from 0 cm to 5 cm folds onto the part from 5 cm to 10 cm, so the point 4 cm from the left end lands exactly on the point 6 cm from it.
(b) The holes are 4 cm and 6 cm from the left end.
Answer: (a) 5 cm; (b) 4 cm and 6 cm
Common mistakes
- Answering 6 cm, the middle of the strip. The fold would be in the middle only if the left end landed on the right end. Here the end lands 2 cm short of it, so the fold is halfway along the 10 cm the end moves: 5 cm.
- Answering 10 cm, the place where the end lands. The end is 5 cm before the fold when it starts and 5 cm past the fold when it lands, so the fold is only half of the 10 cm from the left end.
Worked example: A Bank Shot on a Pool Table: Where to Strike the Cushion
Question A plan of a pool table is drawn on a grid in which one unit is 25 cm. The cushions are the lines x = 0, x = 10, y = 0 and y = 5. The cue ball is at C(1, 2) and a target ball is at T(9, 4). A player plans to roll the cue ball against the cushion y = 5 so that it rebounds onto the target. Assume the ball leaves the cushion at the same angle as it strikes it. (a) Find the coordinates of T', the reflection of the target ball in the cushion y = 5. (b) Find the point on the cushion that the cue ball must strike, and the distance, in meters, that the cue ball rolls to the target.
1.(a) The target T(9, 4) is 1 unit below the cushion y = 5, so its reflection is 1 unit above it, at T'(9, 6).
(a) T(9, 4) is 1 unit below the cushion, so its reflection is T'(9, 6), 1 unit above it. 2.The straight line from C(1, 2) to T'(9, 6) rises 4 for 8 across, so its gradient is 12 and its equation is y − 2 = 12(x − 1).
The straight line from C(1, 2) to T'(9, 6) has gradient 12: y − 2 = 12(x − 1). 3.The line meets the cushion where y = 5: 5 − 2 = 12(x − 1), so x − 1 = 6 and x = 7. The cue ball must strike the cushion at P(7, 5).
It crosses the cushion y = 5 where x = 7: the cue ball must strike the cushion at P(7, 5). 4.The path from C to P to T is as long as the straight line CT', which is √82 + 42 = √80 ≈ 8.944 units. At 25 cm to a unit, that is 8.944 × 0.25 ≈ 2.24 m.
The path C to P to T is as long as CT' = √80 ≈ 8.944 units, which is 8.944 × 0.25 ≈ 2.24 m. 5.(b) The cue ball must strike the cushion at (7, 5), and it rolls about 2.24 m. Check: from C to P the ball goes 6 across and 3 up, and from P to T it goes 2 across and 1 down, so both parts of the path slope at 1 in 2 and make the same angle with the cushion.
(b) Strike the cushion at (7, 5); the ball rolls about 2.24 m, and both parts of its path slope at 1 in 2.
Answer: (a) T'(9, 6); (b) the point (7, 5) on the cushion, and the cue ball rolls √80 units, about 2.24 m
Common mistakes
- Aiming at the point of the cushion straight above the target, (9, 5). The ball arrives there moving to the right and rebounds still moving to the right, so it leaves the cushion heading away from T, which is straight below (9, 5).
- Reflecting the target by changing the sign of its y-coordinate, to (9, −4). That is a reflection in the line y = 0, the opposite cushion; the reflection in y = 5 puts the image as far above y = 5 as T is below it.