📚 Transformations: Translation, Reflection, Rotation and Enlargement | 几何变换：平移、反射、旋转与缩放
Transformations are one of the most visual and frequently tested topics in IGCSE Mathematics. A transformation is a rule that moves every point of a shape to a new position, producing an image. In this article, we will cover the four core transformations you need to master: translation, reflection, rotation and enlargement, along with their coordinate rules, notation and exam strategies.
几何变换是 IGCSE 数学中最直观且高频考查的考点之一。变换是一种规则，它将图形的每一个点移动到新位置，从而生成像图。本文将系统地讲解四种核心变换：平移、反射、旋转和缩放，包括坐标规则、记法和考试策略。
1. What is a Transformation? | 什么是几何变换？
A transformation is a function that maps a shape (the object) onto another shape (the image). The original shape is called the object, and the resulting shape is called the image. Transformations can be described using coordinates, vectors or simple verbal statements.
变换是一种将图形（原图形）映射到另一个图形（像图）的函数。原始图形称为原图形，变换后得到的图形称为像图。变换可以用坐标、向量或文字描述来表达。
There are four standard transformations in the IGCSE syllabus. Translation and rotation preserve both the size and shape of the object, while reflection preserves size but reverses orientation. Enlargement changes the size of the shape according to a scale factor.
IGCSE 考纲中有四种标准变换。平移和旋转保持图形的大小和形状不变；反射保持大小但会反转方向；缩放则根据比例因子改变图形的大小。
You should always be able to describe a transformation fully. This means stating the type of transformation and all the necessary details: the vector for translation, the equation of the mirror line for reflection, the centre, angle and direction for rotation, and the scale factor and centre for enlargement.
你应当始终能够完整地描述一个变换。这意味着要说明变换的类型以及所有必要细节：平移的向量、反射的镜轴方程、旋转的中心、角度和方向，以及缩放的比例因子和缩放中心。
2. Translation | 平移
A translation slides a shape from one position to another without rotating, reflecting or resizing it. Every point of the shape moves the same distance in the same direction. The movement is described by a column vector.
平移是将图形从一个位置滑动到另一个位置，不旋转、不反射、不改变大小。图形的每一个点都沿相同方向移动相同距离。这种移动用列向量表示。
Column vector: (a over b) means move a units horizontally and b units vertically
列向量：𝄆a, b𝄇 表示水平移动 a 个单位，垂直移动 b 个单位
The top number in the column vector represents the horizontal movement: a positive value means moving right, and a negative value means moving left. The bottom number represents the vertical movement: a positive value means moving up, and a negative value means moving down.
列向量中的上方数字代表水平移动：正值表示向右移动，负值表示向左移动。下方数字代表垂直移动：正值表示向上移动，负值表示向下移动。
For a point (x, y) translated by vector (a over b), the image point (x’, y’) is given by:
对于点 (x, y) 按向量 (a over b) 平移后，像点 (x’, y’) 的坐标为：
x’ = x + a, y’ = y + b
For example, the point (2, 3) translated by the vector (3, -1) becomes (5, 2). The triangle with vertices (1, 1), (4, 1) and (2, 3) translated by the same vector would form a congruent triangle at (4, 2), (7, 2) and (5, 4).
例如，点 (2, 3) 按向量 (3, -1) 平移后变为 (5, 2)。顶点为 (1, 1)、(4, 1) 和 (2, 3) 的三角形按同一向量平移后，会形成位于 (4, 2)、(7, 2) 和 (5, 4) 的全等三角形。
To find the translation vector that takes an object to its image, subtract the coordinates of a point on the object from the corresponding point on the image. For instance, if A(1, 2) maps to A'(4, 5), then the vector is (4-1 over 5-2) = (3 over 3).
要求将原图形移到像图的平移向量，用像图上对应点的坐标减去原图形上点的坐标。例如，若 A(1, 2) 映射到 A'(4, 5)，则向量为 (4-1 over 5-2) = (3 over 3)。
3. Reflection | 反射（轴对称）
A reflection produces a mirror image of a shape across a straight line called the mirror line. Each point of the image is the same perpendicular distance from the mirror line as the corresponding point of the object, but on the opposite side. The image is congruent to the object but its orientation is reversed.
反射是在一条称为镜像轴的直线上生成图形的镜像。像图上的每个点到镜像轴的垂直距离与原图形对应点到镜像轴的距离相等，但在镜像轴的另一侧。像图与原图形全等，但方向相反。
In IGCSE, you need to be able to reflect a shape in the x-axis, the y-axis, the line y = x, the line y = -x, and any vertical or horizontal line such as x = 2 or y = -1.
在 IGCSE 中，你需要掌握关于 x 轴、y 轴、直线 y = x、直线 y = -x 以及任意竖直或水平直线（如 x = 2 或 y = -1）的反射。
| Mirror line / 镜像轴 | Rule / 规则 |
| x-axis / x 轴 | (x, y) → (x, -y) |
| y-axis / y 轴 | (x, y) → (-x, y) |
| y = x | (x, y) → (y, x) |
| y = -x | (x, y) → (-y, -x) |
| x = a | (x, y) → (2a – x, y) |
| y = b | (x, y) → (x, 2b – y) |
Let us check the rule for the line x = a. A point (x, y) reflected in the vertical line x = a has the same y-coordinate, but its x-coordinate is transformed so that a is the midpoint of the original and image x-coordinates. Therefore x’ = 2a – x.
我们来验证直线 x = a 的反射规则。点 (x, y) 关于竖直直线 x = a 反射后，y 坐标保持不变，但 x 坐标会变为 x’ = 2a – x，使得 a 成为原坐标与像坐标的中点。
For example, the point (4, 3) reflected in the line x = 2 becomes (-0, 3) since 2 × 2 – 4 = 0. A triangle reflected in the y-axis will have its vertices’ x-coordinates multiplied by -1, so (3, 1) becomes (-3, 1).
例如，点 (4, 3) 关于直线 x = 2 反射后变为 (0, 3)，因为 2 × 2 – 4 = 0。三角形关于 y 轴反射时，各顶点的 x 坐标变为原来的相反数，因此 (3, 1) 变为 (-3, 1)。
When describing a reflection, you must always state the equation of the mirror line. Writing simply “reflection” is not enough to gain full marks.
在描述反射时，必须说明镜像轴的方程。只写”反射”是不足以获得满分的。
4. Rotation | 旋转
A rotation turns a shape around a fixed point, known as the centre of rotation. To describe a rotation fully, you need three pieces of information: the centre of rotation, the angle of rotation, and the direction (clockwise or anticlockwise).
旋转是图形绕一个固定点（称为旋转中心）转动。要完整描述一个旋转，你需要三个信息：旋转中心、旋转角度和旋转方向（顺时针或逆时针）。
The most frequently tested rotations in IGCSE are about the origin (0, 0). The following coordinate rules apply when the centre of rotation is the origin and the angle is measured anticlockwise unless stated otherwise.
IGCSE 中最常考查的是绕原点 (0, 0) 的旋转。以下坐标规则适用于旋转中心为原点的情况，除非特别说明，角度按逆时针方向测量。
| Rotation / 旋转 | Rule / 规则 |
| 90° anticlockwise / 逆时针 90° | (x, y) → (-y, x) |
| 90° clockwise / 顺时针 90° | (x, y) → (y, -x) |
| 180° (either direction) / 180°（任一方向） | (x, y) → (-x, -y) |
To verify the 90° anticlockwise rule, take the point (1, 0). Rotating it 90° anticlockwise about the origin sends it to (0, 1), which matches the rule (-y, x) = (0, 1). Similarly, rotating (0, 1) gives (-1, 0), which is indeed the position of the point after a quarter-turn anticlockwise.
为了验证逆时针旋转 90° 的规则，取点 (1, 0)。将其绕原点逆时针旋转 90° 后到达 (0, 1)，符合规则 (-y, x) = (0, 1)。同样地，旋转点 (0, 1) 得到 (-1, 0)，确实是逆时针旋转四分之一圈后的位置。
For a rotation about a point other than the origin, you must construct the image using tracing paper or carefully apply the perpendicular distance method. For example, to rotate a triangle 90° clockwise about the point (2, 2), measure the horizontal and vertical distances of each vertex from (2, 2), then swap and adjust the signs accordingly.
对于绕非原点的旋转，你必须使用描图纸作图或仔细运用垂直距离法。例如，要将三角形绕点 (2, 2) 顺时针旋转 90°，先测量每个顶点到 (2, 2) 的水平距离和垂直距离，然后交换它们并调整正负号。
Always state the centre of rotation explicitly. For instance, “rotation of 90° clockwise about (1, -2)” is a fully correct description, whereas “rotation of 90°” is incomplete.
始终明确写出旋转中心。例如，”绕 (1, -2) 顺时针旋转 90°” 是一个完整的描述，而”旋转 90°”是不完整的。
5. Enlargement | 缩放（放大与缩小）
An enlargement changes the size of a shape by a scale factor k about a fixed centre of enlargement. If k > 1, the image is larger than the object; if 0 < k < 1, the image is smaller. The image and object are similar shapes: they have the same angles and their corresponding sides are in the ratio k : 1.
缩放是图形围绕一个固定的缩放中心按比例因子 k 改变大小。当 k > 1 时，像图比原图形大；当 0 < k < 1 时，像图比原图形小。像图与原图形是相似形：对应角相等，对应边的比值为 k : 1。
To find the image of a point P(x, y) under an enlargement with centre C(a, b) and scale factor k, use the formula:
要求点 P(x, y) 在缩放中心 C(a, b) 和比例因子 k 下的像点，使用公式：
x’ = a + k(x – a), y’ = b + k(y – b)
Equivalently, the vector from the centre to the image is k times the vector from the centre to the object:
等价地，从中心到像点的向量等于从中心到原图形对应点的向量乘以 k：
(x’ – a, y’ – b) = k(x – a, y – b)
For example, enlarge the point (3, 4) by scale factor 2 with centre (1, 2). The vector from centre to point is (2, 2). Multiplying by 2 gives (4, 4), so the image point is (1 + 4, 2 + 4) = (5, 6).
例如，将点 (3, 4) 以中心 (1, 2) 按比例因子 2 缩放。从中心到该点的向量为 (2, 2)。乘以 2 得到 (4, 4)，因此像点为 (1 + 4, 2 + 4) = (5, 6)。
When asked to find the centre of enlargement, join each object point to its corresponding image point with a straight line. The point where all these lines meet is the centre of enlargement.
当要求找出缩放中心时，将原图形的每个点与其对应像点连成直线。所有直线交汇的点就是缩放中心。
The scale factor k can be found by dividing a side length of the image by the corresponding side length of the object: k = (image length) / (object length).
比例因子 k 可以通过用像图的边长除以原图形对应边长来求得：k =（像图边长）/（原图形边长）。
6. Fractional and Negative Scale Factors | 分数与负比例因子
A fractional scale factor, where 0 < k < 1, produces a reduced image. This is still an enlargement in the mathematical sense, even though the image is smaller than the object. For example, a scale factor of ½ halves every side length.
分数比例因子，即 0 < k < 1，会产生缩小的像图。在数学意义上这仍然是缩放，尽管像图比原图形小。例如，比例因子为 ½ 时，每条边长度变为原来的一半。
A negative scale factor such as -1, -2 or -½ produces an image that is on the opposite side of the centre of enlargement from the object, and the image is also inverted. This is best understood through the vector formula: a negative k reverses the direction of the vector from the centre to the point.
负比例因子，如 -1、-2 或 -½，会使像图位于缩放中心相对于原图形相反的一侧，并且像图同时被倒置。这可以通过向量公式来理解：负的 k 会将中心到点的向量方向反转。
For example, take the point (4, 5) with centre (2, 1) and scale factor -1. The vector from centre to point is (2, 4). Multiplying by -1 gives (-2, -4), so the image is at (2 + (-2), 1 + (-4)) = (0, -3). The image lies directly opposite the object across the centre.
例如，取点 (4, 5)，中心为 (2, 1)，比例因子为 -1。从中心到点的向量为 (2, 4)。乘以 -1 得到 (-2, -4)，所以像点为 (2 + (-2), 1 + (-4)) = (0, -3)。像点位于中心相对于原图形的正对面。
When using a scale factor of -1, the image is congruent to the object but rotated 180° about the centre of enlargement. A negative fractional scale factor combines both reversal and shrinking.
当比例因子为 -1 时，像图与原图形全等，但绕缩放中心旋转了 180°。负分数比例因子同时结合了反向和缩小两种效果。
7. Combined Transformations | 复合变换
In IGCSE, you may be asked to apply two transformations in sequence. For example, first reflect a triangle in the y-axis, then translate it by a given vector. The order of the transformations matters: applying translation first and reflection second usually produces a different result from applying reflection first and translation second.
在 IGCSE 中，你可能会被要求依次施加两个变换。例如，先将三角形关于 y 轴反射，再按给定向量平移。变换的次序很重要：先平移后反射通常与先反射后平移的结果不同。
To perform a combined transformation, apply the first transformation to the object to obtain an intermediate image. Then apply the second transformation to this intermediate image to obtain the final image. Always work step by step and label each intermediate image clearly, for example A’ and then A”.
进行复合变换时，先对原图形施加第一个变换得到中间像图，然后对这个中间像图施加第二个变换得到最终像图。务必逐步操作，并清楚标注每个中间像图，例如 A’ 和 A”。
Here is a worked example. A point P(2, 1) is first translated by vector (3, 2) and then reflected in the x-axis. The translation gives P'(2 + 3, 1 + 2) = P'(5, 3). Reflecting P’ in the x-axis gives P”(5, -3).
下面是一个实例。点 P(2, 1) 先按向量 (3, 2) 平移，再关于 x 轴反射。平移得到 P'(2 + 3, 1 + 2) = P'(5, 3)。将 P’ 关于 x 轴反射得到 P”(5, -3)。
If the order is reversed, the result changes. Reflecting P(2, 1) in the x-axis first gives P'(2, -1). Then translating by (3, 2) gives P”(5, 1). Comparing with the previous result, the final coordinates differ, demonstrating that transformations do not commute.
如果颠倒次序，结果会不同。先将 P(2, 1) 关于 x 轴反射得到 P'(2, -1)，再按向量 (3, 2) 平移得到 P”(5, 1)。与之前的结果比较，最终坐标不同，这表明变换不满足交换律。
Questions sometimes ask you to describe the single transformation that is equivalent to two given transformations. You can determine this by tracking one point through both steps and then identifying the overall mapping. In some simple cases, such as two rotations about the same centre, the angles add together.
有些问题要求你描述两个给定变换等价的单一变换。你可以跟踪一个点经过两步的过程来确定整体映射。在一些简单情况下，例如绕同一中心旋转两次，旋转角度相加即可。
8. Invariant Points and Lines | 不动点与不变直线
An invariant point is a point that maps to itself under a transformation. A line is an invariant line if every point on it maps to a point on the same line, though individual points on the line may move along it.
不动点是在变换下映射到自身的点。不变直线是指该直线上的每个点都映射到同一直线上的点，但直线上的个别点可能会沿该直线移动。
Each transformation has characteristic invariant features. Under a reflection, every point lying exactly on the mirror line is invariant, since its position does not change. For example, reflecting a point in the line y = x leaves the point (3, 3) unchanged.
每种变换都有其典型的不变特征。在反射下，恰好位于镜像轴上的每个点都是不动点，因为其位置不变。例如，关于直线 y = x 反射时，点 (3, 3) 保持不变。
Under a rotation, only the centre of rotation itself is invariant, assuming the angle of rotation is not a multiple of 360°. For a 180° rotation about (2, 2), the point (2, 2) remains fixed while all other points move.
在旋转下，只有旋转中心本身是不动点，前提是旋转角度不是 360° 的倍数。例如，绕 (2, 2) 旋转 180° 时，点 (2, 2
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