Transformations — Reflection, Rotation, Translation, Dilation

Understanding Geometric Transformations

A geometric transformation is a rule that moves, flips, stretches, or turns a shape in the coordinate plane. Every transformation takes each point of a figure (the pre-image) and maps it to a new point (the image).

The four fundamental transformations are: Translation (sliding), Reflection (flipping), Rotation (turning), and Dilation (scaling). The first three preserve size and shape — they are called isometries (or rigid motions). Dilation changes size but preserves shape — it’s a similarity transformation.

In Indian school syllabi, transformations appear in Class 9-10 CBSE through coordinate geometry problems, and in JEE through complex number geometry and matrix applications.

Key Terms and Definitions

Pre-image: The original figure before transformation.

Image: The resulting figure after transformation. Usually denoted with a prime symbol: A maps to A’, B maps to B’.

Isometry: A transformation that preserves distances and angles (congruence). Translation, reflection, and rotation are all isometries.

Invariant point: A point that maps to itself under a transformation. For a reflection in the y-axis, points on the y-axis are invariant.

Composition of transformations: Applying one transformation followed by another.

Translation

A translation slides every point of a figure by the same vector $(h, k)$ . No rotation or reflection occurs.

Rule: $(x, y) \to (x + h, y + k)$

Example: Translate triangle with vertices A(1, 2), B(3, 2), C(2, 4) by the vector $(3, -1)$ .

A(1,2) \to A'(4, 1)

B(3,2) \to B'(6, 1)

C(2,4) \to C'(5, 3)

The shape, size, and orientation are all preserved. The triangle just shifts.

A translation is completely described by the translation vector $(h, k)$ . Positive $h$ moves right, negative $h$ moves left. Positive $k$ moves up, negative $k$ moves down.

Reflection

A reflection flips a figure over a line called the axis of reflection (or mirror line). Each point maps to a point on the opposite side, at the same perpendicular distance from the mirror line.

Key Reflection Rules

Reflection in x-axis: $(x, y) \to (x, -y)$
Reflection in y-axis: $(x, y) \to (-x, y)$
Reflection in line $y = x$ : $(x, y) \to (y, x)$
Reflection in line $y = -x$ : $(x, y) \to (-y, -x)$
Reflection in origin (not a line, but point reflection): $(x, y) \to (-x, -y)$

Example: Reflect the point P(3, 4) in the line $y = x$ .

Using the rule $(x, y) \to (y, x)$ : P(3, 4) maps to P’(4, 3).

Example (CBSE Board style): Find the reflection of triangle A(1, 0), B(4, 0), C(2, 3) in the x-axis.

A(1,0) \to A'(1, 0) \quad \text{(on x-axis, invariant)}

B(4,0) \to B'(4, 0) \quad \text{(on x-axis, invariant)}

C(2,3) \to C'(2, -3)

Rotation

A rotation turns a figure around a fixed point called the centre of rotation, through a given angle.

Rotation About the Origin

90° anticlockwise: $(x, y) \to (-y, x)$
90° clockwise (= 270° anticlockwise): $(x, y) \to (y, -x)$
180°: $(x, y) \to (-x, -y)$
360°: $(x, y) \to (x, y)$ (back to original)

Example: Rotate P(3, 2) by 90° anticlockwise about the origin.

$(3, 2) \to (-2, 3)$

Verification: The distance from the origin remains the same: $\sqrt{9+4} = \sqrt{13}$ and $\sqrt{4+9} = \sqrt{13}$ ✓

Rotation in Complex Numbers (JEE)

In JEE, rotation is elegantly handled using complex numbers. A point $z = x + iy$ rotated by angle $\theta$ anticlockwise about the origin becomes $z \cdot e^{i\theta}$ .

For 90° rotation: multiply by $i$ . $(3 + 2i) \times i = 3i + 2i^2 = -2 + 3i$ , giving point $(-2, 3)$ — same as before!

Dilation (Scaling)

A dilation scales a figure by a scale factor $k$ about a centre of dilation.

Rule (centre at origin): $(x, y) \to (kx, ky)$

If $|k| > 1$ : figure gets larger (enlargement)
If $0 < |k| < 1$ : figure gets smaller (reduction)
If $k < 0$ : figure is scaled and reflected through the centre

Example: Dilate triangle A(2, 1), B(4, 1), C(3, 3) by scale factor 2 about the origin.

A(2,1) \to A'(4, 2)

B(4,1) \to B'(8, 2)

C(3,3) \to C'(6, 6)

The shape is preserved (similar triangles) but all lengths double.

In CBSE Class 10, dilation appears as “similar triangles” — you won’t see the word “dilation” explicitly. But every time you work with similar triangles, you’re applying a dilation. JEE problems on “homothety” use dilation directly.

Solved Examples

Easy — CBSE Class 9

Problem: Write the image of the point (-3, 5) when reflected in the y-axis.

Reflection in y-axis: $(x, y) \to (-x, y)$

Image = $(3, 5)$ . The x-coordinate flips sign, y-coordinate stays.

Medium — CBSE Class 10 / JEE Level

Problem: A point P(2, 3) is reflected in the x-axis to get P’, then P’ is reflected in the y-axis to get P”. Find the coordinates of P”.

Step 1: Reflect P(2, 3) in x-axis → P’(2, -3)

Step 2: Reflect P’(2, -3) in y-axis → P”(-2, -3)

This is equivalent to a 180° rotation about the origin!

Hard — JEE Main Level

Problem: The point P(1, 2) is reflected in the line $y = 3$ to get P’. Find the coordinates of P’.

The reflection of $(x_1, y_1)$ in the horizontal line $y = k$ is $(x_1, 2k - y_1)$ .

P' = (1, 2(3) - 2) = (1, 4)

The x-coordinate is unchanged (the mirror line is horizontal). The y-coordinate: point P is at $y=2$ , which is 1 unit below $y=3$ ; the reflection is 1 unit above $y=3$ , giving $y=4$ .

Exam-Specific Tips

CBSE Class 9/10

CBSE tests reflections heavily — especially in lines $y = x$ , $y = -x$ , and the axes. Board papers often ask: “Find the image of point A in line L, and verify it lies on line L’ such that…” Always show the midpoint calculation to confirm the original and image are equidistant from the mirror line.

JEE Main

JEE uses transformations in the context of finding loci, images of circles/lines, and complex number geometry. The matrix representation of transformations (rotation matrix) appears in linear algebra questions. Reflection of a line or circle in another line is a common JEE problem.

Common Mistakes

Mistake 1: Confusing 90° clockwise and anticlockwise. Anticlockwise (positive direction) takes $(x,y)$ to $(-y, x)$ . Clockwise takes $(x,y)$ to $(y, -x)$ . Students often swap these. Use a specific point like (1, 0) — rotating it 90° anticlockwise should give (0, 1), not (0, -1).

Mistake 2: Wrong formula for reflection in $y = x$ . Many students write $(x, y) \to (-y, -x)$ instead of $(y, x)$ . The reflection in $y = x$ simply swaps coordinates. The negative version $(-y, -x)$ is the reflection in $y = -x$ .

Mistake 3: Not preserving distance from centre in rotation. A rotation preserves all distances from the centre of rotation. If a point is 5 units from the origin before rotation, it’s 5 units after. Use this as a quick sanity check.

Mistake 4: Applying dilation to distances instead of coordinates. If triangle ABC has area 6 sq units and you dilate by factor 3, the new area is $3^2 \times 6 = 54$ sq units (areas scale as the square of the linear scale factor).

Practice Questions

Q1. Find the image of P(-2, 4) under reflection in the x-axis.

Rule: $(x, y) \to (x, -y)$ . Image = $(-2, -4)$ .

Q2. Rotate point A(4, 0) by 90° anticlockwise about the origin.

Rule: $(x, y) \to (-y, x)$ . A(4, 0) → A’(0, 4).

Q3. Translate the point (3, -2) by the vector (-1, 5).

$(3 + (-1), -2 + 5) = (2, 3)$ .

Q4. A triangle has area 10 sq units. After dilation with scale factor 2, find the new area.

Area scales by $k^2 = 4$ . New area = $4 \times 10 = 40$ sq units.

Q5. Find the reflection of P(2, 5) in the line $y = x$ , then reflect the image in the y-axis.

Step 1: Reflection in $y = x$ : $(2, 5) \to (5, 2)$

Step 2: Reflection in y-axis: $(5, 2) \to (-5, 2)$

Final image: $(-5, 2)$ .

Q6. A point P is such that its reflection in $y = 0$ (x-axis) is P’(3, -4). Find P.

If P’ is the reflection of P in the x-axis, then P is the reflection of P’ in the x-axis.

P = $(3, 4)$ .

Additional Worked Examples

Reflection in an Arbitrary Line (JEE Level)

Problem: Find the image of the point P(1, 2) after reflection in the line $y = x + 1$ .

The line $y = x + 1$ can be written as $x - y + 1 = 0$ . The perpendicular from P(1, 2) has slope $-1$ (negative reciprocal of line’s slope 1).

Perpendicular line: $y - 2 = -1(x - 1) \implies y = -x + 3$ .

Solve $y = x + 1$ and $y = -x + 3$ : $x + 1 = -x + 3 \implies 2x = 2 \implies x = 1, y = 2$ .

Wait — the foot is (1, 2) which is P itself. That means P lies on the line $y = x + 1$ ! Check: $2 = 1 + 1 = 2$ . Yes. So P is on the mirror line. Its reflection is P itself: (1, 2).

Let us redo with a different point. Reflect Q(3, 1) in the line $y = x + 1$ .

Perpendicular from Q(3, 1) with slope $-1$ : $y - 1 = -1(x - 3) \implies y = -x + 4$ .

Intersection with $y = x + 1$ : $x + 1 = -x + 4 \implies x = 1.5, y = 2.5$ . Foot = (1.5, 2.5).

If Q’ is the reflection, then (1.5, 2.5) is the midpoint of Q and Q’.

$Q'_x = 2(1.5) - 3 = 0$ , $Q'_y = 2(2.5) - 1 = 4$ .

Q’ = (0, 4).

Reflection of point $(x_1, y_1)$ in the line $ax + by + c = 0$ :

x' = x_1 - \frac{2a(ax_1 + by_1 + c)}{a^2 + b^2}

y' = y_1 - \frac{2b(ax_1 + by_1 + c)}{a^2 + b^2}

This formula handles any line — no need to find foot of perpendicular separately. Memorise for JEE.

Matrix Representation of Transformations (JEE Advanced)

Linear transformations can be represented as matrix multiplication:

Transformation	Matrix
Reflection in x-axis	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
Reflection in y-axis	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
Rotation by $\theta$ anticlockwise	$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$
Dilation by factor $k$	$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

To apply: multiply the matrix by the column vector $\begin{pmatrix} x \\ y \end{pmatrix}$ of the point.

In JEE Advanced, rotation matrices appear in the context of complex numbers and coordinate geometry. The rotation of a complex number $z$ by angle $\theta$ about the origin is $z' = z \cdot e^{i\theta} = z(\cos\theta + i\sin\theta)$ — this is equivalent to the matrix multiplication above.

Q7. A triangle has vertices at (0, 0), (4, 0), (0, 3). Find its area after dilation by scale factor 3 about the origin.

Original area $= \frac{1}{2} \times 4 \times 3 = 6$ sq units. After dilation by factor 3, area scales by $3^2 = 9$ . New area $= 6 \times 9 = 54$ sq units.

FAQs

Q: What is the difference between rotation and revolution? In geometry, rotation is the transformation of a 2D figure about a point. Revolution in physics refers to an object orbiting around another body. In maths, both words can sometimes be used interchangeably for 3D solids of revolution, but for 2D transformations, always use “rotation.”

Q: Do all transformations preserve area? Isometries (translation, reflection, rotation) preserve area. Dilation scales area by $k^2$ , where $k$ is the scale factor. So a dilation with $k = 3$ makes the area 9 times larger.

Q: What is a composite transformation? Applying two or more transformations in sequence. For example, reflecting in the x-axis then in the y-axis is equivalent to a 180° rotation about the origin. The order matters — most pairs of transformations are not commutative.

Q: How are transformations connected to functions? Every transformation is a function: it takes each input point and gives exactly one output point. The transformation $(x, y) \to (x + 3, y - 2)$ is a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ . Matrix multiplication encodes linear transformations (rotations, reflections, dilations) elegantly.