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Geometry - Symmetry and Transformations (Translation, Rotation, Reflection, Enlargement)

Grade 8Cambridge (IGCSE)

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Translation involves sliding a shape a fixed distance in a specific direction without rotating or flipping it. It is described by a translation vector (xy)\begin{pmatrix} x \\ y \end{pmatrix}, where xx denotes the horizontal movement (right positive, left negative) and yy denotes the vertical movement (up positive, down negative).

A triangle translated by vector (3, 2)
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Reflection creates a mirror image of a shape across a line of reflection. Every point on the image is the same distance from the line as the corresponding point on the object. Key reflection lines include x=0x=0 (y-axis), y=0y=0 (x-axis), y=xy=x, and y=−xy=-x.

Reflection of a triangle across the line y=x
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Rotation turns a shape around a fixed point called the center of rotation. A rotation is defined by the angle (e.g., 90∘90^\circ, 180∘180^\circ), the direction (clockwise or anti-clockwise), and the coordinates of the center point.

90 degree anti-clockwise rotation about the origin
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Enlargement changes the size of a shape but preserves its proportions and angles (similarity). It is defined by a center of enlargement and a scale factor kk. If k>1k > 1, the shape grows; if 0<k<10 < k < 1, the shape shrinks.

Enlargement of a square with scale factor 2 from the origin

📐Formulae

Translation Vector: \text{Translation Vector: } (xy)\begin{pmatrix} x \\ y \end{pmatrix}  (where x=horizontal shift, y=vertical shift)\text{ (where } x = \text{horizontal shift, } y = \text{vertical shift)}

Scale Factor (k)=Length of Image sideLength of Object side\text{Scale Factor (k)} = \frac{\text{Length of Image side}}{\text{Length of Object side}}

Reflection in y=x:(x,y)→(y,x)\text{Reflection in } y = x: (x, y) \rightarrow (y, x)

Reflection in y=−x:(x,y)→(−y,−x)\text{Reflection in } y = -x: (x, y) \rightarrow (-y, -x)

Rotation 180∘ about origin: (x,y)→(−x,−y)\text{Rotation } 180^\circ \text{ about origin: } (x, y) \rightarrow (-x, -y)

Distance from center to image=k×Distance from center to object\text{Distance from center to image} = k \times \text{Distance from center to object}

💡Examples

Problem 1:

Translate the point A(2,3)A(2, 3) by the vector (−45)\begin{pmatrix} -4 \\ 5 \end{pmatrix}. Find the coordinates of the image A′A'.

Solution:

A′=(−2,8)A' = (-2, 8)

Explanation:

Add the xx-component of the vector to the xx-coordinate: 2+(−4)=−22 + (-4) = -2. Add the yy-component to the yy-coordinate: 3+5=83 + 5 = 8.

Problem 2:

A square with side length 5 cm is enlarged by a scale factor of 3. What is the side length and area of the new square?

Solution:

Side length = 15 cm; Area = 225 cm²

Explanation:

The new side length is 5×3=155 \times 3 = 15 cm. The area of the new square is 152=22515^2 = 225 cm² (or original area 25×k225 \times k^2).

Problem 3:

Describe the single transformation that maps the point (3,4)(3, 4) onto (4,3)(4, 3).

Solution:

Reflection in the line y=xy = x.

Explanation:

When the xx and yy coordinates are swapped, it indicates a reflection across the diagonal line where yy equals xx.

Problem 4:

What is the order of rotational symmetry for a regular hexagon?

Solution:

6

Explanation:

A regular hexagon can be rotated by 60∘60^\circ (360/6) six times within a full circle and look identical each time.

Problem 5:

Shape PP is a triangle with vertices at (1,1)(1, 1), (3,1)(3, 1), and (1,2)(1, 2). Reflect shape PP in the line x=−1x = -1 to form shape QQ. List the coordinates of QQ.

Reflection of a triangle across the line x=-1

Solution:

The line of reflection is the vertical line x=−1x = -1. Distance of vertex (1,1)(1, 1) from x=−1x = -1 is 1−(−1)=21 - (-1) = 2 units. The image will be 2 units to the left of x=−1x = -1: −1−2=−3-1 - 2 = -3. So, (−3,1)(-3, 1). Distance of vertex (3,1)(3, 1) from x=−1x = -1 is 3−(−1)=43 - (-1) = 4 units. The image will be 4 units to the left: −1−4=−5-1 - 4 = -5. So, (−5,1)(-5, 1). Distance of vertex (1,2)(1, 2) from x=−1x = -1 is 1−(−1)=21 - (-1) = 2 units. The image will be 2 units to the left: −1−2=−3-1 - 2 = -3. So, (−3,2)(-3, 2). The coordinates of QQ are (−3,1)(-3, 1), (−5,1)(-5, 1), and (−3,2)(-3, 2).

Explanation:

To reflect in a vertical line x=cx = c, the yy-coordinate remains the same, and the new xx-coordinate is x′=2c−xx' = 2c - x. Here, x′=2(−1)−x=−2−xx' = 2(-1) - x = -2 - x.

Problem 6:

Triangle TT has vertices A(0,0)A(0, 0), B(2,0)B(2, 0), and C(0,1)C(0, 1). It is enlarged by a scale factor of k=−2k = -2 with the center of enlargement at the origin (0,0)(0, 0). Find the coordinates of the image T′T'.

Enlargement of a triangle by scale factor -2 from the origin

Solution:

Multiply each coordinate of the vertices of TT by the scale factor k=−2k = -2 since the center is the origin. A′(0×−2,0×−2)=(0,0)A'(0 \times -2, 0 \times -2) = (0, 0) B′(2×−2,0×−2)=(−4,0)B'(2 \times -2, 0 \times -2) = (-4, 0) C′(0×−2,1×−2)=(0,−2)C'(0 \times -2, 1 \times -2) = (0, -2) The vertices of T′T' are (0,0)(0, 0), (−4,0)(-4, 0), and (0,−2)(0, -2).

Explanation:

A negative scale factor kk means the image is on the opposite side of the center of enlargement and is inverted. The size is increased by the absolute value ∣k∣|k|.

Symmetry and Transformations (Translation, Rotation, Reflection, Enlargement) Grade 8 Notes &…