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Geometry and Trigonometry - Geometric transformations: translation, reflection, rotation, and enlargement

Grade 9IB

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

🔑Concepts

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Translation: This transformation slides a shape across a coordinate plane without rotating it or changing its size. It is defined by a translation vector (xy)\begin{pmatrix} x \\ y \end{pmatrix}, where xx represents the horizontal shift (right is positive, left is negative) and yy represents the vertical shift (up is positive, down is negative). Visually, the shape moves in a straight line, and every point of the object moves the same distance in the same direction.

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Reflection: A reflection flips a shape over a specific mirror line. Each point of the image is the same perpendicular distance from the mirror line as the corresponding point of the original object. Visually, the orientation of the shape is reversed (like looking in a mirror), but the size and shape remain identical (congruent). Common mirror lines include the xx-axis, yy-axis, y=xy = x, and y=−xy = -x.

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Rotation: This involves turning a shape around a fixed point called the center of rotation. To fully describe a rotation, you need the center coordinates, the angle of rotation (e.g., 90∘,180∘90^\circ, 180^\circ), and the direction (clockwise or anti-clockwise). Visually, every point of the shape moves along an arc of a circle centered at the rotation point.

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Enlargement: Unlike other transformations, enlargement changes the size of the shape but keeps its proportions, making the image similar rather than congruent to the object. It is defined by a center of enlargement and a scale factor kk. Visually, if k>1k > 1, the shape grows and moves further from the center; if 0<k<10 < k < 1, the shape shrinks and moves closer to the center.

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Negative Scale Factors: In an enlargement, if the scale factor kk is negative, the image is formed on the opposite side of the center of enlargement and appears inverted (turned upside down). The distance from the center to the image is ∣k∣|k| times the distance to the original object.

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Invariant Points and Lines: These are points or lines that do not change position after a transformation. For example, any point sitting directly on the mirror line during a reflection is an invariant point. In rotation and enlargement, the center of the transformation is the only invariant point.

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Congruence vs. Similarity: Translation, reflection, and rotation are 'isometries' because they preserve the side lengths and angles of the original shape, resulting in a congruent image. Enlargement preserves angles but changes side lengths, resulting in a similar image.

📐Formulae

Translation: P(x,y)→P′(x+a,y+b)P(x, y) \rightarrow P'(x+a, y+b) using vector (ab)\begin{pmatrix} a \\ b \end{pmatrix}

Reflection in xx-axis: (x,y)→(x,−y)(x, y) \rightarrow (x, -y)

Reflection in yy-axis: (x,y)→(−x,y)(x, y) \rightarrow (-x, y)

Reflection in y=xy = x: (x,y)→(y,x)(x, y) \rightarrow (y, x)

Rotation 90∘90^\circ clockwise about (0,0)(0,0): (x,y)→(y,−x)(x, y) \rightarrow (y, -x)

Rotation 90∘90^\circ anti-clockwise about (0,0)(0,0): (x,y)→(−y,x)(x, y) \rightarrow (-y, x)

Rotation 180∘180^\circ about (0,0)(0,0): (x,y)→(−x,−y)(x, y) \rightarrow (-x, -y)

Scale Factor: k=Image Side LengthObject Side Lengthk = \frac{\text{Image Side Length}}{\text{Object Side Length}}

Enlargement from origin (0,0)(0,0): (x,y)→(kx,ky)(x, y) \rightarrow (kx, ky)

💡Examples

Problem 1:

Triangle ABCABC has vertices A(1,2)A(1, 2), B(4,2)B(4, 2), and C(1,6)C(1, 6). Apply a translation using the vector (−34)\begin{pmatrix} -3 \\ 4 \end{pmatrix} and find the new coordinates of the vertices.

Solution:

  1. Identify the xx-shift and yy-shift from the vector: x=−3x = -3, y=4y = 4.
  2. Add the shifts to each vertex: A′(1+(−3),2+4)=(−2,6)A'(1 + (-3), 2 + 4) = (-2, 6) B′(4+(−3),2+4)=(1,6)B'(4 + (-3), 2 + 4) = (1, 6) C′(1+(−3),6+4)=(−2,10)C'(1 + (-3), 6 + 4) = (-2, 10)
  3. The translated vertices are A′(−2,6)A'(-2, 6), B′(1,6)B'(1, 6), and C′(−2,10)C'(-2, 10).

Explanation:

To translate a point, we add the top value of the vector to the xx-coordinate and the bottom value to the yy-coordinate.

Problem 2:

A square has a vertex at P(2,3)P(2, 3). It undergoes an enlargement with a scale factor k=3k = 3 centered at the origin (0,0)(0, 0). Determine the coordinates of the image vertex P′P'. If the original square had an area of 5 cm25 \text{ cm}^2, what is the area of the enlarged square?

Solution:

  1. For the coordinates: Use the rule (x,y)→(kx,ky)(x, y) \rightarrow (kx, ky). P′(3×2,3×3)=(6,9)P'(3 \times 2, 3 \times 3) = (6, 9).
  2. For the area: The area of an enlarged shape is the original area multiplied by the scale factor squared (k2k^2). New Area=5×32=5×9=45 cm2\text{New Area} = 5 \times 3^2 = 5 \times 9 = 45 \text{ cm}^2.

Explanation:

When the center of enlargement is the origin, we simply multiply coordinates by kk. Note that while side lengths increase by kk, area increases by k2k^2.