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Vectors and Transformations - Enlargements and Shear

Grade 12IGCSE

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

🔑Concepts

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Enlargement: A transformation defined by a center of enlargement and a scale factor kk. It changes the size of the object but preserves its shape (similarity).

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Scale Factor (kk): If k>1k > 1, the image is larger; if 0<k<10 < k < 1, the image is smaller; if kk is negative, the image is inverted and on the opposite side of the center.

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Shear: A transformation where all points on a specific line (the invariant line) remain fixed, while other points move parallel to that line by a distance proportional to their perpendicular distance from the line.

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Area Invariance: Enlargement changes area by a factor of k2k^2, whereas a Shear preserves the original area of the shape.

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Transformation Matrices: Linear transformations (centered at the origin) can be represented by 2×22 \times 2 matrices where the columns represent the images of unit vectors (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix}.

📐Formulae

Enlargement Matrix (Center at origin): M=(k00k)\mathbf{M} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}

Shear Matrix (xx-axis invariant): M=(1k01)\mathbf{M} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}, where kk is the shear factor.

Shear Matrix (yy-axis invariant): M=(10k1)\mathbf{M} = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}, where kk is the shear factor.

Area of Image: Areaimage=∣det⁡(M)∣×Areaobject\text{Area}_{\text{image}} = |\det(M)| \times \text{Area}_{\text{object}}

Vector Mapping: (x′y′)=M(xy)\begin{pmatrix} x' \\ y' \end{pmatrix} = \mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix}

💡Examples

Problem 1:

Find the image of the point P(2,3)P(2, 3) under an enlargement with center (0,0)(0,0) and scale factor k=−3k = -3.

Solution:

(x′y′)=(−300−3)(23)=(−6−9)\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -6 \\ -9 \end{pmatrix}. The image is P′(−6,−9)P'(-6, -9).

Explanation:

Since the center is the origin, we multiply the coordinate vector by the enlargement matrix. The negative scale factor reflects the point through the origin and triples its distance.

Problem 2:

A shear maps the point (1,1)(1, 1) to (4,1)(4, 1) while the xx-axis remains invariant. Determine the transformation matrix.

Solution:

For an xx-axis invariant shear, the matrix is (1k01)\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}. Applying this to (1,1)(1,1): (1k01)(11)=(1+k1)\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1+k \\ 1 \end{pmatrix}. Given the image is (4,1)(4,1), 1+k=4  ⟹  k=31+k = 4 \implies k = 3. Matrix is (1301)\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}.

Explanation:

Because the xx-axis is invariant, the yy-coordinate remains unchanged. The 'shear factor' kk represents how much the xx-coordinate shifts per unit of yy.

Problem 3:

Triangle TT has an area of 5 cm25 \text{ cm}^2. It undergoes an enlargement with scale factor k=4k=4 followed by a shear. What is the area of the final image?

Solution:

Area after enlargement: 5×42=5×16=80 cm25 \times 4^2 = 5 \times 16 = 80 \text{ cm}^2. Area after shear: A shear has a determinant of 11 (e.g., 1(1)−k(0)=11(1) - k(0) = 1), so it preserves area. Final area =80 cm2= 80 \text{ cm}^2.

Explanation:

Enlargement scales area by k2k^2. Shear does not change the area of a shape, only its displacement/tilt.