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Vectors and Transformations - Enlargements and Shear

Grade 12A Level

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

🔑Concepts

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An enlargement is a transformation that changes the size of an object while preserving its shape and orientation. It is defined by a center of enlargement and a scale factor kk. If k>1k > 1, the object expands; if 0<k<10 < k < 1, it shrinks; if kk is negative, the image is inverted and appears on the opposite side of the center.

Enlargement of a triangle from the origin with scale factor 2
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A shear transformation shifts points parallel to a fixed line called the invariant line. The distance a point moves is proportional to its distance from that line. For an xx-axis invariant shear, points (x,y)(x, y) map to (x+ky,y)(x + ky, y), meaning the yy-coordinate remains unchanged while the xx-coordinate shifts based on the shear factor kk.

A rectangle being sheared parallel to the x-axis
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The determinant of the transformation matrix M\mathbf{M} tells us how the area changes. For an enlargement with scale factor kk, the area increases by k2k^2 because det⁡(k00k)=k2\det \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} = k^2. For a shear, the area remains invariant because det⁡(1k01)=1\det \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} = 1.

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Negative scale factors in enlargement result in a point reflection through the center. A scale factor of k=−1k = -1 is equivalent to a rotation of 180∘180^{\circ} about the center of enlargement.

📐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.

Problem 4:

A square with vertices A(0,0)A(0,0), B(1,0)B(1,0), C(1,1)C(1,1), and D(0,1)D(0,1) is transformed by a shear with the yy-axis invariant and a shear factor k=2k=2. Determine the coordinates of the image C′C' and draw the resulting shape.

A square and its image under a y-axis invariant shear

Solution:

The transformation matrix for a shear with the yy-axis invariant is M=(10k1)\mathbf{M} = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}. Given k=2k=2, the matrix is M=(1021)\mathbf{M} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}. To find C′C', we multiply: (1021)(11)=((1×1)+(0×1)(2×1)+(1×1))=(13)\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} (1 \times 1) + (0 \times 1) \\ (2 \times 1) + (1 \times 1) \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}. The coordinates of the image vertices are A′(0,0)A'(0,0), B′(1,2)B'(1,2), C′(1,3)C'(1,3), and D′(0,1)D'(0,1).

Explanation:

In a yy-axis invariant shear, the xx-coordinates stay the same while the yy-coordinates change by kk times the xx-distance. Since CC is at x=1x=1, its yy-coordinate increases by 2×1=22 \times 1 = 2, moving from 11 to 33.

Problem 5:

A triangle with area 6 units26 \text{ units}^2 undergoes an enlargement with center (0,0)(0,0) and scale factor k=−2k = -2. Find the area of the image and the coordinates of the image of vertex P(3,−1)P(3, -1).

Negative enlargement mapping P to P' through the origin

Solution:

  1. Area of Image: The area scale factor is k2=(−2)2=4k^2 = (-2)^2 = 4. Areaimage=4×6=24 units2\text{Area}_{\text{image}} = 4 \times 6 = 24 \text{ units}^2. 2. Coordinates of P′P': Using the matrix M=(−200−2)\mathbf{M} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}: (−200−2)(3−1)=(−62)\begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} -6 \\ 2 \end{pmatrix}. The image P′P' is at (−6,2)(-6, 2).

Explanation:

A negative scale factor enlarges the object and rotates it 180∘180^{\circ} about the center. The area is always positive as it depends on ∣k∣2|k|^2.