Exam practice: Grade 10 Optional Math Unit 9

a.

Translate △ABC by the vector a.

Figure 9.8 Exercise 2(a): △ABC and the vector a.

Hint

Read a off the grid: how far right, how far up? Then move each vertex by that much.

Answer

a = $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ : 4 units right, none up. So A(−4, 4) → A′(0, 4), B(−5, 1) → B′(−1, 1), C(−1, 1) → C′(3, 1).

Figure 9.9 The answer to 2(a): △A′B′C′ is △ABC moved 4 units to the right. B′ lands on C.

b.

Translate the parallelogram KLMN by the vector a.

Figure 9.10 Exercise 2(b): parallelogram KLMN and the vector a.

Answer

a = $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ . K(−4, 1) → K′(−2, 4), L(−5, −2) → L′(−3, 1), M(−1, −2) → M′(1, 1), N(0, 1) → N′(2, 4).

Figure 9.11 The answer to 2(b): the image K′L′M′N′ is still a parallelogram, of the same size, moved 2 right and 3 up.

c.

Translate the pentagon PQRST by the vector a.

Figure 9.12 Exercise 2(c): pentagon PQRST and the vector a.

Answer

a = $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ . P(−1, 4) → P′(0, 6), Q(−3, 3) → Q′(−2, 5), R(−3, 1) → R′(−2, 3), S(0, 1) → S′(1, 3), T(0, 3) → T′(1, 5).

Figure 9.13 The answer to 2(c): every vertex moved 1 right and 2 up.

3.

What is meant by translation? Explain with an example.

Answer

A translation moves every point of a figure the same distance in the same direction. It is described by a translation vector $\begin{pmatrix} a \\ b \end{pmatrix}$ : a units along the X-axis (right +, left −) and b units along the Y-axis (up +, down −), and P(x, y) → P′(x + a, y + b). The image is congruent to the object and faces the same way.
Example: translating A(2, 3) by $\begin{pmatrix} 4 \\ -1 \end{pmatrix}$ gives A′(6, 2). In daily life, a drawer sliding out, or a lift going straight up, is a translation.

4.

Using the translation vector $T = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$ , find the images of:

a.A(−1, 3)
b.B(2, −2)
c.C(5, 6)
d.D(−4, −4)
e.E(7, 2)
f.F(2, −4)

Answer

Add 2 to x and −2 to y:

a.A′(1, 1)
b.B′(4, −4)
c.C′(7, 4)
d.D′(−2, −6)
e.E′(9, 0)
f.F′(4, −6)

5.

P(2, 3), Q(−2, 4) and R(4, −2) are the vertices of △PQR. It is translated by $T = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ to △P′Q′R′. Find the vertices of △P′Q′R′.

Answer

P(2, 3) → P′(5, 5), Q(−2, 4) → Q′(1, 6), R(4, −2) → R′(7, 0).

6.

For the points A(3, 5), B(−3, 5) and C(3, −5):

a.To which point is A translated by AB?
b.To which point is B translated by BC?
c.To which point is C translated by CA?

Answer

a. $\overrightarrow{AB} = \begin{pmatrix} -3 - 3 \\ 5 - 5 \end{pmatrix} = \begin{pmatrix} -6 \\ 0 \end{pmatrix}$ , so A(3, 5) → A′(−3, 5).
b. $\overrightarrow{BC} = \begin{pmatrix} 3 + 3 \\ -5 - 5 \end{pmatrix} = \begin{pmatrix} 6 \\ -10 \end{pmatrix}$ , so B(−3, 5) → B′(3, −5).
c. $\overrightarrow{CA} = \begin{pmatrix} 3 - 3 \\ 5 + 5 \end{pmatrix} = \begin{pmatrix} 0 \\ 10 \end{pmatrix}$ , so C(3, −5) → C′(3, 5).

Each point lands on the next one, A on B, B on C, C on A: translating a point by the vector from it to another point always lands on that point.

7.

A translation takes A(−5, 6) to A′(−3, 3). Find the translation vector.

Answer

Image minus object: $T = \begin{pmatrix} -3 - (-5) \\ 3 - 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ .

8.

Find the translation vector that moves R(4, 7) to R′(−3, −5). Use it to find the image of S(4, 1).

Answer

$T = \begin{pmatrix} -3 - 4 \\ -5 - 7 \end{pmatrix} = \begin{pmatrix} -7 \\ -12 \end{pmatrix}$ .
S(4, 1) → S′(4 − 7, 1 − 12) = S′(−3, −11).

9.

$U(5, 3) \xrightarrow{\ T = \begin{pmatrix} a \\ b \end{pmatrix}\ } U'(4, 2)$ . Find a and b, and use the vector to find the image of V(−5, 4).

Answer

5 + a = 4, so a = −1; 3 + b = 2, so b = −1. $T = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$ .
V(−5, 4) → V′(−6, 3).

10.

The vertices of △ABC are A(1, −1), B(−2, 2) and C(3, 3). Translate it by $T = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ , and draw △ABC and △A′B′C′ on one graph.

Answer

A(1, −1) → A′(3, 2), B(−2, 2) → B′(0, 5), C(3, 3) → C′(5, 6).

Figure 9.14 The answer to question 10: △ABC and its image △A′B′C′.

11.

A(1, 8), B(−3, 9), C(0, 13) and D(4, 12) are the vertices of a quadrilateral ABCD.

a.Find the vector AB.
b.Translate ABCD by AB, and draw ABCD and its image on one graph.

Answer

a. $\overrightarrow{AB} = \begin{pmatrix} -3 - 1 \\ 9 - 8 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \end{pmatrix}$ .
b.A(1, 8) → A′(−3, 9), B(−3, 9) → B′(−7, 10), C(0, 13) → C′(−4, 14), D(4, 12) → D′(0, 13).

(ABCD is in fact a parallelogram: AB = DC and $\overrightarrow{BC} = \overrightarrow{AD} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ . Its image is a parallelogram too. Note that A′ is B: translating by AB moves A onto B.)

Figure 9.15 The answer to question 11: ABCD (blue) and its image A′B′C′D′ (red), moved by AB. A′ falls on B, and D′ on C.

12.

The vertices of △ABC are A(−4, 6), B(3, −2) and C(1, 2). It is translated by $T_1 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ to △A′B′C′, which is then translated by $T_2 = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to △A′′B′′C′′.

a.Find the vertices of △A′B′C′ and △A′′B′′C′′.
b.Draw all three triangles on one graph.

Answer

a. Under T₁: A′(−3, 4), B′(4, −4), C′(2, 0).
Under T₂: A′′(−5, 7), B′′(2, −1), C′′(0, 3).
The single vector $T_1 + T_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$ takes ABC straight to A′′B′′C′′: check, A(−4, 6) → (−5, 7).

Figure 9.16 The answer to question 12: △ABC (blue), △A′B′C′ (red) and △A′′B′′C′′ (green). The coordinates are in the answer above.

13.

A point P(x, y) is reflected in the line x = 1, and its image is then reflected in the line x = 3. Compare the final image with the image of P under the translation $T = \begin{pmatrix} 4 \\ 0 \end{pmatrix}$ . What can you conclude?

Hint

Reflection in x = h sends (x, y) to (2h − x, y).

Answer

Reflect in x = 1: P(x, y) → P′(2 − x, y).
Reflect in x = 3: P′(2 − x, y) → P′′(6 − (2 − x), y) = P′′(x + 4, y).
The translation by $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ also gives (x + 4, y). The two results are the same for every point.
Conclusion: two reflections in parallel lines, one after the other, have the same effect as a single translation. The translation is perpendicular to the lines, its distance is twice the distance between them (2 × 2 = 4), and it points from the first line towards the second. Section 9.3.1 proves this in general.

a.

The vertices of △ABC are A(2, 2), B(6, 2) and C(2, 5). Reflect it first in the X-axis and then in the Y-axis. Find the vertices of the final image, and draw both triangles on one graph. What is meant by a combined transformation?

Answer

In the X-axis, (x, y) → (x, −y): A′(2, −2), B′(6, −2), C′(2, −5).
In the Y-axis, (x, y) → (−x, y): A′′(−2, −2), B′′(−6, −2), C′′(−2, −5).
(The two reflections together are a half turn about O: (x, y) → (−x, −y).)
A combined transformation is the single transformation that has the same effect as two (or more) transformations done one after the other.

Figure 9.28 The answer to 2(a): △ABC and its image △A′′B′′C′′, a half turn about O.

b.

The triangle with vertices A(1, 2), B(4, −1) and C(2, 5) is reflected in the line x = −3 and then in the line y = 4. Draw △ABC and the final image △A′′B′′C′′ on one graph.

Answer

In x = −3, (x, y) → (−6 − x, y): A′(−7, 2), B′(−10, −1), C′(−8, 5).
In y = 4, (x, y) → (x, 8 − y): A′′(−7, 6), B′′(−10, 9), C′′(−8, 3).
The mirrors cross at (−3, 4) at right angles, so the two together are a rotation of 180° about (−3, 4).

Figure 9.29 The answer to 2(b): △ABC and △A′′B′′C′′ are opposite each other through (−3, 4), where the mirrors cross.

c.

The triangle with vertices A(1, 2), B(4, −1) and C(2, 5) is reflected in the line x = 4 (R₁) and then in the line x = −1 (R₂). Find R₂ ∘ R₁, and draw △ABC and its image △A′B′C′ on one graph.

Answer

The lines are parallel, 5 units apart, so R₂ ∘ R₁ is a translation:
$R_2 \circ R_1 = 2\begin{pmatrix} -1 - 4 \\ 0 \end{pmatrix} = \begin{pmatrix} -10 \\ 0 \end{pmatrix}$
A(1, 2) → A′(−9, 2), B(4, −1) → B′(−6, −1), C(2, 5) → C′(−8, 5).

Figure 9.30 The answer to 2(c): reflecting in x = 4 and then x = −1 moves △ABC 10 units to the left.

a.

△ABC has vertices A(4, 0), B(3, 3) and C(1, −1). It is rotated about the origin through 90° and then, in the same direction, through 180°. Which single transformation is this? Use it to find the image △A′B′C′, and draw both triangles on one graph.

Answer

90° + 180° = 270°: a rotation of +270° about (0, 0), the same as −90°, so (x, y) → (y, −x).
A(4, 0) → A′(0, −4), B(3, 3) → B′(3, −3), C(1, −1) → C′(−1, −1).

Figure 9.31 The answer to 3(a): a turn of 270 anticlockwise, or 90 clockwise, about O.

b.

P(1, 2), Q(4, −1) and R(2, 5) are the vertices of △PQR. It is rotated about the origin through −90° and then through −180°. Which single transformation is this? Use it to find the image, and draw both triangles on one graph.

Answer

(−90°) + (−180°) = −270°, a clockwise turn of 270°, which ends in the same place as +90° about (0, 0): (x, y) → (−y, x).
P(1, 2) → P′(−2, 1), Q(4, −1) → Q′(1, 4), R(2, 5) → R′(−5, 2).
(Check P step by step: −90° takes (1, 2) to (2, −1); −180° takes that to (−2, 1). ✓)

Figure 9.32 The answer to 3(b): −270 is the same as a quarter turn anticlockwise.

c.

R₁ is a rotation about the origin through 90° and R₂ a rotation about the origin through −270°. What is R₂ ∘ R₁? Use it to transform the quadrilateral with vertices A(−4, 0), B(−6, 2), C(−4, 3) and D(2, 5), and draw both on one graph.

Answer

90° + (−270°) = −180°, a half turn about (0, 0) (the same as +180°): (x, y) → (−x, −y).
A(−4, 0) → A′(4, 0), B(−6, 2) → B′(6, −2), C(−4, 3) → C′(4, −3), D(2, 5) → D′(−2, −5).

Figure 9.33 The answer to 3(c): ABCD and its image after a half turn about O.

d.

A(3, 0), B(4, 2), C(2, 3) and D(1, 2) form a quadrilateral. It is rotated about the origin through 180°, and the image is rotated about the origin through 90° in the positive direction. Draw ABCD and the final image A′B′C′D′ on one graph.

Answer

180° + 90° = 270°: (x, y) → (y, −x).
A(3, 0) → A′(0, −3), B(4, 2) → B′(2, −4), C(2, 3) → C′(3, −2), D(1, 2) → D′(2, −1).

Figure 9.34 The answer to 3(d): ABCD after a turn of 270 anticlockwise about O.

4.

P(2, 1), Q(−1, 4) and R(−2, 2) are the vertices of △PQR. $T_1 = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ and $T_2 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ are two translations.

a.Use the combined transformation T₁ ∘ T₂ (or T₂ ∘ T₁) to find the image of △PQR.
b.Draw △PQR and △P′Q′R′ on one graph.

Answer

a. $T_1 \circ T_2 = T_2 \circ T_1 = \begin{pmatrix} 2 + 3 \\ -3 + 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}$ .
P(2, 1) → P′(7, 0), Q(−1, 4) → Q′(4, 3), R(−2, 2) → R′(3, 1).
b.

Figure 9.35 The answer to 4(b): △PQR moved 5 right and 1 down.

5.

△BUN has vertices B(5, 3), U(1, 4) and N(2, 1). $T_1 = \begin{pmatrix} -3 \\ -2 \end{pmatrix}$ and $T_2 = \begin{pmatrix} -1 \\ -4 \end{pmatrix}$ are two translations.

a.Find the combined transformation T₁ ∘ T₂.
b.Use T₂ ∘ T₁ to find the image of △BUN.
c.Draw △BUN and △B′U′N′ on one graph.

Answer

a. $T_1 \circ T_2 = \begin{pmatrix} -3 - 1 \\ -2 - 4 \end{pmatrix} = \begin{pmatrix} -4 \\ -6 \end{pmatrix}$ .
b.T₂ ∘ T₁ is the same translation. B(5, 3) → B′(1, −3), U(1, 4) → U′(−3, −2), N(2, 1) → N′(−2, −5).

c.

Figure 9.36 The answer to 5(c): △BUN moved 4 left and 6 down.

6.

The line segment PQ has ends P(−3, 5) and Q(7, −4). E₁[(0, 0), 2] and E₂[(0, 0), 3/2] are two enlargements, and PQ is transformed by E₁ ∘ E₂.

a.Find the center and the scale factor of E₁ ∘ E₂.
b.Find the image P′Q′.
c.Draw PQ and P′Q′ on one graph.

Answer

a.The same center, (0, 0); scale factor $2 \times \frac{3}{2} = 3$ . So E₁ ∘ E₂ = E[(0, 0), 3].
b.P(−3, 5) → P′(−9, 15), Q(7, −4) → Q′(21, −12).

Figure 9.37 The answer to 6(c): PQ, from P(−3, 5) to Q(7, −4), and P′Q′, from P′(−9, 15) to Q′(21, −12). P′Q′ is three times as long as PQ, parallel to it, on the same lines from O.

7.

△ABC has vertices A(2, 3), B(4, 5) and C(6, 2). E₁[(0, 0), 4] and E₂[(0, 0), 1/2] are two enlargements.

a.Find the scale factor of E₂ ∘ E₁.
b.Use E₂ ∘ E₁ to find the image △A′B′C′.
c.Draw △ABC and △A′B′C′ on one graph.

Answer

a. $k = 4 \times \frac{1}{2} = 2$ , with center (0, 0).
b.A(2, 3) → A′(4, 6), B(4, 5) → B′(8, 10), C(6, 2) → C′(12, 4).

Figure 9.38 The answer to 7(c): every length of △A′B′C′ is twice the matching length of △ABC.

2.

△ABC has vertices A(2, −1), B(2, 1) and C(4, −1). Reflect it first in the line y − x = 0, then rotate the image about the origin through 180°. Draw △ABC and the final image △A′′B′′C′′ on one graph.

Answer

The line y − x = 0 is y = x, so (x, y) → (y, x): A′(−1, 2), B′(1, 2), C′(−1, 4).
A half turn, (x, y) → (−x, −y): A′′(1, −2), B′′(−1, −2), C′′(1, −4).
(Together: (x, y) → (−y, −x), the reflection in y = −x.)

Figure 9.42 The answer to question 2: △ABC and △A′′B′′C′′ are mirror images in the line y = −x.

a.

△BIN has vertices B(1, 2), I(−1, −2) and N(5, 0). It is translated by $T = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and then reflected in the X-axis. Find the image, and draw △BIN and its image on one graph.

Answer

Translate: B′(2, 4), I′(0, 0), N′(6, 2).
Reflect in the X-axis: B′′(2, −4), I′′(0, 0), N′′(6, −2).

Figure 9.43 The answer to 3(a): △BIN and △B′′I′′N′′. I′′ lands on the origin.

b.

△MAN has vertices M(−1, 2), A(2, 1) and N(1, 0). It is translated by $T = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ ; find the image. Then reflect the image in the line x = −2. Draw △MAN and the final image △M′′A′′N′′ on one graph.

Answer

Translate: M′(0, 4), A′(3, 3), N′(2, 2).
Reflect in x = −2, (x, y) → (−4 − x, y): M′′(−4, 4), A′′(−7, 3), N′′(−6, 2).

Figure 9.44 The answer to 3(b): △MAN and △M′′A′′N′′.

a.

T is a translation and R a rotation. △ABC has vertices A(2, 1), B(2, 4) and C(5, 2). It is translated by AB and then rotated 90° anticlockwise about the origin. Find the image under R ∘ T, and draw △ABC and its image on one graph.

Answer

$\overrightarrow{AB} = \begin{pmatrix} 2 - 2 \\ 4 - 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \end{pmatrix}$ . Translate: A′(2, 4), B′(2, 7), C′(5, 5).
Rotate +90°, (x, y) → (−y, x): A′′(−4, 2), B′′(−7, 2), C′′(−5, 5).

Figure 9.45 The answer to 4(a): △ABC and its image under R ∘ T.

b.

△ABC has vertices A(1, 4), B(4, 1) and C(7, 5). It is translated by AC and then rotated 90° clockwise about the origin. With T for the translation and R for the rotation, find the image under R ∘ T, and draw △ABC and its image on one graph.

Answer

$\overrightarrow{AC} = \begin{pmatrix} 7 - 1 \\ 5 - 4 \end{pmatrix} = \begin{pmatrix} 6 \\ 1 \end{pmatrix}$ . Translate: A′(7, 5), B′(10, 2), C′(13, 6).
Rotate −90°, (x, y) → (y, −x): A′′(5, −7), B′′(2, −10), C′′(6, −13).

Figure 9.46 The answer to 4(b): △ABC and its image under R ∘ T.

a.

△XYZ has vertices X(2, 3), Y(2, 6) and Z(3, 4). It is translated by $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ and then enlarged by E[(0, 0), 2]. Find the image, and draw both triangles on one graph.

Answer

Translate: X′(0, 6), Y′(0, 9), Z′(1, 7).
Enlarge, (x, y) → (2x, 2y): X′′(0, 12), Y′′(0, 18), Z′′(2, 14).

Figure 9.47 The answer to 5(a): △XYZ and △X′′Y′′Z′′, twice as big.

b.

△MNP has vertices M(1, 1), N(3, 1) and P(2, 3). Rotate it 90° anticlockwise about the origin. Then enlarge the image with center (−1, 2) and scale factor 2. Draw △MNP and the final image △M′′N′′P′′ on one graph.

Answer

Rotate +90°, (x, y) → (−y, x): M′(−1, 1), N′(−1, 3), P′(−3, 2).
Enlarge about (−1, 2) with k = 2, (x, y) → (2(x + 1) − 1, 2(y − 2) + 2):
M′′(−1, 0), N′′(−1, 4), P′′(−5, 2).

Figure 9.48 The answer to 5(b): △MNP and △M′′N′′P′′. C(−1, 2) is the center of the enlargement.

a.

E is the enlargement with center (3, 1) and scale factor 2, and R the reflection in the line y = x. △ABC has vertices A(2, 3), B(4, 5) and C(1, −2). Find the image of △ABC under E ∘ R, and draw both triangles on one graph.

Answer

R first, (x, y) → (y, x): A′(3, 2), B′(5, 4), C′(−2, 1).
Then E, (x, y) → (2(x − 3) + 3, 2(y − 1) + 1) = (2x − 3, 2y − 1): A′′(3, 3), B′′(7, 7), C′′(−7, 1).

Figure 9.49 The answer to 6(a): △ABC and its image under E ∘ R.

b.

G₁ is the enlargement with center (3, 4) and scale factor 2, and G₂ the reflection in the line y = x. △PQR has vertices P(−3, 5), Q(7, 4) and R(6, −2). Under G₂ ∘ G₁, find the vertices of △P′Q′R′, and draw both triangles on one graph.

Answer

G₁ first, (x, y) → (2(x − 3) + 3, 2(y − 4) + 4) = (2x − 3, 2y − 4): (−9, 6), (11, 4), (9, −8).
Then G₂, (x, y) → (y, x): P′(6, −9), Q′(4, 11), R′(−8, 9).

Figure 9.50 The answer to 6(b): △PQR and its image △P′Q′R′ under G₂ ∘ G₁; the coordinates are in the answer above.

a.

△CUT has vertices C(2, 5), U(−1, 3) and T(4, 1). Rotate it 90° clockwise about the origin, then enlarge it by E[(0, 0), 2]. Find the image, and draw both triangles on one graph.

Answer

Rotate −90°, (x, y) → (y, −x): C′(5, −2), U′(3, 1), T′(1, −4).
Enlarge by 2: C′′(10, −4), U′′(6, 2), T′′(2, −8).

Figure 9.51 The answer to 7(a): △CUT and △C′′U′′T′′.

b.

The parallelogram GITA has vertices G(2, 2), I(6, 2), T(7, 4) and A(3, 4). Rotate it 90° anticlockwise about the origin, then enlarge it by E[(0, 0), 3]. Find the image, and draw both on one graph.

Answer

Rotate +90°, (x, y) → (−y, x): G′(−2, 2), I′(−2, 6), T′(−4, 7), A′(−4, 3).
Enlarge by 3: G′′(−6, 6), I′′(−6, 18), T′′(−12, 21), A′′(−12, 9).

Figure 9.52 The answer to 7(b): GITA and G′′I′′T′′A′′, three times as big and turned a quarter turn; the coordinates are in the answer above.

a.

△BAT has vertices B(4, 6), A(2, 2) and T(8, 2). It is rotated 90° anticlockwise about the point P(1, 3), then reflected in the X-axis. Find the image, and draw both triangles on one graph.

Hint

To rotate about P(1, 3): find each point’s position from P, turn that +90°, then add P back.

Answer

Rotate +90° about (a, b) = (1, 3): (x, y) → (a − (y − b), b + (x − a)) = (4 − y, x + 2).
B(4, 6) → B′(−2, 6), A(2, 2) → A′(2, 4), T(8, 2) → T′(2, 10).
Reflect in the X-axis: B′′(−2, −6), A′′(2, −4), T′′(2, −10).

Figure 9.53 The answer to 8(a): △BAT and △B′′A′′T′′. P(1, 3) is the center of the rotation.

b.

The parallelogram DUCK has vertices D(2, 3), U(1, 1), C(4, 1) and K(5, 3). It is rotated 90° anticlockwise about the point A(4, 0), then reflected in the line x = y. Find the image, and draw both on one graph.

Answer

Rotate +90° about (4, 0): (x, y) → (4 − y, x − 4).
D(2, 3) → D′(1, −2), U(1, 1) → U′(3, −3), C(4, 1) → C′(3, 0), K(5, 3) → K′(1, 1).
Reflect in y = x: D′′(−2, 1), U′′(−3, 3), C′′(0, 3), K′′(1, 1).
(K′ lies on the mirror, so K′′ = K′.)

Figure 9.54 The answer to 8(b): DUCK (blue) and D′′U′′C′′K′′ (red). K′′ lands on U. A(4, 0) is the center of the rotation.

2.

Define: unit square; matrix transformation.

Answer

A unit square is a square whose sides are 1 unit long; the one usually used has vertices O(0, 0), A(1, 0), B(1, 1) and C(0, 1), written $\begin{pmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}$ .
A matrix transformation moves a point or a figure by working with matrices: the coordinates are written as columns and either a 2 × 1 matrix is added (a translation) or the columns are multiplied by a 2 × 2 matrix (image = matrix × object).

3.

Transform the point A(−4, 6) by each of these matrices:

a. $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$
b. $\begin{pmatrix} 6 \\ 0 \end{pmatrix}$
c. $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$
d. $\begin{pmatrix} 1/2 \\ 1/3 \end{pmatrix}$

Answer

They are 2 × 1 matrices, so add:

a.(−4 + 3, 6 + 0) = (−1, 6)
b.(−4 + 6, 6 + 0) = (2, 6)
c.(−4 + 3, 6 + 2) = (−1, 8)
d. $\left(-4 + \frac{1}{2},\ 6 + \frac{1}{3}\right) = \left(-\frac{7}{2},\ \frac{19}{3}\right)$

a.

Find the 2 × 2 matrix that transforms (x, y) to (2x + y, 3x + 4y).

Answer

The numbers in front of x and y in each new coordinate are the rows: $\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ . Check: $\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2x + y \\ 3x + 4y \end{pmatrix}$ . ✓

b.

Find the matrix that transforms (x, y) to (x − 2y, 2x − 3y).

Answer

$\begin{pmatrix} 1 & -2 \\ 2 & -3 \end{pmatrix}$

a.

P(3, 4) and Q(8, 4) are the ends of PQ. Find the image of PQ under the matrix $\begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix}$ .

Answer

$\begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 3 & 8 \\ 4 & 4 \end{pmatrix} = \begin{pmatrix} 9 + 8 & 24 + 8 \\ 3 + 4 & 8 + 4 \end{pmatrix} = \begin{pmatrix} 17 & 32 \\ 7 & 12 \end{pmatrix}$
P′(17, 7) and Q′(32, 12).

b.

A rectangle has vertices A(2, 3), B(2, 6), C(5, 6) and D(5, 3). The 2 × 2 matrix $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$ transforms it to A′B′C′D′. Find A′, B′, C′ and D′.

Answer

$\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 2 & 2 & 5 & 5 \\ 3 & 6 & 6 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 2 & 5 & 5 \\ 7 & 10 & 16 & 13 \end{pmatrix}$
A′(2, 7), B′(2, 10), C′(5, 16), D′(5, 13). The rectangle becomes a parallelogram: each point slides up by twice its x-coordinate (a shear).

c.

The unit square $\begin{pmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}$ is transformed by the matrix $\begin{pmatrix} 3 & 2 \\ 1 & 5 \end{pmatrix}$ into the quadrilateral O′A′B′C′. Find its vertices.

Answer

$\begin{pmatrix} 3 & 2 \\ 1 & 5 \end{pmatrix} \begin{pmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 3 & 5 & 2 \\ 0 & 1 & 6 & 5 \end{pmatrix}$
O′(0, 0), A′(3, 1), B′(5, 6), C′(2, 5): the first column of the matrix is A′, the second C′, as Figure 9.56 shows.

a.

Find the 2 × 2 matrix that transforms the unit square $\begin{pmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}$ into the parallelogram $\begin{pmatrix} 0 & 3 & 4 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix}$ .

Hint

The image of A(1, 0) is the matrix’s first column, and the image of C(0, 1) its second.

Answer

A(1, 0) → (3, 0) and C(0, 1) → (1, 1), so the matrix is $\begin{pmatrix} 3 & 1 \\ 0 & 1 \end{pmatrix}$ .
Check B(1, 1): (3 + 1, 0 + 1) = (4, 1). ✓

b.

Find the 2 × 2 matrix that transforms the unit square into the parallelogram $\begin{pmatrix} 0 & 3 & 4 & 1 \\ 0 & 1 & 3 & 2 \end{pmatrix}$ .

Answer

A(1, 0) → (3, 1) and C(0, 1) → (1, 2), so the matrix is $\begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$ .
Check B(1, 1): (3 + 1, 1 + 2) = (4, 3). ✓
(Or the long way: with $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the image of A(1, 0) is (a, c) = (3, 1) and of C(0, 1) is (b, d) = (1, 2).)

c.

Find the 2 × 2 matrix that transforms the unit square into the parallelogram $\begin{pmatrix} 0 & 2 & 5 & 3 \\ 0 & 3 & 4 & 1 \end{pmatrix}$ .

Answer

A(1, 0) → (2, 3) and C(0, 1) → (3, 1), so the matrix is $\begin{pmatrix} 2 & 3 \\ 3 & 1 \end{pmatrix}$ .
Check B(1, 1): (2 + 3, 3 + 1) = (5, 4). ✓

a.

Prove with matrices that a rotation of −90° about the origin followed by the reflection in the line x = 0 is the same as the reflection in the line y = −x.

Answer

Rotation of −90°: $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ . Reflection in x = 0 (the Y-axis): $B = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ .
The rotation comes first, so the combined matrix is
$BA = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$
which is the matrix of the reflection in y = −x. Proved.

b.

Prove with matrices that a rotation of +90° about the origin followed by the reflection in the Y-axis is the same as the reflection in the line y = x.

Answer

Rotation of +90°: $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ . Reflection in the Y-axis: $B = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ .
$BA = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
which is the matrix of the reflection in y = x. Proved.

1.

A circle passes through the points (−2, 0) and (0, −2), and its center lies on the line 2x − 3y + 1 = 0.

a.Define the circle as a conic section.
b.Find the center of the circle.
c.Find the equation of the line through the center perpendicular to the line 2x − 3y + 1 = 0.

Answer

a.A circle is the curve in which a plane cuts a cone when the plane is perpendicular to the cone’s axis (parallel to its base).
b.Let the center be (h, k). It is the same distance from both points:

$(h + 2)^2 + k^2 = h^2 + (k + 2)^2$ , so 4h + 4 = 4k + 4, and h = k.
It lies on the line: 2h − 3h + 1 = 0, so h = 1 and k = 1. The center is (1, 1).
c. The slope of 2x − 3y + 1 = 0 is 2/3, so a perpendicular line has slope −3/2. Through (1, 1):
y − 1 = −3/2(x − 1), 2y − 2 = −3x + 3, so 3x + 2y − 5 = 0.

2.

A line segment PQ joins P(2, −3) and Q(−4, 9).

a.Find the equation of the line PQ.
b.Find the equation of the line parallel to PQ through M(−1, 1).
c.Find the equation of the line perpendicular to PQ through N(1, 3).
d.Translate P and Q by the vector PQ.

Answer

a.Slope = (9 − (−3)) / (−4 − 2) = 12/(−6) = −2. Through P: y + 3 = −2(x − 2), so 2x + y − 1 = 0.
b.The same slope, −2, through (−1, 1): y − 1 = −2(x + 1), so 2x + y + 1 = 0.
c.Slope 1/2 through (1, 3): $y - 3 = \frac{1}{2}(x - 1)$ , 2y − 6 = x − 1, so x − 2y + 5 = 0.
d. $\overrightarrow{PQ} = \begin{pmatrix} -6 \\ 12 \end{pmatrix}$ : P(2, −3) → P′(−4, 9) (which is Q) and Q(−4, 9) → Q′(−10, 21).

3.

A cone has a vertex angle of 70°. A plane cuts it in a parabola.

a.Define the parabola as a conic.
b.Find the angle between the axis of the cone and the cutting plane.

Answer

a.A parabola is the curve in which a plane cuts a cone when the plane is parallel to one of the cone’s slant sides (a generator).
b.The vertex angle is 70°, so each slant side makes 70° ÷ 2 = 35° with the axis. The plane is parallel to a slant side, so it also makes 35° with the axis.

4.

A circle has the equation $4x^2 + 4y^2 - 24x - 20y - 3 = 0$ .

a.If a plane cuts a cone parallel to its base, which conic is formed?
b.Find the center and the radius of the circle.
c.One end of a diameter of the circle is (1, 2). Find the equation of that diameter.

Answer

a.A circle.
b.Divide by 4: $x^2 + y^2 - 6x - 5y - 3/4 = 0$ . Compare with $x^2 + y^2 + 2gx + 2fy + c = 0$ : g = −3, f = −5/2, c = −3/4.

Center (−g, −f) = (3, 5/2); radius = √(9 + 25/4 + 3/4) = √16 = 4 units.
c. The diameter passes through (1, 2) and the center (3, 5/2): slope = (5/2 − 2)/(3 − 1) = 1/4.
$y - 2 = \frac{1}{4}(x - 1)$ , 4y − 8 = x − 1, so x − 4y + 7 = 0. (Its other end is (2 × 3 − 1, 2 × 5/2 − 2) = (5, 3).)

5.

A circle has the equation $x^2 + y^2 - 2x + 4y - 4 = 0$ .

a.Define the circle, and write its equation in standard form.
b.Find the center of the circle.
c.One end of a diameter is (1, 1). Find the equation of the line through B(−3, 1) perpendicular to this diameter.

Answer

a.A circle is the set of all points in a plane at a fixed distance (the radius) from a fixed point (the center). Completing the squares: $(x - 1)^2 + (y + 2)^2 = 9$ , the standard form $(x - h)^2 + (y - k)^2 = r^2$ with radius 3.
b.(1, −2), radius 3 units.
c.The diameter through (1, 1) and the center (1, −2) is the upright line x = 1. A line perpendicular to it is level, and through B(−3, 1) it is y = 1, that is y − 1 = 0.

6.

R₁ is the reflection in the X-axis and R₂ the reflection in the Y-axis. A(1, 5), B(5, 1) and C(8, 3) are the vertices of △ABC.

a.Find the equations of the sides AB and BC.
b.Which single transformation is R₁ ∘ R₂?
c.Use it to find the image △A′B′C′.
d.Draw △ABC and △A′B′C′ on one graph.

Answer

a. AB: slope (1 − 5)/(5 − 1) = −1; y − 5 = −(x − 1), so x + y − 6 = 0.
BC: slope (3 − 1)/(8 − 5) = 2/3; $y - 1 = \frac{2}{3}(x - 5)$ , 3y − 3 = 2x − 10, so 2x − 3y − 7 = 0.

b.The axes cross at O at 90°: a rotation of 180° about (0, 0).
c.(x, y) → (−x, −y): A′(−1, −5), B′(−5, −1), C′(−8, −3).

Figure 9.60 The answer to 6(d): △ABC and its image after a half turn about O.

7.

P(1, 2), Q(5, 2) and R(1, 5) are the vertices of △PQR.

a.Write the matrix of the reflection in the X-axis.
b.Find the image of △PQR under the matrix $\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix}$ .
c.Find the equation of the perpendicular PS from P to the side QR.

Answer

a. $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
b. $\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 5 & 1 \\ 2 & 2 & 5 \end{pmatrix} = \begin{pmatrix} 7 & 11 & 16 \\ 4 & 12 & 7 \end{pmatrix}$

P′(7, 4), Q′(11, 12), R′(16, 7).
c. Slope of QR = (5 − 2)/(1 − 5) = −3/4, so PS has slope 4/3. Through P(1, 2): y − 2 = 4/3(x − 1), 3y − 6 = 4x − 4, so 4x − 3y + 2 = 0.

8.

△MUN has vertices M(−2, 2), U(2, 2) and N(2, 8). It is transformed by the vectors $T_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $T_2 = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ .

a.Find the combined transformation T₁ ∘ T₂.
b.Use T₁ ∘ T₂ to find the image △M′U′N′.
c.Draw △MUN and △M′U′N′ on one graph.

Answer

a. $T_1 \circ T_2 = \begin{pmatrix} 1 + 3 \\ 2 + 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ .
b.M′(2, 5), U′(6, 5), N′(6, 11).

Figure 9.61 The answer to 8(c): △MUN moved 4 right and 3 up.

9.

R₁ is the reflection in the line y = x and R₂ the reflection in the line x = 0. A(3, 2), B(1, 2) and C(2, 3) are the vertices of △ABC.

a.Which single transformation is R₁ ∘ R₂?
b.Use it to find the image △A′B′C′.
c.Draw △ABC and △A′B′C′ on one graph.
d.Find the equation of the altitude AD from A to the side BC.

Answer

a.R₂ first: (x, y) → (−x, y); then R₁: → (y, −x). That is a rotation of −90° (the same as +270°) about (0, 0). (The mirrors cross at O at 45°, and the turn of 90° goes from the first mirror, x = 0, towards the second, y = x: clockwise.)
b.(x, y) → (y, −x): A′(2, −3), B′(2, −1), C′(3, −2).
d.Slope of BC = (3 − 2)/(2 − 1) = 1, so AD has slope −1. Through A(3, 2): y − 2 = −(x − 3), so x + y − 5 = 0.

Figure 9.62 The answer to 9(c): △ABC and its image after a quarter turn clockwise about O.

10.

△BUT has vertices B(3, 2), U(1, −1) and T(5, −5).

a.Two straight lines have slopes m₁ and m₂, and the angle between them is θ. Write the relation between m₁, m₂ and θ.
b.Find the equation of the line through the centroid G of △BUT parallel to the side UT.
c.Find the image △B′U′T′ under the matrix $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ .

Answer

a. $\tan\theta = \pm\,\frac{m_1 - m_2}{1 + m_1 m_2}$
b.G = ((3 + 1 + 5)/3, (2 − 1 − 5)/3) = (3, −4/3). Slope of UT = (−5 + 1)/(5 − 1) = −1.

y + 4/3 = −(x − 3); multiply by 3: 3y + 4 = −3x + 9, so 3x + 3y − 5 = 0.
c. The matrix is a half turn about O, (x, y) → (−x, −y): B′(−3, −2), U′(−1, 1), T′(−5, 5).

Exam practice

Unit 9 · Transformation

Exercise 9.1

Question 2

Questions 3 to 13

Exercise 9.2.1

Question 2

Question 3

Questions 4 to 7

Exercise 9.2.2

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Exercise 9.3

Questions 2 and 3

Question 4

Question 5

Question 6

Question 7

Miscellaneous exercise