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Lesson Worksheet: Linear Transformations in Planes: Reflection Mathematics • 10th Grade

In this worksheet, we will practice finding the matrix of linear transformation of reflection along the x- or y-axis or the line of a given equation and the image of a vector under the reflection.

Q1:

Which of the following are necessary and sufficient conditions on π‘Ž, 𝑏, 𝑐, and 𝑑 for the matrix ο”π‘Žπ‘π‘π‘‘ο  to represent a reflection?

  • A𝑏=𝑐, 𝑑=βˆ’π‘Ž, π‘Žβˆ’π‘=1
  • B𝑏=βˆ’π‘, 𝑑=βˆ’π‘Ž, π‘Žβˆ’π‘=1
  • C𝑏=βˆ’π‘, 𝑑=βˆ’π‘Ž, π‘Ž+𝑐=1
  • D𝑏=𝑐, 𝑑=π‘Ž, π‘Ž+𝑐=1
  • E𝑏=𝑐, 𝑑=βˆ’π‘Ž, π‘Ž+𝑐=1

Q2:

A reflection in a line through the origin sends the vector 34 to 43. Find the matrix representation of this reflection.

  • A⎑⎒⎒⎣2425725βˆ’7252425⎀βŽ₯βŽ₯⎦
  • B1101
  • C0110
  • D⎑⎒⎒⎣4535βˆ’3545⎀βŽ₯βŽ₯⎦
  • E⎑⎒⎒⎣2425725725βˆ’2425⎀βŽ₯βŽ₯⎦

Q3:

A reflection in a line through the origin sends the vector 34 to 4βˆ’3. Find the matrix representation of this reflection.

  • A⎑⎒⎒⎣2425725βˆ’7252425⎀βŽ₯βŽ₯⎦
  • B⎑⎒⎒⎣453535βˆ’45⎀βŽ₯βŽ₯⎦
  • C⎑⎒⎒⎣4535βˆ’3545⎀βŽ₯βŽ₯⎦
  • D0110
  • E⎑⎒⎒⎣2425725725βˆ’2425⎀βŽ₯βŽ₯⎦

Q4:

Consider the reflection in the line 𝑦=12π‘₯.

Find the matrix that represents this transformation.

  • A⎑⎒⎒⎣35454535⎀βŽ₯βŽ₯⎦
  • B⎑⎒⎒⎣12βˆ’121212⎀βŽ₯βŽ₯⎦
  • C⎑⎒⎒⎣354545βˆ’35⎀βŽ₯βŽ₯⎦
  • D⎑⎒⎒⎣453535βˆ’45⎀βŽ₯βŽ₯⎦
  • E⎑⎒⎒⎣121212βˆ’12⎀βŽ₯βŽ₯⎦

What is the image of the point (12,5) under this reflection?

  • Aο€Ό172,72
  • Bο€Ό565,335
  • Cο€Ό72,172
  • Dο€Ό565,635
  • Eο€Ό635,165

Q5:

Consider the linear transformation which maps a point to its reflection in the π‘₯-axis.

Find the matrix 𝐴 which represents this transformation.

  • A𝐴=10βˆ’11
  • B𝐴=1001
  • C𝐴=ο”βˆ’101βˆ’1
  • D𝐴=1101
  • E𝐴=100βˆ’1

Where does this transformation map the point (2,βˆ’3)?

  • A(βˆ’2,3)
  • B(2,3)
  • C(βˆ’2,3)
  • D(βˆ’2,βˆ’3)
  • E(2,βˆ’3)

Q6:

Consider the given figure.

The points 𝑂(0,0), 𝐴(1,0), 𝐡(1,1), and 𝐢(0,1) are corners of the unit square. This square is reflected in the line 𝑂𝐷 with equation 𝑦=12π‘₯ to form the image π‘‚π΄π΅πΆβˆ—βˆ—βˆ—.

As π΄βˆ— is the image of 𝐴 in the line through 𝑂 and 𝐷, π‘šβˆ π΄π‘‚π΄=2π‘šβˆ π·π‘‚π΄βˆ—. Use this fact and the identity tantantan2πœƒ=2πœƒ1βˆ’πœƒοŠ¨ to find the gradient and hence equation of βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π‘‚π΄βˆ— from the gradient of ⃖⃗𝑂𝐷.

  • A𝑦=43π‘₯
  • B𝑦=βˆ’23π‘₯
  • C𝑦=34π‘₯
  • D𝑦=βˆ’43π‘₯
  • E𝑦=23π‘₯

Using the fact that βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π‘‚πΆβˆ— is perpendicular to βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π‘‚π΄βˆ—, find the equation of βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π‘‚πΆβˆ—.

  • A𝑦=βˆ’43π‘₯
  • B𝑦=βˆ’32π‘₯
  • C𝑦=43π‘₯
  • D𝑦=34π‘₯
  • E𝑦=βˆ’34π‘₯

Using the fact that 𝑂𝐢=𝑂𝐴=1βˆ—βˆ—, find the coordinates of πΆβˆ— and π΄βˆ—.

  • A𝐢=ο€Ό1625,βˆ’925οˆβˆ—, 𝐴=ο€Ό925,1625οˆβˆ—
  • B𝐢=ο€Όβˆ’35,βˆ’45οˆβˆ—, 𝐴=ο€Ό45,35οˆβˆ—
  • C𝐢=ο€Ό45,βˆ’35οˆβˆ—, 𝐴=ο€Ό35,45οˆβˆ—
  • D𝐢=ο€Ό47,βˆ’37οˆβˆ—, 𝐴=ο€Ό37,47οˆβˆ—
  • E𝐢=ο€Όβˆ’37,47οˆβˆ—, 𝐴=ο€Ό47,37οˆβˆ—

Using the fact that a reflection in a line through the origin is a linear transformation, find the matrix which represents reflection in the line 𝑦=12π‘₯.

  • A⎑⎒⎒⎣453535βˆ’45⎀βŽ₯βŽ₯⎦
  • B⎑⎒⎒⎣473737βˆ’47⎀βŽ₯βŽ₯⎦
  • C⎑⎒⎒⎣92516251625βˆ’925⎀βŽ₯βŽ₯⎦
  • D⎑⎒⎒⎣354545βˆ’35⎀βŽ₯βŽ₯⎦
  • E⎑⎒⎒⎣374747βˆ’37⎀βŽ₯βŽ₯⎦

Q7:

Consider the matrix 𝑀=ο“π›Όπ›Όπ›Όβˆ’π›ΌοŸ where 𝛼=√22.

Find π‘€οŠ¨.

  • A100βˆ’1
  • B1001
  • Cο”βˆ’1001
  • D1111
  • E⎑⎒⎒⎣12121212⎀βŽ₯βŽ₯⎦

Find det(𝑀).

  • A2
  • B1
  • C12
  • D0
  • Eβˆ’1

By drawing the image of the unit square under the transformation, identify the geometrical transformation this matrix corresponds to.

  • Aa projection onto the line 𝑦=π‘₯
  • Ba rotation by 45∘ clockwise about the point (1,0)
  • Ca reflection in the line 𝑦=(22.5)π‘₯tan∘
  • Da rotation of 45∘ clockwise about the origin
  • Ea reflection in the line 𝑦=π‘₯

Q8:

Consider the given figure.

The points 𝑂(0,0), 𝐴(1,0), 𝐡(1,1), and 𝐢(0,1) are corners of the unit square. This square is reflected in the line 𝑂𝐷 with equation 𝑦=π‘˜π‘₯ to form the image π‘‚π΄π΅πΆβˆ—βˆ—βˆ—.

As π΄βˆ— is the image of 𝐴 in the line through 𝑂 and 𝐷, π‘šβˆ π΄π‘‚π΄=2π‘šβˆ π·π‘‚π΄βˆ—. Use this fact and the identity tantantan2πœƒ=2πœƒ1βˆ’πœƒοŠ¨ to find the gradient and hence equation of βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π‘‚π΄βˆ— from the gradient of ⃖⃗𝑂𝐷.

  • A𝑦=2π‘˜1βˆ’π‘˜π‘₯
  • B𝑦=2π‘˜1+π‘˜π‘₯
  • C𝑦=2π‘˜π‘˜βˆ’1π‘₯
  • D𝑦=π‘˜π‘˜βˆ’1π‘₯
  • E𝑦=π‘˜1βˆ’π‘˜π‘₯

Using the fact that βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π‘‚πΆβˆ— is perpendicular to βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π‘‚π΄βˆ—, find the equation of βƒ–οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©οƒ©βƒ—π‘‚πΆβˆ—.

  • A𝑦=π‘˜βˆ’12π‘˜π‘₯
  • B𝑦=1βˆ’π‘˜2π‘˜π‘₯
  • C𝑦=2π‘˜1βˆ’π‘˜π‘₯
  • D𝑦=π‘˜βˆ’12π‘˜π‘₯
  • E𝑦=2π‘˜π‘˜βˆ’1π‘₯

Using the fact that 𝑂𝐢=𝑂𝐴=1βˆ—βˆ—, find the coordinates of πΆβˆ— and π΄βˆ—.

  • A𝐢=ο€Ύ2π‘˜1+π‘˜,π‘˜βˆ’11+π‘˜οŠβˆ—οŠ¨, 𝐴=ο€Ύ1βˆ’π‘˜1+π‘˜,2π‘˜1+π‘˜οŠβˆ—οŠ¨
  • B𝐢=ο€Ύπ‘˜βˆ’11+π‘˜,2π‘˜1+π‘˜οŠβˆ—οŠ¨οŠ¨οŠ¨, 𝐴=ο€Ύ2π‘˜1+π‘˜,1βˆ’π‘˜1+π‘˜οŠβˆ—οŠ¨οŠ¨οŠ¨
  • C𝐢=ο€Ύπ‘˜1+π‘˜,π‘˜βˆ’11+π‘˜οŠβˆ—οŠ¨οŠ¨οŠ¨, 𝐴=ο€Ύ1βˆ’π‘˜1+π‘˜,π‘˜1+π‘˜οŠβˆ—οŠ¨οŠ¨οŠ¨
  • D𝐢=ο€Ύ2π‘˜1+π‘˜,π‘˜βˆ’11+π‘˜οŠβˆ—οŠ¨οŠ¨οŠ¨, 𝐴=ο€Ύ1βˆ’π‘˜1+π‘˜,2π‘˜1+π‘˜οŠβˆ—οŠ¨οŠ¨οŠ¨
  • E𝐢=ο€Ύπ‘˜βˆ’11+π‘˜,2π‘˜1+π‘˜οŠβˆ—οŠ¨, 𝐴=ο€Ύ2π‘˜1+π‘˜,1βˆ’π‘˜1+π‘˜οŠβˆ—οŠ¨

Using the fact that a reflection in a line through the origin is a linear transformation, find the matrix which represents reflection in the line 𝑦=π‘˜π‘₯.

  • A⎑⎒⎒⎒⎣1βˆ’π‘˜1+π‘˜2π‘˜1+π‘˜2π‘˜1+π‘˜π‘˜βˆ’11+π‘˜βŽ€βŽ₯βŽ₯βŽ₯⎦
  • B⎑⎒⎒⎒⎣1βˆ’π‘˜1+π‘˜2π‘˜1+π‘˜2π‘˜1+π‘˜π‘˜βˆ’11+π‘˜βŽ€βŽ₯βŽ₯βŽ₯⎦
  • C⎑⎒⎒⎒⎣1βˆ’π‘˜1+π‘˜2π‘˜1+π‘˜2π‘˜1+π‘˜1βˆ’π‘˜1+π‘˜βŽ€βŽ₯βŽ₯βŽ₯⎦
  • D⎑⎒⎒⎒⎣1+π‘˜1βˆ’π‘˜1+π‘˜2π‘˜1+π‘˜2π‘˜1+π‘˜π‘˜βˆ’1⎀βŽ₯βŽ₯βŽ₯⎦
  • E⎑⎒⎒⎒⎣1βˆ’π‘˜1+π‘˜2π‘˜1+π‘˜βˆ’2π‘˜1+π‘˜1βˆ’π‘˜1+π‘˜βŽ€βŽ₯βŽ₯βŽ₯⎦

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