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Shape and Space - Combining Transformations

Grade 6IB

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

🔑Concepts

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A combined transformation is a sequence of two or more transformations applied one after the other. The image produced by the first transformation becomes the object for the second transformation.

A coordinate plane showing a triangle reflected first across the y-axis and then across the x-axis.
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The order of transformations matters. Applying a translation then a reflection may result in a different final position than applying the reflection then the translation.

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Multiple translations can be combined by adding their vectors. For example, a translation by (ab)\begin{pmatrix} a \\ b \end{pmatrix} followed by (cd)\begin{pmatrix} c \\ d \end{pmatrix} is equivalent to a single translation of (a+cb+d)\begin{pmatrix} a + c \\ b + d \end{pmatrix}.

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When reflecting an object twice across two parallel lines, the result is equivalent to a single translation. When reflecting across two intersecting lines, the result is a rotation.

📐Formulae

Translation by vector \text{Translation by vector } \begin{pmatrix} a \ b \end{pmatrix}:(x,y)→(x+a,y+b): (x, y) \rightarrow (x + a, y + b) roller

Reflection in the x-axis:(x,y)→(x,−y)\text{Reflection in the } x\text{-axis}: (x, y) \rightarrow (x, -y)

Reflection in the y-axis:(x,y)→(−x,y)\text{Reflection in the } y\text{-axis}: (x, y) \rightarrow (-x, y)

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

💡Examples

Problem 1:

A triangle has a vertex at P(1,2)P(1, 2). It is first translated by the vector (4−3)\begin{pmatrix} 4 \\ -3 \end{pmatrix} and then reflected across the yy-axis. Find the final coordinates of point PP.

Solution:

Step 1 (Translation): Add the vector components to the coordinates: (1+4,2+(−3))=(5,−1)(1 + 4, 2 + (-3)) = (5, -1) Step 2 (Reflection): Reflect (5,−1)(5, -1) across the yy-axis by changing the sign of the xx-coordinate: (−5,−1)(-5, -1) The final coordinates are P′′(−5,−1)P''(-5, -1).

Explanation:

We first apply the translation to the original point to get an intermediate point P′(5,−1)P'(5, -1). We then apply the reflection rule for the yy-axis, which is (x,y)→(−x,y)(x, y) \rightarrow (-x, y), to P′P' to find the final position.

Problem 2:

Point QQ is at (3,4)(3, 4). It is reflected across the xx-axis and then rotated 180∘180^\circ about the origin. Determine the final position of QQ.

Solution:

Step 1 (Reflection in xx-axis): (3,4)→(3,−4)(3, 4) \rightarrow (3, -4) Step 2 (Rotation 180∘180^\circ): (3,−4)→(−3,−(−4))=(−3,4) (3, -4) \rightarrow (-3, -(-4)) = (-3, 4) The final coordinates are Q′′(−3,4)Q''(-3, 4).

Explanation:

Reflecting across the xx-axis negates the yy-value. Rotating 180∘180^\circ negates both the xx and yy values of the intermediate point.

Problem 3:

Calculate the total vertical shift if a shape is translated by (015)\begin{pmatrix} 0 \\ 15 \end{pmatrix} and then by (0−7)\begin{pmatrix} 0 \\ -7 \end{pmatrix}.

Solution:

15−78\begin{array}{r} 15 \\ -7 \\ \hline 8 \end{array} The final vertical translation is 88 units up.

Explanation:

When combining translations, we can simply add the corresponding components of the vectors. 15+(−7)=815 + (-7) = 8.

Problem 4:

A square has a vertex at A(2,1)A(2, 1). It is rotated 90∘90^{\circ} counter-clockwise about the origin O(0,0)O(0,0) and then translated by the vector (−12)\begin{pmatrix} -1 \\ 2 \end{pmatrix}. Determine the final coordinates of vertex AA.

Graph showing point A rotating to A' and then translating to A''.

Solution:

  1. Rotation: The rule for 90∘90^{\circ} counter-clockwise rotation about (0,0)(0,0) is (x,y)→(−y,x)(x, y) \rightarrow (-y, x). Applying this to A(2,1)A(2, 1) gives A′(−1,2)A'(-1, 2).
  2. Translation: Apply the vector (−12)\begin{pmatrix} -1 \\ 2 \end{pmatrix} to A′(−1,2)A'(-1, 2). (−1+(−1),2+2)=(−2,4)(-1 + (-1), 2 + 2) = (-2, 4). The final coordinates are (−2,4)(-2, 4).

Explanation:

First, we apply the rotation rule to find the intermediate position. Then, we add the translation components to the intermediate xx and yy values to find the final position.

Problem 5:

A shape is reflected in the line y=0y = 0 (xx-axis) and then reflected again in the line x=0x = 0 (yy-axis). If a point on the shape was at (4,3)(4, 3), where is it now?

Diagram showing the movement of a point through two successive reflections in the axes.

Solution:

  1. Reflection in xx-axis: The rule is (x,y)→(x,−y)(x, y) \rightarrow (x, -y). Point (4,3)(4, 3) becomes (4,−3)(4, -3).
  2. Reflection in yy-axis: The rule is (x,y)→(−x,y)(x, y) \rightarrow (-x, y). Point (4,−3)(4, -3) becomes (−4,−3)(-4, -3). Final position: (−4,−3)(-4, -3).

Explanation:

Reflecting across the x-axis flips the sign of the y-coordinate. Reflecting that result across the y-axis flips the sign of the x-coordinate. This combined transformation is equivalent to a 180∘180^{\circ} rotation about the origin.