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Geometry and Trigonometry - Identical representation of transformations-extended

Grade 9IB

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

🔑Concepts

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Transformations are functions that map an initial shape (the object) onto a new shape (the image). The four basic types are Translation, Reflection, Rotation, and Enlargement.

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In the extended IB Grade 9 curriculum, transformations can be represented using mapping notation: (x,y)→(x′,y′)(x, y) \rightarrow (x', y'). This describes how each coordinate of a point is altered.

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Identical representation involves showing that two different transformations, or a sequence of transformations, can be expressed by a single transformation or a matrix multiplication.

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Geometric transformations can be represented using 2×22 \times 2 matrices where the image point P′(x′y′)P' \begin{pmatrix} x' \\ y' \end{pmatrix} is found by multiplying the transformation matrix MM by the object point P(xy)P \begin{pmatrix} x \\ y \end{pmatrix}: P′=M×PP' = M \times P.

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A sequence of transformations T1T_1 followed by T2T_2 is represented by the matrix product M2×M1M_2 \times M_1. Note the order: the first transformation is on the right.

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Invariant points are points that remain fixed under a specific transformation. For any invariant point, MM (xy)=(xy)\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}.

📐Formulae

Translation by vector (hk):(x,y)→(x+h,y+k)Translation \text{ by vector } \begin{pmatrix} h \\ k \end{pmatrix}: (x, y) \rightarrow (x + h, y + k)

Reflection in x-axis:(100−1)Reflection \text{ in } x\text{-axis}: \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

Reflection in y-axis:(−1001)Reflection \text{ in } y\text{-axis}: \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}

Reflection in line y=x:(0110)Reflection \text{ in line } y = x: \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

Rotation 90∘ Counter-clockwise about origin:(0−110)Rotation \text{ } 90^\circ \text{ Counter-clockwise about origin}: \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

Rotation 180∘ about origin:(−100−1)Rotation \text{ } 180^\circ \text{ about origin}: \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}

Enlargement (Dilation) by factor k from origin:(k00k)Enlargement \text{ (Dilation) by factor } k \text{ from origin}: \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}

💡Examples

Problem 1:

Find the image of the point A(3,−2)A(3, -2) after a reflection in the yy-axis followed by a translation of (25)\begin{pmatrix} 2 \\ 5 \end{pmatrix}.

Solution:

  1. Reflection in yy-axis: (x,y)→(−x,y)(x, y) \rightarrow (-x, y). Point A(3,−2)A(3, -2) becomes A′(−3,−2)A'(-3, -2).
  2. Translation by (25)\begin{pmatrix} 2 \\ 5 \end{pmatrix}: (−3+2,−2+5)=(−1,3)(-3 + 2, -2 + 5) = (-1, 3). Final image is A′′(−1,3)A''(-1, 3).

Explanation:

We first apply the reflection mapping and then add the translation vector components to the resulting coordinates.

Problem 2:

Determine the 2×22 \times 2 matrix that represents a rotation of 90∘90^\circ counter-clockwise about the origin followed by a reflection in the line y=xy=x.

Solution:

Let R=(0−110)R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} be the rotation and M=(0110)M = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} be the reflection. Combined Matrix C=M×RC = M \times R: C=(0110)(0−110)C = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} C=((0×0+1×1)(0×−1+1×0)(1×0+0×1)(1×−1+0×0))C = \begin{pmatrix} (0 \times 0 + 1 \times 1) & (0 \times -1 + 1 \times 0) \\ (1 \times 0 + 0 \times 1) & (1 \times -1 + 0 \times 0) \end{pmatrix} C=(100−1)C = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

Explanation:

Matrix multiplication is used to combine transformations. The matrix for the second transformation is written to the left of the first. The resulting matrix (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} is identical to a reflection in the xx-axis.

Problem 3:

Calculate the result of the following coordinate adjustment: 150−8565\begin{array}{r} 150 \\ - 85 \\ \hline 65 \end{array} If these represent xx-coordinates of a shape moving left by 8585 units, find the new xx if the original was 150150.

Solution:

The new xx-coordinate is 6565.

Explanation:

A horizontal translation left is represented by subtracting from the xx-coordinate: x′=x−85x' = x - 85.