Home / IB Mathematics AHL 3.9 Geometric transformations AI HL Paper 1- Exam Style Questions

IB Mathematics AHL 3.9 Geometric transformations AI HL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AI HL Paper 1 - All Topics

Question

A triangle has vertices at the coordinates \( R(-2,2) \), \( S(3,-5) \), and \( T(5,1) \). This triangle is subjected to two consecutive geometric transformations:
I. A transformation defined by the matrix \( A = \begin{pmatrix} \cos \frac{\pi}{3} & -\sin \frac{\pi}{3} \\ \sin \frac{\pi}{3} & \cos \frac{\pi}{3} \end{pmatrix} \).
II. A transformation defined by the matrix \( B = \begin{pmatrix} -0.8 & 0.6 \\ 0.6 & 0.8 \end{pmatrix} \).
(a) Characterize the specific geometric nature of the transformation represented by matrix \( A \).
It is noted that the final position of triangle \( RST \) can be reached using only a single reflection across a line passing through the origin with equation \( y = a x \).
(b) Determine the value of the constant \( a \), given that \( a > 0 \).

Most-appropriate topic codes (IB Mathematics: Applications and Interpretation HL):

• AHL 3.9: Matrix transformations in two dimensions; rotations and reflections — parts (a), (b)
• AHL 1.14: Matrix multiplication and its properties — part (b) 
▶️ Answer/Explanation

(a)
Matrix \( A \) corresponds to the standard rotation matrix form. It represents an anticlockwise rotation about the origin by an angle of \( \frac{\pi}{3} \) radians, which is equivalent to \(60^\circ\).
\( \boxed{\text{Anticlockwise rotation of } 60^\circ \text{ about the origin}} \).

(b)
The single matrix representing the combined transformation is found by calculating the product \( C = BA \).
First, evaluate matrix \( A \): \( A = \begin{pmatrix} 0.5 & -0.866025\ldots \\ 0.866025\ldots & 0.5 \end{pmatrix} \).
Next, perform matrix multiplication: \( C = \begin{pmatrix} -0.8 & 0.6 \\ 0.6 & 0.8 \end{pmatrix} \begin{pmatrix} 0.5 & -0.866025\ldots \\ 0.866025\ldots & 0.5 \end{pmatrix} = \begin{pmatrix} 0.119615\ldots & 0.992820\ldots \\ 0.992820\ldots & -0.119615\ldots \end{pmatrix} \).
A reflection matrix in the line \( y = x\tan\theta \) has the general form \( \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix} \).
By comparison: \( \cos 2\theta \approx 0.119615 \) and \( \sin 2\theta \approx 0.992820 \).
Calculating the angle: \( \tan 2\theta \approx \frac{0.992820}{0.119615} \approx 8.300 \), which gives \( 2\theta \approx 1.443 \) radians.
Thus, \( \theta \approx 0.7215 \) radians.
The gradient of the reflection line is \( a = \tan\theta \).
\( a = \tan(0.7215) \approx 0.88675 \).
\( \boxed{0.887} \) (to 3 s.f.)

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