Category: Mathematics Compulsory and Extended – Hong Kong Diploma of Secondary Education (HKDSE)

Posted on

202402161017 Solution to 2017-DSE-MATH-I-6

The coordinates of the points A and B are (-3,4) and (9,-9) respectively.

\begin{aligned} \mathbf{OA} & = (-3,4) \\ \mathbf{OB} & = (9,-9) \\ \end{aligned}

A is rotated anticlockwise about the origin through 90^\circ to A'.

f(r,\theta )=(r,\theta +\frac{\pi}{2})

B' is the reflection image of B with respect to the x-axis.

g(x,y)=(x,-y)

(a) Write down the coordinates of A' and B'.

Hint. Conversion between Cartesian and polar coordinates:

\begin{aligned} r(x,y) & = \sqrt{x^2+y^2} \\ \theta (x,y) & = \tan^{-1}\bigg(\frac{y}{x}\bigg) \\ x(r,\theta ) & = r\cos\theta \\ y(r,\theta ) & = r\sin\theta \\ \end{aligned}

(b) Prove that AB is perpendicular to A'B'.

Hint. Show that \mathbf{AB}\cdot\mathbf{A'B'}=0.

\begin{aligned} \mathbf{AB} & = \mathbf{AO} + \mathbf{OB} \\ & = \mathbf{OB} - \mathbf{OA} \\ \mathbf{A'B'} & = \mathbf{A'O} + \mathbf{OB'} \\ & = \mathbf{OB'} - \mathbf{OA'} \\ \end{aligned}

\begin{aligned} & \quad \mathbf{AB}\cdot\mathbf{A'B'} \\ & = (\mathbf{OB}-\mathbf{OA})\cdot (\mathbf{OB'}-\mathbf{OA'}) \\ & = \mathbf{OB}\cdot\mathbf{OB'} - \mathbf{OB}\cdot\mathbf{OA'} - \mathbf{OA}\cdot\mathbf{OB'} + \mathbf{OA}\cdot\mathbf{OA'} \\ \end{aligned}

Scalar (/dot) product of two vectors:

\begin{aligned} \mathbf{X}\cdot\mathbf{Y} & =|\mathbf{X}||\mathbf{Y}|\cos\measuredangle{(\mathbf{X},\mathbf{Y})} \\ \textrm{\scriptsize{OR}}\quad\mathbf{X}\cdot\mathbf{Y} & =\sum_{i=1}^{n}x_iy_i \\ \end{aligned}

(modified with hints added)

Roughwork.

We give segment AB the equation L:

\begin{aligned} \frac{y-4}{x-(-3)}& = \frac{4-(-9)}{-3-9} = -\frac{13}{12} \\ L:\quad 0 & = 13x+12y-9 \\ \textrm{where } & (x,y)\in [-3,9]\times [-9,4] \\ \end{aligned}

and segment A'B' the equation L':

\begin{aligned} \frac{y-(-3)}{x-(-4)}& = \frac{9-(-3)}{9-(-4)} = \frac{12}{13} \\ L':\quad 0 & = 12x-13y+9 \\ \textrm{where } & (x,y)\in [-4,9]\times [-3,9] \\ \end{aligned}

The intersection point C(a,b) of lines L and L' can be obtained by solving their simultaneous equations:

\begin{aligned} \quad\enspace & \left\{ \begin{aligned} 13a+12b-9 & = 0 \\ 12a-13b+9 & = 0 \\ \end{aligned}\right\} \\ &\Rightarrow \left\{ \begin{aligned} a & = \frac{9}{313} \\ b & = \frac{225}{313} \\ \end{aligned}\right\} \\ \end{aligned}

Should L\perp L', the line L rotated by \pi /2 about point C(a,b) on the xy-plane would equal line L', the rotation matrix being

\begin{pmatrix} \cos\theta & -\sin\theta & -a\cos\theta + b\sin\theta + a\\ \sin\theta & \cos\theta & -a\sin\theta - b\cos\theta + b \\ 0 & 0 & 1 \\ \end{pmatrix}\Bigg|_{\theta =\frac{\pi}{2}}

to be applied to

\begin{pmatrix} x \\ y \\ 1 \\ \end{pmatrix}\quad\textrm{ for } x,y\in L(x,y)

such that

\begin{aligned} & \quad \begin{pmatrix} 0 & -1 & a+b \\ 1 & 0 & -a+b \\ 0 & 0 & 1 \\\end{pmatrix}\begin{pmatrix} x \\ y \\ 1 \\ \end{pmatrix} \\ & = \begin{pmatrix} a+b-y \\ x-a+b \\ 1 \\\end{pmatrix} \\ & \stackrel{\textrm{def}}{=} \begin{pmatrix} x' \\ y' \\ 1 \\\end{pmatrix}\quad \textrm{ for }x',y'\in L'(x',y') \\ \end{aligned}

the centre of rotation being invariant, i.e.,

(x,y)|_{(a,b)}=(x',y')|_{(a,b)}.

The rest is left the reader as an exercise.