Tag: Geometry Transformation

Translation rule is explained as follows. Suppose a point $(x, y)$ is to be translated by $T = \begin{pmatrix} a \\ b \end{pmatrix}$, so that its image’s coordinates are given by
$\begin{pmatrix} x’ \\ y’ \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix}$
Given: $P'(3,-13)$ is to be translated by $\begin{pmatrix}-10 \\ 7 \end{pmatrix}$, so we have
$\begin{aligned} \begin{pmatrix} 3 \\-13 \end{pmatrix} & = \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix}-10 \\ 7 \end{pmatrix} \\ \begin{pmatrix} x \\ y \end{pmatrix} & = \begin{pmatrix} 3 \\-13 \end{pmatrix}- \begin{pmatrix}-10 \\ 7 \end{pmatrix} \\ \begin{pmatrix} x \\ y \end{pmatrix} & = \begin{pmatrix} 13 \\-20 \end{pmatrix} \end{aligned}$
Thus, the coordinates of point $P$ are $\boxed{(13,-20)}$
(Answer A)

The rotation rule is given as follows.
The coordinates of the image of point $(x, y)$ after being rotated about the origin by $\theta$ counter clockwise are given by
$\begin{pmatrix} x’ \\ y’ \end{pmatrix} = \begin{pmatrix} \cos \theta &-\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$
For $(x’, y’) = (-10,-2)$ and $\theta =-90^{\circ}$, we have
$$\begin{aligned} \begin{pmatrix}-10 \\-2 \end{pmatrix} & = \begin{pmatrix} \cos (-90^{\circ}) &-\sin (-90^{\circ}) \\ \sin (-90^{\circ}) & \cos (-90^{\circ}) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ \begin{pmatrix}-10 \\-2 \end{pmatrix} & = \begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ \begin{pmatrix}-10 \\-2 \end{pmatrix} & = \begin{pmatrix} y \\-x \end{pmatrix} \end{aligned}$$It is obtained that $y =-10$ and $x = 2$. Thus, the coordinates of point $P$ are $(2,-10$). As of that, $a=2$ dan $b=-10$, hence $\boxed{a+2b=2+2(-10)=-18}$
(Answer A)

If a point $A(x, y)$ is reflected across the line $y=-x$, then the image is given by $A’ = (-y,-x)$.
By following this fact, the coordinates of the image of point $A(-1,4)$ are $A'(-4, 1)$.
(Answer B)

Given $P(x, y) = P(5,4)$. The center of dilation is $(a, b) = (-2,-3)$ and $k =-4$.
Suppose the image point lies on $(x’, y’)$. By following the dilation rule, this yields
$\begin{aligned} x’ & = k(x-a) + a \\ & =-4(5-(-2)) + (-2) \\ & =-4(7)-2 =-30 \end{aligned}$
$\begin{aligned} y’ & = k(y-b) + b \\ & =-4(4-(-3))- 3 \\ & =-4(7)-3=-31 \end{aligned}$
Thus, the coordinates of the image of point $P(5,4)$ under said dilation are $(-30, -31)$.
(Answer A)

Rotation rule is given as follows.
The coordinates of the image of point $(x, y)$ after being rotated counter clockwise by an angle of $\theta$ about the center point $(a, b)$ are given by
$$\begin{pmatrix} x’ \\ y’ \end{pmatrix} = \begin{pmatrix} \cos \theta &-\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x-a \\ y-b \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix}$$For $(x, y) = (3,-2)$ and rotation of $\theta = 90^{\circ}$ about the center point $(-1, 1)$, we have
$$\begin{aligned} \begin{pmatrix} x’ \\ y’ \end{pmatrix} & = \begin{pmatrix} \cos 90^{\circ} &-\sin 90^{\circ} \\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix} \begin{pmatrix} 3-(-1) \\-2-1 \end{pmatrix} + \begin{pmatrix}-1 \\ 1 \end{pmatrix} \\ & = \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 4 \\-3 \end{pmatrix} + \begin{pmatrix}-1 \\ 1 \end{pmatrix} \\ & = \begin{pmatrix} 3 \\ 4 \end{pmatrix} + \begin{pmatrix}-1 \\ 1 \end{pmatrix} \\ & = \begin{pmatrix} 2 \\ 5 \end{pmatrix} \end{aligned}$$Thus, the coordinates of the image of point $B$ are $\boxed{B'(2,5)}$
(Answer E)

Rotation rule is given as follows.
The coordinates of the image of point $(x, y)$ after being rotated counter clockwise by an angle of $\theta$ about the origin are given by
$\begin{pmatrix} x’ \\ y’ \end{pmatrix} = \begin{pmatrix} \cos \theta &-\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $
For $(x, y) = (2,-3)$ and rotation of $\theta = 90^{\circ}$ about the origin, we have
$\begin{aligned} \begin{pmatrix} x’ \\ y’ \end{pmatrix} & = \begin{pmatrix} \cos 90^{\circ} &-\sin 90^{\circ} \\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix} \begin{pmatrix} 2 \\-3 \end{pmatrix} \\ & = \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\-3 \end{pmatrix} \\ & = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \end{aligned}$
Thus, the coordinates of the image of point $P$ are $\boxed{P'(3,2)}$
(Answer A)

The dilation rule is given as follows. If a point $(x,y)$ is dilated around center point $(a,b)$ by a scale factor of $k$, then the image point lies in $(k(x-a)+a, k(y-b)+b)$.
This means that the coordinates of the image of the point $(-4, 8)$ after being dilated around the center point $(-8, 12)$ by a scale factor of $1$ are
$(1(-4-(-8))+(-8), 1(8-12)+12)$ $= (-4, 8)$
Dilation $[P, 1]$ maps point $(-4,8)$ to $\boxed{(-4, 8)}$
(Answer A)

Dilation and reflection rule is given as follows.
If a point $(x, y)$ is dilated in the origin by a scale factor of $k$, then its image point lies on the coordinates of $(kx, ky)$.
If a point $(x, y)$ is reflected over $X$-axis, then its image point lies on the coordinates of $(x, -y)$.
Alternatively, we can make an arrow scheme to represent the reflection process as follows.
$B(4, 8) \xrightarrow{R_X} B'(4,-8)$
After it, make the similar scheme for the dilation process.
$\begin{aligned} B'(4,-8) \xrightarrow{D\left[O, \dfrac{1}{2}\right]} & P'(\dfrac{1}{2} \times 4, \dfrac{1}{2} \times-8) \\ & = P^{\prime \prime}(2,-4) \end{aligned}$
Thus, the image point lies on the coordinates of $(2,-4)$.
(Answer B)

The image point $T$ after being transformed by the given matrix can be expressed as follows.
$$\begin{aligned} T\begin{pmatrix}-1 \\ 5 \end{pmatrix} \xrightarrow{\begin{pmatrix}-4 & 3 \\ 2 &-1 \end{pmatrix}} & T’\left[\begin{pmatrix}-4 & 3 \\ 2 &-1 \end{pmatrix}\begin{pmatrix}-1 \\ 5 \end{pmatrix} \right] \\ & = T’ \begin{pmatrix}-4(-1) + 3(5) \\ 2(-1) + (-1)(5) \end{pmatrix} \\ & = T’\begin{pmatrix} 19 \\-7 \end{pmatrix} \end{aligned}$$The transformation process is then followed by reflection across the line $\color{red} {x=8}$ such that we have
$\begin{aligned} T’\begin{pmatrix} 19 \\-7 \end{pmatrix} \xrightarrow{R_{x =-8}} & T^{\prime \prime}\begin{pmatrix} 2(\color{red}{8})- 19 \\-7 \end{pmatrix} \\ & = T^{\prime \prime}\begin{pmatrix}-3 \\-7 \end{pmatrix} \end{aligned}$
Thus, the image point $T$ under the transformations is given by $\boxed{(-3,-7)}$
(Answer E)

The dilation rule is given as follows. If point $(x,y)$ is dilated with center $(a,b)$ and scale factor of $k$, then the image point will be given by $(k(x-a)+a, k(y-b)+b)$.
The image point $K(6, 4)$ after being dilated with the center $(-2, 3)$ by a scale factor of $4$ is
$K'(4(6+2)-2, 4(4-3)+3)$ $= K'(30, 7)$
The image point $L(-3, 1)$ after being dilated with the center $(-2, 3)$ by a scale factor of $4$ is
$L'(4(-3+2)-2, 4(1-3)+3)$ $= L'(-6,-5)$
The image point $M(2, -2)$ after being dilated with the center $(-2, 3)$ by a scale factor of $4$ is
$M'(4(2+2)-2, 4(-2-3)+3)$ $= M'(14,-17)$
Thus, the images of the triangle’s vertices are $\boxed{K(30, 7), L(-6,-5), M(14,-17)}$
(Answer A)

The dilation rule is given as follow. If point $(x,y)$ is dilated in the origin with a scale factor of $k$, then the image point will be given by $(kx, ky)$.
The image point $A(-2, 3)$ after being dilated in the origin by a scale factor of $4$ is
$A'(4(-2), 4(3)) = (-8, 12)$
The image point $B(2, 3)$ after being dilated in the origin by a scale factor of $4$ is
$B'(4(2), 4(3)) = (8, 12)$
The image point $C(0, -4)$ after being dilated in the origin by a scale factor of $4$ is
$C'(4(0), 4(-4)) = (0,-16)$
Sketch the image points in the Cartesian coordinate system and connect them by lines such that they form a triangle (pertaining the following graph).

The triangle has an area of
$A = \dfrac{a \times t}{2} = \dfrac{16 \times 28}{2} = 224$
(Answer B)

Suppose the initial vector is $(x, y)$. The reflection across the line $y = x$ can be expressed in the following diagram.
$\begin{pmatrix} a \\ b \end{pmatrix} \xrightarrow{M_{y=x}} \begin{pmatrix} b \\ a \end{pmatrix}$
The transformation is followed by rotation of $90^{\circ}$ clockwise, which is same as $270^{\circ}$ anti clockwise. Hence, we can make a following transformation scheme.
$$\begin{aligned} \begin{pmatrix} b \\ a \end{pmatrix} \xrightarrow{R[O, 270^{\circ}]} & \begin{pmatrix} \cos 270^{\circ} &-\sin 270^{\circ} \\ \sin 270^{\circ} & \cos 270^{\circ} \end{pmatrix}\begin{pmatrix} b \\ a \end{pmatrix} \\ & = \begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix} \begin{pmatrix} b \\ a \end{pmatrix} \\ & = \begin{pmatrix} a \\-b \end{pmatrix} \end{aligned}$$We have found that the vector’s image is in the form of $(a, -b)$. Since it is given that vector $\overline{a} = (-3,4)$ be the vector’s image, so we obtain $a=-3$ and $b=-4$.
Thus, the initial vector is $\boxed{\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix}-3 \\-4 \end{pmatrix}}$
(Answer B)

Suppose the line $y = 2x+3$ passes through the point $(x, y)$. After being translated by $T(3, 2)$, we found its image as shown in the diagram below.
$(x, y) \xrightarrow{T(3, 2)} (x+3, y+2)$
Hence, we have $x’ = x + 3$ and $y’ = y + 2$, or in the same manner,
$\begin{cases} x = x’-3 \\ y = y’-2 \end{cases}$
Substitute $x$ and $y$ in terms of $x’$ and $y’$ to the initial line equation.
$\begin{aligned} y & = 2x + 3 \\ y’-2 & = 2(x’-3) + 3 \\ y’ & = 2x’-6 + 3 + 2 \\ y’ & = 2x’-1 \end{aligned}$
Thus, the image of the line $y = 2x+3$ after being translated by $T(3,2)$ is given by $\boxed{y=2x-1}$
(Answer E)

The image of the point $(x, y)$ by the matrix transformation can be expressed as the scheme below shown.
$\begin{aligned} \begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}} \begin{pmatrix} x’ \\ y’ \end{pmatrix} & = \left[\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} \right] \\ & = \begin{pmatrix} x + y \\ x + 2y \end{pmatrix} \end{aligned}$
The transformation to the point is followed by the reflection over $X$-axis, so we have
$\begin{aligned} \begin{pmatrix} x + y \\ x + 2y \end{pmatrix} \xrightarrow{R_{x}} \begin{pmatrix} x + y \\-x-2y \end{pmatrix} \end{aligned}$ 
We got $x^{\prime \prime} = x + y$ dan $y^{\prime \prime} =-x-2y$. 
By recalling the concept of system of linear equations in two variables, we have
$\begin{cases}-y = x^{\prime \prime} + y^{\prime \prime} \\ x = 2x^{\prime \prime} + y^{\prime \prime} \end{cases}$
Substitute them into the equation $2x+y-1=0$, and we will have
$\begin{aligned} 2(2x^{\prime \prime} + y^{\prime \prime})-(x^{\prime \prime} + y^{\prime \prime})-1 & = 0 \\ 3x^{\prime \prime} + y^{\prime \prime}- 1 & = 0 \end{aligned}$
By removing the sign of double accents, we found the line’s image equation, that is $\boxed{3x+y-1=0}$
(Answer A)

The image of point $(x, y)$ after being reflected across the line $y=x$ can be expressed as follows.
$\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{R_{y=x}} \begin{pmatrix} y \\ x \end{pmatrix} $ 
The transformation to the point is followed by rotation by $90^{\circ}$ about the origin.
$\begin{aligned} \begin{pmatrix} y \\ x \end{pmatrix} \xrightarrow{R[O, 90^{\circ}]} & \begin{pmatrix} \cos 90^{\circ} &-\sin 90^{\circ} \\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix} \begin{pmatrix} y \\ x \end{pmatrix} \\ & = \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} y \\ x \end{pmatrix} \\ & = \begin{pmatrix}-x \\ y \end{pmatrix} \end{aligned}$
We have $x^{\prime \prime} =-x$ and $y^{\prime \prime}= y$. 
Substitute them into the equation $3x-y+2=0$ to get
$3(-x^{\prime \prime})-y^{\prime \prime} + 2 = 0 \Leftrightarrow 3x^{\prime \prime} + y^{\prime \prime}-2 = 0$
By removing the sign of double accents, we have the image equation, that is $\boxed{3x+y-2=0}$
(Answer B)

Suppose a point $(x, y)$ is translated by $T(3,-4)$, so we have
$\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{T(3,-4)} \begin{pmatrix} x + 3 \\ y- 4 \end{pmatrix}$
The transformation to this point is followed by dilation in the origin by a scale factor of $2$, so we have
$\begin{pmatrix} x + 3 \\ y-4 \end{pmatrix} \xrightarrow{D[O, 2]} \begin{pmatrix} 2x + 6 \\ 2y- 8 \end{pmatrix}$
This leads to
$\begin{cases} x^{\prime \prime} = 2x + 6 \Leftrightarrow x = \dfrac{x^{\prime \prime}-6}{2} \\ y^{\prime \prime} = 2y-8 \Leftrightarrow y = \dfrac{y^{\prime \prime}+8}{2} \end{cases}$
Substitute $x, y$ in terms of $x^{\prime \prime}$ and $y^{\prime \prime}$ into the equation $3x+2y=6$, so we will have
$\begin{aligned} 3\left(\dfrac{x^{\prime \prime}-6}{2}\right) + 2\left(\dfrac{y^{\prime \prime}+8}{2}\right) & = 6 \\ \text{Multiply each side by 2}& \\ 3(x^{\prime \prime}-6) + 2(y^{\prime \prime} + 8) & = 12 \\ 3x^{\prime \prime}-18 + 2y^{\prime \prime} + 16 & = 12 \\ 3x^{\prime \prime} + 2y^{\prime \prime} & = 14 \end{aligned}$
By removing the sign of double accents, the image of the line is $\boxed{3x+2y=14}$
(Answer A)

Since $T = \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix}-2 \\ 3 \end{pmatrix}$, We have
$\begin{aligned} \begin{pmatrix} x’ \\ y’ \end{pmatrix} & = \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} \\ \begin{pmatrix} x’ \\ y’ \end{pmatrix} & = \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix}-2 \\ 3 \end{pmatrix} \\ \begin{pmatrix} x \\ y \end{pmatrix} & = \begin{pmatrix} x’ + 2\\ y’- 3\end{pmatrix} \end{aligned}$
Substitute $x = x’+2$ and $y =y’-3$ into the equation $y=2x-3$, so we have
$\begin{aligned} y’-3 & = 2(x’+2)-3 \\ y’-3 & =2x’+1 \\ y’ & = 2x’+4 \end{aligned}$
Thus, the image of the line is expressed as $\boxed{y=2x+4}$
(Answer A)

The reflection process over $X$-axis is shown in the following scheme.
$\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{M_{\text{Sumbu}~X}} \begin{pmatrix} x \\-y \end{pmatrix}$ 
The dilation process in the origin by a scale factor of $3$ is shown in the following scheme. 
$\begin{pmatrix} x \\-y \end{pmatrix} \xrightarrow{D[O, 3]} \begin{pmatrix} 3x \\-3y \end{pmatrix}$
We have $x^{\prime \prime} = 3x$ and $y^{\prime \prime} =-3y$, so it can be written as follows.
$\begin{cases} x = \dfrac13x^{\prime \prime} \\ y =-\dfrac13y^{\prime \prime} \end{cases}$
Substitute them into equation $y=x^2+3x+3$, so we have
$\begin{aligned}&-\dfrac13y^{\prime \prime} = \left(\dfrac13x^{\prime \prime}\right)^2 + 3\left(\dfrac13x^{\prime \prime}\right) + 3 \\ & \text{Multiply each side by 9} \\ &-3y^{\prime \prime} = (x^{\prime \prime})^2 + 9x^{\prime \prime} + 27 \\ & (x^{\prime \prime})^2 + 9x^{\prime \prime} + 3y^{\prime \prime} + 27 = 0 \end{aligned}$
By removing the sign of double accents, the curve’s image is expressed as $\boxed{x^2 + 9x + 3y + 27 = 0}$
(Answer B)

Suppose a point $(x, y)$ is dilated with center $P(-1, 2)$ by a scale factor of $3$, then the scheme can be made as follows.
$T\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{D[P(-1,2), 3]} \begin{pmatrix} x’ \\ y’ \end{pmatrix}$
as
$\begin{aligned} \begin{pmatrix} x’ \\ y’ \end{pmatrix} & = k\begin{pmatrix} x-a \\ y-b \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} \\ & = 3 \begin{pmatrix} x-(-1)\\ y- 2 \end{pmatrix} + \begin{pmatrix}-1 \\ 2 \end{pmatrix} \\ & = \begin{pmatrix} 3x + 2\\ 3y- 4\end{pmatrix} \end{aligned}$
The transformation of point $(x’, y’)$ is followed by rotation about the origin by $- \dfrac12\pi$ radian or $-90^{\circ}$, so the scheme can be made as follows.
$\begin{pmatrix} x’ \\ y’ \end{pmatrix} \xrightarrow{R[O,-90^{\circ}]} \begin{pmatrix} x^{\prime \prime} \\ y^{\prime \prime} \end{pmatrix}$
as
$$\begin{aligned} \begin{pmatrix} x^{\prime \prime} \\ y^{\prime \prime} \end{pmatrix} & = \begin{pmatrix} \cos (-90^{\circ}) &-\sin (-90^{\circ}) \\ \sin (-90^{\circ}) & \cos (-90^{\circ}) \end{pmatrix} \begin{pmatrix} x’ \\ y’ \end{pmatrix} \\ & = \begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}\begin{pmatrix} 3x+2 \\ 3y-4 \end{pmatrix} \\ & = \begin{pmatrix} 3y-4 \\-3x-2\end{pmatrix} \end{aligned}$$Hence, we have
$\begin{cases} x^{\prime \prime} = 3y-4 \Leftrightarrow y = \dfrac{x^{\prime \prime} + 4}{3} \\ y^{\prime \prime} =-3x-2 \Leftrightarrow x =\dfrac{y^{\prime \prime} + 2}{-3} \end{cases}$
Substitute $x, y$ in terms of $x^{\prime \prime}$ and $y^{\prime \prime}$ into the curve equation $y = x^2+3$, so we have
$\begin{aligned} \dfrac{x^{\prime \prime}+4}{3} & = \left(\dfrac{y^{\prime \prime}+2}{-3}\right)^2+3 \\ \text{Multiply each}&~\text{side by 9} \\ 3(x^{\prime \prime}+4) & = (y^{\prime \prime}+2)^2 + 27 \\ 3x^{\prime \prime} + 12 & = (y^{\prime \prime})^2 + 4y^{\prime \prime} + 31 \\ 3x^{\prime \prime} & = (y^{\prime \prime})^2 + 4y^{\prime \prime} + 19 \end{aligned}$
By removing the sign of double accents, we have the equation of the image’s curve, that is $\boxed{3x = y^2+4y+19}$
(Answer B)

The reflection across the line $y = mx$ can be represented by the matrix in form of $\begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}$.
For the reflection across the line $y=\dfrac13x$, the representation matrix is given by
$T_1 = \begin{pmatrix} \dfrac13 & 0 \\ 0 & \dfrac13 \end{pmatrix}$
For the reflection across the line $y=-3x$, the representation matrix is given by
$T_1 = \begin{pmatrix}-3 & 0 \\ 0 &-3 \end{pmatrix}$
Hence, the representation matrix for $T$ can be expressed as follows.
$\begin{aligned} T & = T_2 \cdot T_1 \\ & = \begin{pmatrix} \dfrac13 & 0 \\ 0 & \dfrac13 \end{pmatrix} \cdot \begin{pmatrix}-3 & 0 \\ 0 &-3 \end{pmatrix} \\ & = \begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix} \end{aligned}$
(Answer B)

Note that the matrix representation for dilation centered at the origin by a scale factor of $3$ is given by $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$.
Given:
$T_1 =\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}~~~~T_2 =\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ 
The transformation by the two matrices is expressed by
$\begin{aligned} T_2 \cdot T_1 & = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix} \\ & = \begin{pmatrix} 6 & 3 \\ 12 & 9 \end{pmatrix} \end{aligned}$
The area of the transformed image is as follows. 
$\begin{aligned} L & = \left|\begin{pmatrix} 6 & 3 \\ 12 & 9 \end{pmatrix}\right| \times~\text{Original Area} \\ & = \left|54-36\right| \times~\text{Original Area} \\ & = 18 \times~\text{Original Area} \end{aligned}$
Thus, the area of the rectangular image will be $\boxed{18 \times~\text{Original Area}}$ 
(Answer B)

Given:
$T_1 = \begin{pmatrix} \dfrac14 & \dfrac58 \\ \frac12 & 2 \end{pmatrix}~~~~T_2 = \begin{pmatrix} 4 & 1 \\ 8 & 1 \end{pmatrix}$
The transformation by the two matrices is expressed by
$\begin{aligned} T_2 \cdot T_1 & = \begin{pmatrix} 4 & 1 \\ 8 & 1 \end{pmatrix}\begin{pmatrix} \dfrac14 & \dfrac58 \\ \frac12 & 2 \end{pmatrix} \\ & = \begin{pmatrix} \dfrac32 & \dfrac92 \\ \dfrac52 & 7 \end{pmatrix} \end{aligned}$
The area of the captured object is as follows.
$\begin{aligned} L & = \left|\begin{vmatrix} \dfrac32 & \dfrac92 \\ \dfrac52 & 7 \end{vmatrix}\right| \times~\text{The original area} \\ & = \left|\dfrac{21}{2}- \dfrac{45}{4}\right| \times 32~\text{cm}^2 \\ & = \left|-\dfrac34\right| \times 32~\text{cm}^2 = 24~\text{cm}^2 \end{aligned}$
Thus, the area of the captured object is $\boxed{24~\text{cm}^2}$ 
(Answer A)

As shown in the figure that the process of dilation takes center at the bottom leftmost point. Assume this as point $(0, 0)$, so it can be assumed that $A (0, 1)$, $B(3, 1)$, $C(3, 3)$, and $D(1, 3)$. The coordinates of the dilated points are $A'(3, 0)$, $B'(9, 3)$, $C'(9, 9)$, and $ D'(3, 9)$.
From this, we know that there is a number that multiplies for each coordinate value. For example, just take the point $B(3,1) $ to $B'(9, 3)$. The multiplier is $3$, which means the scale factor for the dilation is $\boxed{3}$
(Answer B)

Geometrically, we can perform a translation from a point to the another one that is passed by each of the lines.
Start from point $(-2, 0)$. It is moved to the right by $5$ units $(+5)$ towards point $(3, 0)$, so the appropriate translation is
$\begin{bmatrix} 5 \\ 0 \end{bmatrix}$.
In addition, it can also be from the point $(0, 4)$, then moved down as far as $4$ units $(-4)$ and $3$ units to the right of $(+3)$ towards point $(3, 0)$, so the appropriate translation is $\begin{bmatrix} 3 \\ -4 \end{bmatrix}$.
(Answer E)

By pertaining the given alphabet line, the distance between letter E and I is $4$. Since the value of scale factor is $\dfrac12$ (just value, ignore the sign first), the distance between the image of letter E and I is $\dfrac12 \times 4 = 2$. Two letters with the distance of $2$ from I can be either G or K. But, the scale factor is negative, so the image must be inverted (or opposite) from the dilation center point (letter I). This means that the image of letter E must be letter K.
(Answer D)