# Problems of Geometry Transformation with Solutions

Geometry transformation is one of thing you’ll learn at the phase of middle and high school. You will learn about the change of position and size of objects by using mathematical principles. There are $5$ types of transformations to study: translation, reflection, rotation, dilatation, and transformation by matrix.
Here we provide you a set of solved problems about geometry transformation to make you learn more and test your understanding. We hope it helps you.

### Quote by Paulo Coelho

If you lose something/someone but you found yourself, you won.

Problem Number 1
Given that point $P'(3,-13)$ is the image of point $P$ under translation $T = \begin{pmatrix}-10 \\ 7 \end{pmatrix}$. The coordinates of point $P$ are $\cdots \cdot$
A. $(13,-20)$               D. $(-5,-4)$
B. $(13,-4)$                 E. $(-5,-20)$
C. $(4,20)$

Solution

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 $\boxed{(13,-20)}$

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Problem Number 2
The coordinates of the image of point $P(a,b)$ under rotation about the origin $(0,0)$ by $-90^{\circ}$ are $P'(-10,-2)$. The value of $a+2b = \cdots \cdot$
A. $-18$               C. $8$                  E. $22$
B. $-8$                 D. $18$

Solution

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 $\boxed{a+2b=2+2(-10)=-18}$

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Problem Number 3
The coordinates of the image of point $A$ with $A(-1,4)$ after being reflected across the line $y=-x$ are $\cdots \cdot$
A. $A'(4,1)$                   D. $A'(4,3)$
B. $A'(-4,1)$                E. $A'(-4,-1)$
C. $A'(4,-1)$

Solution

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 $A'(-4, 1)$.

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Problem Number 4
The coordinates of the image of point $P(5,4)$ after being dilated around $(-2, -3)$ by a scale factor of $-4$ are $\cdots \cdot$
A. $(-30,-31)$             D. $(-14,-1)$
B. $(-30,7)$                  E. $(-14,-7)$
C. $(-26,-1)$

Solution

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 $(-30, -31)$.

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Problem Number 5
The point $B(3,-2)$ is rotated by $90^{\circ}$ about the center point $P(-1,1)$. The coordinates of the image of point $B$ are $\cdots \cdot$
A. $B’(-4,3)$                   D. $B’(1,4)$
B. $B’(-2,1)$                   E. $B’(2,5)$
C. $B’(-1,2)$

Solution

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 $\boxed{B'(2,5)}$

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Problem Number 6
The coordinates of the image of point $P(2, -3)$ under rotation that is given by $R[O,90^{\circ}]$ are $\cdots \cdot$
A. $P’(3,2)$                D. $P’(-3,2)$
B. $P’(2,3)$                E. $P’(-3,-2)$
C. $P’(-2,3)$

Solution

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 $\boxed{P'(3,2)}$

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Problem Number 7
Given a point $P(-8,12)$. The dilation $[P,1]$ maps point $(-4,8)$ to point $\cdots \cdot$
A. $(-4,8)$                 D. $(4,-16)$
B. $(-4,16)$               E. $(4,-8)$
C. $(-4,-8)$

Solution

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 $\boxed{(-4, 8)}$

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Problem Number 8
The image of point $B(4,8)$ under reflection over $X$-axis, followed by dilation $\left[O, \dfrac{1}{2}\right]$ lies on the coordinates of $\cdots \cdot$
A. $(-2, 4)$                 D. $(-8, 4)$
B. $(2,-4)$                 E. $(-8,-4)$
C. $(8,-2)$

Solution

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 $(2,-4)$.

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Problem Number 9
Given the coordinates of point $T(-1,5)$. The image point $T$ after being transformed by matrix $\begin{pmatrix}-4 & 3 \\ 2 &-1 \end{pmatrix}$, followed by the reflection across the line $x = 8$, is $\cdots \cdot$
A. $T'(30,-7)$                  D. $T'(3,-7)$
B. $T'(19, 23)$                   E. $T'(-3,-7)$
C. $T'(19,-22)$

Solution

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 $\boxed{(-3,-7)}$

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Problem Number 10
Triangle $KLM$ with vertices $K(6,4), L(-3, 1), M(2,-2)$ is to be transformed under dilation with center $(-2, 3)$ by a scale factor of $4$. What are the images of the triangle’s vertices?
A. $K'(30, 7), L'(-6,-5), M'(14,-17)$
B. $K'(30, 7), L'(-6,-5), M'(10,-12)$
C. $K'(30, 7), L'(-3,-7), M'(14,-17)$
D. $K'(7, 24), L'(-5,-6), M'(14, 8)$
E. $K'(7, 24), L'(-6,-5), M'(7, 30)$

Solution

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 $\boxed{K(30, 7), L(-6,-5), M(14,-17)}$

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Problem Number 11
A dilation in the origin and scale factor of $4$ is applied to $\triangle ABC$ with vertices $A(-2,3), B(2,3)$, and $C(0,-4)$. The area of the dilated triangle is $\cdots \cdot$
A. $120$                  D. $280$
B. $224$                  E. $480$
C. $240$

Solution

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
$A = \dfrac{a \times t}{2} = \dfrac{16 \times 28}{2} = 224$

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Problem Number 12
A vector $\overline{a} = (-3,4)$ is the image after being reflected across the line $y=x$, and then being rotated by $90^{\circ}$ clockwise about the origin. The initial vector before the transformations occur is $\cdots \cdot$
A. $(3,4)$                        D. $(4,-3)$
B. $(-3,-4)$                 E. $(-3,4)$
C. $(-4,3)$

Solution

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 $\boxed{\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix}-3 \\-4 \end{pmatrix}}$

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Problem Number 13
If given a line equation $y = 2x+3$, then line’s image under the translation $T = (3, 2)$ is represented as $\cdots \cdot$
A. $y = 3x$                    D. $y = 2x-4$
B. $y = 2x + 6$             E. $y = 2x-1$
C. $y = 2x-6$

Solution

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 $\boxed{y=2x-1}$

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Problem Number 14
The equation of the image of line $2x+y-1=0$ after being transformed by matrix $\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ and then continued by the reflection over $X$-axis, is $\cdots \cdot$
A. $3x+y-1=0$
B. $5x-y+1=0$
C. $3x+y+1=0$
D. $5x+y-1=0$
E. $5x+y+1=0$

Solution

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 $\boxed{3x+y-1=0}$

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Problem Number 15
The image of the line $3x-y+2=0$ after being reflected across the line $y=x$ and then rotated by $90^{\circ}$ about the origin is $\cdots \cdot$
A. $3x+y+2=0$
B. $3x+y-2=0$
C. $-3x+y+2=0$
D. $-x+3y+2=0$
E. $x-3y+2=0$

Solution

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 $\boxed{3x+y-2=0}$

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Problem Number 16
The line $3x+2y=6$ is translated by $T(3,-4)$, then followed by dilation in the origin by a scale factor of $2$. The image of the line is expressed as $\cdots \cdot$
A. $3x+2y=14$
B. $3x+2y=7$
C. $3x+y=14$
D. $3x+y=7$
E. $x+3y=14$

Solution

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
$\begin{pmatrix} x + 3 \\ y-4 \end{pmatrix} \xrightarrow{D[O, 2]} \begin{pmatrix} 2x + 6 \\ 2y- 8 \end{pmatrix}$
$\begin{cases} x^{\prime \prime} = 2x + 6 \Leftrightarrow x = \dfrac{x^{\prime \prime}-6}{2} \\ y^{\prime \prime} = 2y-8 \Leftrightarrow y = \dfrac{y^{\prime \prime}+8}{2} \end{cases}$
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 $\boxed{3x+2y=14}$

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Problem Number 17
The line $y=2x-3$ is translated by $T = \begin{pmatrix}-2 \\ 3 \end{pmatrix}$. The line of the image is expressed as $\cdots \cdot$
A. $y=2x+4$
B. $y=2x-4$
C. $y=2x-3$
D. $y=-2x+4$
E. $y=-2x-3$

Solution

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 $\boxed{y=2x+4}$

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Problem Number 18
The image of the curve $y=x^2+3x+3$ after being reflected over $X$-axis, followed by the dilation in the origin by a scale factor of $3$, is $\cdots \cdot$
A. $x^2+9x-3y+27=0$
B. $x^2+9x+3y+27=0$
C. $3x^2+9x-y+27=0$
D. $3x^2+9x+y+27=0$
E. $3x^2+9x+27=0$

Solution

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 $\boxed{x^2 + 9x + 3y + 27 = 0}$

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Problem Number 19
The curve $y = x^2 + 3$ is dilated with the center $P (-1,2)$ by a scale factor of $3$, then rotated by $-\dfrac12\pi$ about the origin. The equation of the curve’s image is given by $\cdots \cdot$
A. $3y=x^2+4x+19$
B. $3x=y^2+4y+19$
C. $y=x^2+4x+19$
D. $x=y^2 + 4y + 19$
E. $x=y^2+19$

Solution

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 $\boxed{3x = y^2+4y+19}$

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Problem Number 20
Transformation $T$ is a reflection across the line $y=\dfrac13x$, followed by a reflection across the line $y=-3x$. The matrix representation for $T$ is $\cdots \cdot$
A. $\begin{pmatrix}-1 & 0 \\ 0 & 1 \end{pmatrix}$             D. $\begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}$
B. $\begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix}$          E. $\begin{pmatrix} 0 &-1 \\-1 & 0 \end{pmatrix}$
C. $\begin{pmatrix} 1 & 0 \\ 0 &-1 \end{pmatrix}$

Solution

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}

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Problem Number 21
A photocopier can make copies of images/writing of different sizes. A rectangular image is copied with certain settings. If this setting can be compared to the transformation process of the matrix $\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$, then dilated in the origin by scale factor of $3$, then the area of ​​the rectangular image will be $\cdots$ times the original area.

A. $12$                   C. $24$                 E. $36$
B. $18$                   D. $30$

Solution

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 $\boxed{18 \times~\text{Original Area}}$

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Problem Number 22
A camera processes the image by transforming it into the matrix $\begin {pmatrix} \dfrac14 & \dfrac58 \\ \dfrac12 & 2 \end{pmatrix}$. Next, the image is transformed again to the matrix $\begin {pmatrix} 4 & 1 \\ 8 & 1 \end{pmatrix}$. If the camera takes an image of an object with an area of ​​$32~\text{cm}^2$, then the area of ​​the captured object is $\cdots \cdot$
A. $24~\text{cm}^2$                    D. $36~\text{cm}^2$
B. $28~\text{cm}^2$                    E. $40~\text{cm}^2$
C. $34~\text{cm}^2$

Solution

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 $\boxed{24~\text{cm}^2}$

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Problem Number 23
If the rectangle $ABCD$ is dilated to $A’B’C’D’$ as shown, then the appropriate scale factor is $\cdots \cdot$ A. $2$              B. $3$              C. $4$               D. $9$

Solution

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 $\boxed{3}$

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Problem Number 24
Look at the following graph. One of the translation that moves the line $g$ to the line $l$ is given by $\cdots \cdot$
A. $\begin{bmatrix} 0 \\ 5 \end{bmatrix}$                      D. $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$
B. $\begin{bmatrix} 0 \\ -5 \end{bmatrix}$                   E. $\begin{bmatrix} 3 \\ -4 \end{bmatrix}$
C. $\begin{bmatrix} -5 \\ 0 \end{bmatrix}$

Solution

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 $\begin{bmatrix} 3 \\ -4 \end{bmatrix}$.

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Problem Number 25
Look at the following alphabet line. The image of letter E after being dilated with the center of letter I by a scale factor of $-\dfrac12$ is $\cdots \cdot$
A. letter A                        C. letter G
B. letter C                        D. letter K

Solution

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.