# Long Question

A triangle ABC with vertices A(2, 3), B(2, 6) and C(3, 4) is translated by T = $\begin{pmatrix} -2 \\ \text{3} \end{pmatrix}$ and the image so obtained is enlarged by E[(0, 0), 2]. Write the co-ordinates of the vertices of the images so formed and represent △ABC and its both images in the same graph.

A △ABC have vertices A(2, 3), B(2, 6) and C(3, 4). Let △A'B'C' is formed when they are translated by vector $\begin{pmatrix} -2 \\ \text{3} \end{pmatrix}$, so,
(x, y) $\xrightarrow{\text{T.V.}\begin{pmatrix} -2 \\ \text{3} \end{pmatrix}}$ (x - 2, y + 3)
A(2, 3) $\rightarrow$ A'(0, 6)
B(2, 6) $\rightarrow$ B'(0, 9)
C(3, 4) $\rightarrow$ C'(1, 7)

Again, △A'B'C' is enlarged by E[(0, 0), 2] to form A"B"C",
(x, y) $\xrightarrow{\text{E[(0, 0), 2]}}$ (2x, 2y)
A'(0, 6) $\rightarrow$ A"(0, 12)
B'(0, 9) $\rightarrow$ B"(0, 18)
C'(1, 7) $\rightarrow$ C"(2, 14)

Finally, plotting △ABC and its images on graph,

E denotes the enlargement about the centre (3, 1) with a scale factor of 2 and R denotes a reflection on the line y = x. Find the image of △ABC having the vertices A(2, 3), B(4, 5) and C(1, -2) under the combined transformation E∘R. Draw both figure on the same graph paper.

△ABC have vertices A(2, 3), B(4, 5) and C(1, -2). Let △A'B'C' is formed when they are reflection on the line y = x, so,
(x, y) $\xrightarrow{\text{reflection on y = x}}$ (y, x)
A(2, 3) $\rightarrow$ A'(3, 2)
B(4, 5) $\rightarrow$ B'(5, 4)
C(1, -2) $\rightarrow$ C'(-2, 1)

Again, △A'B'C' is enlarged by E[(3, 1), 2] to form A"B"C",
(x, y) $\xrightarrow{\text{E[(a, b), k]}}$ {k(x - a) + a, k(y - b) + b}
(x, y) $\xrightarrow{\text{E[(3, 1), 2]}}$ {2(x - 3) + 3, 2(y - 1) + 1}
A'(3, 2) $\rightarrow$ A"{2(3 - 3) + 3, 2(2 - 1) + 1} = A"(3, 3)
B'(5, 4) $\rightarrow$ B"{2(5 - 3) + 3, 2(4 - 1) + 1} = (7, 7)
C'(-2, 1) $\rightarrow$ C"{2(-2 - 3) + 3, 2(1 - 1) + 1} = (-7, 1)

Finally, plotting △ABC and △A"B"C" which is formed under the combined transformation E∘R on graph,

Triangle PQR having vertices P(2, 1), Q(5, 3) and R(7, -1) is reflected on the y-axis. The image so formed is enlarged bt E[(0, 0), 2]. Write the co-ordinates of the vertices images thus obtained and present the △PQR and its image in same graph paper.

△PQR have vertices P(2, 1), Q(5, 3), and R(7, -1). Let △P'Q'R' is formed when they are reflected on y-axis . so,
(x, y) $\xrightarrow{\text{y-axis}}$ (-x, y)
P(2, 1) $\rightarrow$ P'(-2, 1)
Q(5, 3) $\rightarrow$ Q'(-5, 3)
R(7, -1) $\rightarrow$ R'(-7, -1)

Again, △P'Q'R' is enlarged by E[(0, 0), 2] to form △P"Q"R",
(x, y) $\xrightarrow{\text{E[(0, 0), 2]}}$ (2x, 2y)
P'(-2, 1) $\rightarrow$ P"(-4, 2)
Q'(-5, 3) $\rightarrow$ Q"(-10, 6)
R'(-7, -1) $\rightarrow$ R"(-14, -2)

Finally, plotting △PQR and its images on graph,

L(2, 2), O(6, 2), V(7,4) and E(3, 4) are the vertices of a parallelogram LOVE. Find the coordinates of the vertices of the images of ▱LOVE under the rotation of positive 90° about origin followed by enlargement E[(0,0), 3]. Represent the object and image on the same graph paper.

▱LOVE have vertices L(2, 2), O(6, 2), V(7,4), and E(3, 4). Let ▱L'O'V'E' is formed when they are rotated at +90° about origin. so,
(x, y) $\xrightarrow{+90°}$ (-y, x)
L(2, 2) $\rightarrow$ L'(-2, 2)
O(6, 2) $\rightarrow$ O'(-2, 6)
V(7, 4) $\rightarrow$ V'(-4, 7)
E(3, 4) $\rightarrow$ E'(-4, 3)

Again, ▱L'O'V'E' is enlarged by E[(0, 0), 3] to form ▱L"O"V"E",
(x, y) $\xrightarrow{\text{E[(0, 0), 3]}}$ (3x, 3y)
L'(-2, 2) $\rightarrow$ L"(-6, 6)
O'(-2, 6) $\rightarrow$ O"(-6, 18)
V'(-4, 7) $\rightarrow$ V"(-12, 21)
E'(-4, 3) $\rightarrow$ E"(-12. 9)

Finally, plotting ▱LOVE and its images on graph,

P(1, 2), Q(2, 1) and R(4, 3) are the vertices of the triangle PQR. Find the coordinates of the images of the triangle PQR under the reflection line y = x followed by the enlargement E[(0, 0), 2]. Present the △PQR and its images on the same graph.

△PQR have vertices P(1, 2), Q(2, 1) and R(4, 3). Let △P'Q'R is formed when they are reflected under the line y = x. so,
(x, y) $\xrightarrow{\text{y = x}}$ (x, y)
P(1, 2) $\rightarrow$ P'(2, 1)
Q(2, 1) $\rightarrow$ Q'(1, 2)
R(4, 3) $\rightarrow$ R'(3, 4)

Again, △P'Q'R' is enlarged by E[(0, 0), 2] to form P"Q"R",
(x, y) $\xrightarrow{\text{E[(0, 0), 2]}}$ (2x, 2y)
P'(2, 1) $\rightarrow$ P"(4, 2)
Q'(1, 2) $\rightarrow$ Q"(2, 4)
R'(3, 4) $\rightarrow$ R"(6, 8)

Finally, plotting △PQR and its images on graph,

The vertices of △ABC are A(2, 0), B(3, 1) and C(1, 1). △ABC is translated by vector $\begin{pmatrix} 2 \\ \text{-3} \end{pmatrix}$. The image so formed is enlarged by E[(0, 0), 2]. Write the coordinates of the image thus obtained and represent the object and the image on the same graph paper.

△ABC have vertices A(2, 0), B(3, 1) and C(1, 1). Let △A'B'C' is formed when they are translated by vector $\begin{pmatrix} 2 \\ \text{-3} \end{pmatrix}$, so,
(x, y) $\xrightarrow{\text{T.V.}\begin{pmatrix} 2 \\ \text{-3} \end{pmatrix}}$ (x + 2, y - 3)
A(2, 0) $\rightarrow$ A'(4, -3)
B(3, 1) $\rightarrow$ B'(5, -2)
C(1, 1) $\rightarrow$ C'(3, -2)

Again, △A'B'C' is enlarged by E[(0, 0), 2] to form A"B"C",
(x, y) $\xrightarrow{\text{E[(0, 0), 2]}}$ (2x, 2y)
A'(4, -3) $\rightarrow$ A"(8, -6)
B'(5, -2) $\rightarrow$ B"(10, -4)
C'(3, -2) $\rightarrow$ C"(6, -4)

Finally, plotting △ABC and its images on graph,

C(2, 5), A(-1, 3) and T(4, 1) are the vertices Triangle CAT. Find the coordinates of the vertices of the image of △CAT under the rotation of positive 90° about origin followed by enlargement E[(0,0), 2]. Represent the object and images on the same graph paper.

△CAT have vertices C(2, 5), A(-1, 3) and T(4, 1). Let △C'A'T' is formed when they are rotated under positive 90° about origin, so,
(x, y) $\xrightarrow{+90°}$ (-y, x)
C(2, 5) $\rightarrow$ C'(-5, 2)
A(-1, 3) $\rightarrow$ A'(-3, -1)
T(4, 1) $\rightarrow$ T'(-1, 4)

Again, △C'A'T' is enlarged by E[(0, 0), 2] to form C"A"T",
(x, y) $\xrightarrow{\text{E[(0, 0), 2]}}$ (2x, 2y)
C'(-5, 2) $\rightarrow$ A"(-10, 4)
A'(-3, -1) $\rightarrow$ B"(-6, -2)
T'(-1, 4) $\rightarrow$ C"(-2, 8)

Finally, plotting △CAT and its images on graph,

State the single transformation equivalent to the combination of reflections on the X-axis and Y-axis respectively. Using this single transformations find the coordinates of the vertices of the image of △PQR having vertices P(4, 3), Q(1, 1) and R(5, -1). Also draw the object and image on the same graph.

Let A be any point. It is reflected on x-axis to form point A'. so,
A(x, y) $\xrightarrow{\text{x - axis}}$ A'(x, -y)

Again A' is reflected on y-axis to form A",
A'(x, -y) $\xrightarrow{\text{y - axis}}$ A"(-x, -y)

From above,
A(x, y) $\xrightarrow{±180°}$ A"(-x, -y)

So, the single transformation equivalent to the combination of reflections on the X-axis and Y-axis respectively is rotation of ±180° through origin.

Now, using this combination to get the image of △PQR as △P'Q'R',

P(4, 3) $\xrightarrow{±180°}$ P'(-4, -3)
Q(1, 1) $\rightarrow$ Q'(-1, -1)
R(5, -1) $\rightarrow$ R'(-5, 1)

Finally, plotting △PQR and its image on graph,

State the single transformation equivalent to the combination of reflections on the X-axis and Y-axis respectively. Using this single transformation find the coordinates of the vertices of the image of △ having vertices A(2, 3), B(3, -4) and C(1, 2). Also draw the object and image on the same graph.

Let P be any point. It is reflected on x-axis to form point P'. so,
P(x, y) $\xrightarrow{\text{x - axis}}$ P'(x, -y)

Again P' is reflected on y-axis to form P",
P'(x, -y) $\xrightarrow{\text{y - axis}}$ P"(-x, -y)

From above,
P(x, y) $\xrightarrow{±180°}$ P"(-x, -y)

So, the single transformation equivalent to the combination of reflections on the X-axis and Y-axis respectively is rotation of ±180° through origin.

Now, using this combination to get the image of △ABC as △A'B'C',
A(2, 3) $\xrightarrow{±180°}$ A'(-2, -3)
B(3, -4) $\rightarrow$ B'(-3, 4)
C(1, 2) $\rightarrow$ C'(-1, -2)

Finally, plotting △ABC and its image on graph,

A △ABC with vertices A(1, 2), B(4 -1) and C(2, 5) is reflected successively in the line y = x and the X-axis. Find the co-ordinates and graphically represent the image under these transformations. State also the single transformation given by the combinations of these transformations.

A(1, 2), B(4 -1) and C(2, 5) are the vertices of △ABC. Let they are reflected in the line y = x to form △A'B'C',
(x, y) $\xrightarrow{\text{y = x}}$ (y, x)
A(1, 2) $\rightarrow$ A'(2, 1)
B(4, -1) $\rightarrow$ B'(-1, 4)
C(2, 5) $\rightarrow$ C'(5, 2)

Again, △A'B'C' is reflected in the x-axis to form △A"B"C",
(x, y) $\xrightarrow{\text{x-axis}}$ (x, -y)
A'(2, 1) $\rightarrow$ A"(2, -1)
B'(-1, 4) $\rightarrow$ B"(-1, -4)
C'(5, 2) $\rightarrow$ C"(5, -2)

Now, plotting △ABC and its images on graph,

Finally, taking A and A" to find single transformation,
A(1, 2) $\xrightarrow{-90°}$ A"(2, -1)

So, single transformation is rotation of -90° through origin.

The vertices of △ABC are A(2, 5), B(-2, 3)and C(4, 1). △ABC is rotated through negative quarter turn about the origin and the image so obtained is enlarged by taking origin as the centre and scale factor 2. Hence, find the co-ordinates of the the image of △ABC and show the object and image on the same graph.

A(2, 5), B(-2, 3)and C(4, 1) are the vertices of △ABC. Let they are rotated through negative quarter turn about the origin to form △A'B'C',
(x, y) $\xrightarrow{\text{-90°}}$ (y, -x)
A(2, 5) $\rightarrow$ A'(5, -2)
B(-2, 3) $\rightarrow$ B'(3, 2)
C(4, 1) $\rightarrow$ C'(1, -4)

Again, △A'B'C' is enlarged by taking origin as the centre and scale factor 2 to form △A"B"C",
(x, y) $\xrightarrow{\text{E[(0, 0), 2]}}$ (2x, 2y)
A'(5, -2) $\rightarrow$ A"(10, -4)
B'(3, 2) $\rightarrow$ B"(6, 4)
C'(1, -4) $\rightarrow$ C"(2, -8)

Finally, plotting △ABC and its images on graph, <img src="/images/comb-of-transformation/30.svg" alt="" width="600" loading="lazy" height="371"

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K(2, 5), L(-1, 3) and M(4, 1) are the vertices of a triangle KLM. Find the co-ordinates of the vertices of the images of △KLM under the rotation of negative 90° about the origin followed by the enlargement E[(0, 0), 2]. Present object and its images on the same graph paper.

K(2, 5), L(-1, 3) and M(4, 1) are the vertices of a triangle KLM. Let they are rotated about negative 90° to form △K'L'M',
(x, y) $\xrightarrow{-90°}$ (y, -x)
K(2, 5) $\rightarrow$ K'(5, -2)
L(-1, 3) $\rightarrow$ L'(3, 1)
M(4, 1) $\rightarrow$ M'(1, -4)

Again, △K'L'M' is enlarged by E[(0, 0), 2] to form △K"L"M",
(x, y) $\xrightarrow{\text{E[(0, 0), 2]}}$ (2x, 2y)
K'(5, -2) $\rightarrow$ K"(10, -4)
L'(3, 1) $\rightarrow$ L"(6, 2)
M'(1, -4) $\rightarrow$ M"(2, -8)

Finally, plotting △KLM and its images on graph,

Translate the △PQR with vertices P(2, 4), Q(-1, 2) and R(5, -1) by translation vector T = $\begin{pmatrix} \text{-2} \\ 3 \end{pmatrix}$ and then reflect the image so obtained in the line x + y = 0. Write the coordinates of the vertices of the image and draw the object and images on the same graph paper.

P(2, 4), Q(-1, 2) and R(5, -1) are the vertices of △PQR. Let △P'Q'R' is formed when they are translated by the translation vector T = $\begin{pmatrix} \text{-2} \\ 3 \end{pmatrix}$,
(x, y) $\xrightarrow{\text{T.V.}\begin{pmatrix} \text{-2} \\ 3 \end{pmatrix}}$ (x - 2, y + 3)
P(2, 4) $\rightarrow$ P'(0, 7)
Q(-1, 2) $\rightarrow$ Q'(-3, 5)
R(5, -1) $\rightarrow$ R'(3, 2)

Again. △P'Q'R' is reflected in the line x + y = 0 to form △P"Q"R",
(x, y) $\xrightarrow{\text{x = -y}}$ (-y, -x)
P'(0, 7) $\rightarrow$ P"(-7, 0)
Q'(-3, 5) $\rightarrow$ Q"(-5, 3)
R'(3, 2) $\rightarrow$ R"(-2, -3)

Finally, plotting △PQR and its images on graph,

Draw the image in graph that is formed when a triangle with vertices A(1, 0), B(2, 1) and C(3, -1) is translated by $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. Reflect the image in the line x = 2 and also draw the image in the same graph.

A(1, 0), B(2, 1) and C(3, -1) are the vertices of △ABC. Let △A'B'C' is formed when they are translated by the translation vector T = $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$,
(x, y) $\xrightarrow{\text{T.V.}\begin{pmatrix} 1 \\ 2 \end{pmatrix}}$ (x + 1, y + 2)
A(1, 0) $\rightarrow$ A'(2, 2)
B(2, 1) $\rightarrow$ B'(3, 3)
C(3, -1) $\rightarrow$ C'(4, 1)

Again. △A'B'C' is reflected in the line x = 2 to form △A"B"C",
(x, y) $\xrightarrow{\text{x = 2}}$ {2(2) - x, y}
A'(2, 2) $\rightarrow$ A"(2, 2)
B'(3, 3) $\rightarrow$ B"(1, 3)
C'(4, 1) $\rightarrow$ C"(0, 1)

Finally, plotting △ABC and its images on graph,

A triangle with vertices N(1, 2), V(4, -1) and M(2, 5) is reflected successively in the lines x = -1 and y = 2. Find by stating coordinates and represent the images graphically under these transformations. Also state the single transformation given by these transformations.

N(1, 2), V(4, -1) and M(2, 5) are the vertices of △NVM. Let they are reflected in the line x = -1 to form △N'V'M',
(x, y) $\xrightarrow{\text{x = -1}}$ {2(-1) - x, y}
N(1, 2) $\rightarrow$ N'(-3, 2)
V(4, -1) $\rightarrow$ V'(-6, -1)
M(2, 5) $\rightarrow$ M'(-4, 5)

Again, △N'V'M' is reflected in the line y = 2 to form △N"V"M",
(x, y) $\xrightarrow{\text{y = 2}}$ {x, 2(2) - y}
N'(-3, 2) $\rightarrow$ N"(-3, 2)
V'(-6, -1) $\rightarrow$ V"(-6, 5)
M'(-4, 5) $\rightarrow$ M"(-4, -1)

Now, plotting △NVM and its images on graph,

Finally, taking N and N" to find single transformation,
N(1, 2) $\xrightarrow{\text{x = -1}}$ N"(-3, 2)
(x, y) $\xrightarrow{\text{x = -1}}$ {2(-1) - x, y}

So, single transformation is reflection on line x = -1.