Long Question

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,

△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,

△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,

▱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,

△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,

△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,

△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,

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,

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(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.

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. 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,

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,

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,

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.