Rotation
Rotation
Rotation is a circular movement that has a central point that stays fixed and everything else moves around that point in a circle. A full rotation is 360°. The fixed point about which an object is rotated is called the centre of rotation and the angle through which every point of the object is rotated is called the angle of rotation.
The angle of rotation is positive is the rotation is made in an anticlockwise direction and it is negative if the rotation is made in a clockwise direction.
Rotation using Coordinates
 Rotation through 90° in an anticlockwise direction about the centre at the origin(Positive Quater Turn): When we rotate the point along the 90°, the xcoordinate and ycoordinate changes the place and the signs are changed.
We write this as,
P(x, y) → P'(−y, x)  Rotation through 90° in a clockwise direction about the centre at the origin (Negative Quarter Turn): When we rotate the point along the 90° the yaxis, then ycoordinates remain the same, but xcoordinator are transformed into its opposite.
We can write it as,
P(x, y) → P'(y,−x)  Rotation through 180° about the centre at the origin (Half Turn): If the image is obtained by the rotation through 180° in an anticlockwise direction about the centre at the origin is same or the image obtained by the rotation through 180° in clockwise direction.
We can write it as,
P(x, y) → P'(−x,−y)
 Rotation through 90° in an anticlockwise direction about the centre at the origin (Positive Quater Turn) = P(x, y) → P'(−y, x)
 Rotation through 90 in a clockwise direction about the centre at the origin (Negative Quarter Turn) = P(x, y) → P'(y, −x)
 Rotation through 180° about the centre at the origin (Half Turn) = P(x, y) → P'(−x, −y)
Solution:
A(1, 4), B(3, 2) and C(4, 5)
Rotation through 90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
A(1, 4) \(\rightarrow\) A'(4, 1)
B(3, 2) \(\rightarrow\) B'(2, 3)
C(4, 5) \(\rightarrow\) C'(5, 4)
The image formed by rotaion through 90^{0} is shown below in graph:
Solution:
Here,
D(2, 4), E(6, 8) and F(5, 3)
Rotation through 180^{o}
P(x, y) \(\rightarrow\) P'(x, y)
D(2, 4) \(\rightarrow\) D'(2, 4)
E(6, 8) \(\rightarrow\) E'(6, 8)
F(5, 3) \(\rightarrow\) F'(5, 3)
Plotting the \(\triangle\)DEF in graph
Solution:
When rotated through 180° anticlockwise or clockwise about the origin, the new position of the above points is.
 The new position of the point A(3, 5) will be A'(3, 5)
 The new position of the point B(2, 7) will be B'(2, 7)
 The new position of the point C(5, 8) will be C'(5, 8)
 The new position of the point D(9, 4) will be D'(9, 4)
Solution:
A(2, 3), B(5, 3) and C(5, 6)
Rotation through 90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
A(2, 3) \(\rightarrow\) A'(3, 2)
B(5, 3) \(\rightarrow\) B'(3, 5)
C(5, 6) \(\rightarrow\) C'(6, 5)
The imgae formed by ratation through 90^{0} is shown below:
Solution:
A(2, 2), B(5, 5) and C(4, 1)
Rotation through +90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
A(2, 2) \(\rightarrow\) A'(2, 2)
B(5, 5) \(\rightarrow\) B'(5, 5)
C(4, 1) \(\rightarrow\) C'(1, 4)
The image formed by rotation through +90^{o} is shown below:
Solution:
When rotated through 180° anticlockwise or clockwise about the origin, the new position of the above points is.
 The new position of the point A(2, 3) will be A'(2, 3)
 The new position of the point B(4, 6) will be B'(4, 6)
 The new position of the point C(5, 8) will be C'(5, 8)
 The new position of the point D(4, 3) will be D'(4, 3)
Solution:
Rotation through 180^{o}
P(1, 4) \(\rightarrow\) P'(1, 4)
Q(3, 1) \(\rightarrow\) Q'(3, 1)
R(2, 1) \(\rightarrow\) R'(2, 1)
Crossing the \(\triangle\)PQR and \(\triangle\)P'Q'R and plotting them in graph.
Solution:
On plotting the points P(3, 1) and Q(2, 3) on the graph paper to get the line segment PQ.
Now rotate PQ through 180° about the origin O in anticlockwise direction, the new position of points P and Q is:
P(3, 1) → P'(3, 1)
Q(2, 3) → Q'(2, 3)
Thus, the new position of line segment PQ is P'Q'.
Solution:
Rotation through +90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
A(4, 6) \(\rightarrow\) P'(6, 4)
Rotation through 90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
A(4, 6) \(\rightarrow\) P'(6, 4)
Rotation through 180^{o}
P(x, y) \(\rightarrow\) P'(x, y)
A(4, 6) \(\rightarrow\) A'(4, 6)
Solution:
Under rotation about origin through 90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
A(1, 2) \(\rightarrow\) A'(2, 1)
B(4, 5) \(\rightarrow\) B'(5, 4)
C(5, 1) \(\rightarrow\) C'(1, 5)
\(\triangle\)ABC and its image \(\triangle\)A'B'C' are shown on the graph alongside.
Solution:
Under rotation about origin through +90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
P(2, 4) \(\rightarrow\) P'(4, 2)
Q(6, 8) \(\rightarrow\) Q'(8, 6)
R(5, 3) \(\rightarrow\) R'(3, 5)
\(\triangle\)PQR and its image \(\triangle\)P'Q'R' are shown on the graph.
Solution:
Given,
P(2, 1), Q(5, 2) and R(3,3)
Rotation through 90^{o}
P(2, 1) \(\rightarrow\) P'(1, 2)
Q(5, 2) \(\rightarrow\) Q'(2, 5)
R(3, 3) \(\rightarrow\) R'(3, 3)
Plotting the \(\triangle\)PQR and \(\triangle\)P'Q'R' on same graph
Solution:
Under rotaion about origin through 90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
P(5, 1) \(\rightarrow\) P'(1, 5)
Q(1, 2) \(\rightarrow\) Q'(2, 1)
R(4, 5) \(\rightarrow\) R'(5, 4)
\(\triangle\)PQR and its image \(\triangle\)P'Q'R' are shown on the graph alongside.
Solution:
Under rotaion about origin through 90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
A(2, 1) \(\rightarrow\) A'(1, 2)
B(1, 4) \(\rightarrow\) B'(4, 1)
C(3, 2) \(\rightarrow\) C'(2, 3)
\(\triangle\)ABC and its image\(\triangle\)A'B'C' are shown on the graph alongside.
Solution:
P(0, 1), Q(2, 4), R(5, 2) and S(6, 5)
Rotation through +90^{o}
P(x, y) \(\rightarrow\) P'(y, x)
P(0, 1) \(\rightarrow\) P'(1, 0)
Q(2, 4) \(\rightarrow\) Q'(4, 2)
R(5, 2) \(\rightarrow\) R'(2, 5)
S(6, 5) \(\rightarrow\) P'(5, 6)
The parallelogram PQRS and image formed by it when rotating at +90^{o} is given below,

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