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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 anti-clockwise direction and it is negative if the rotation is made in a clockwise direction.
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.
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.
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)
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