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

Rotation using Co-ordinates

• Rotation through 90° in an anti-clockwise direction about the centre at the origin(Positive Quater Turn): When we rotate the point along the 90°, the x-coordinate and y-coordinate 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 y-axis, then y-coordinates remain the same, but x-coordinator 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 anti-clockwise 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 anti-clockwise 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 -90o
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 -900 is shown below in graph:

Solution:

Here,
D(2, 4), E(6, 8) and F(5, -3)
Rotation through 180o
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.

1. The new position of the point A(3, 5) will be A'(-3, -5)
2. The new position of the point B(-2, 7) will be B'(2, -7)
3. The new position of the point C(-5, -8) will be C'(5, 8)
4. 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 -90o
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 -900 is shown below:

Solution:

A(2, 2), B(5, 5) and C(4, 1)
Rotation through +90o
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 +90o is shown below:

Solution:

When rotated through 180° anticlockwise or clockwise about the origin, the new position of the above points is.

1. The new position of the point A(2, 3) will be A'(-2, -3)
2. The new position of the point B(4, -6) will be B'(-4, 6)
3. The new position of the point C(-5, -8) will be C'(5, 8)
4. The new position of the point D(4, -3) will be D'(-4, 3)

Solution:

Rotation through 180o
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 +90o
P(x, y) $$\rightarrow$$ P'(-y, x)
A(4, 6) $$\rightarrow$$ P'(-6, 4)

Rotation through -90o
P(x, y) $$\rightarrow$$ P'(y, -x)
A(4, 6) $$\rightarrow$$ P'(6, -4)

Rotation through 180o
P(x, y) $$\rightarrow$$ P'(-x, -y)
A(4, 6) $$\rightarrow$$ A'(-4, -6)

Solution:

Under rotation about origin through -90o
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 +90o
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 -90o
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 -90o
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 -90o
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 +90o
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 +90o is given below,

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