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.

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)