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,