Soln:

Procedure"

(i) Join O and C and rotate the point C with centre at O and radius OC in the positive direction of 90^{o} and mark it as C'.

(ii) Join O and B and rotate the point B with centre at O and radius OB in the positive direction of 90^{o} and mark it as B'.

(iii) Join O and A and rotate the point A with centre at O and radius OA in the positive direction of 90^{o} and mark it as A'.

Then the triangle A'B'C' formed by joining A' B' and C' by the ruler is the image of triangle ABC under the rotation of 90^{o} about O.

Soln:

Procedure:

^{o}and then the points are marked as E', F' and D' respectively. Also join the points E', F' and D' so that the triangle E' F' D' is the image of triangle EFD under rotation of -90

^{o}about O.

Soln:

Procedure:

Here, join J, K, L and M with O. Then rotate the points J, K, L and M by taking O as the centre and OJ, OK, OL and OM as the radius in the positive direction of 180^{o} and then mark the points J', K', L' and M'. Now join the points J', K', L' and M' so that the quadrilateral J' K' L' M' is the image of quadrilateral JKLM.

Soln:

Procedure:

Here, taking O as the centre and OA, OB and OC on the radius rotate the points A, B and C in the positive direction of 270^{o} and then the points are marked as A', B' and C' respectively. Then join A', B', C' and O so that the shaded portion A'B'C'O is the image of the figure OABC under the rotation of 270^{o} about O.

Soln:

Q(P) = Q(3, 4) =(-4, 3) [ Q = Positive quarter turn, so Q(a, b) = (-b, a)]

Again, Q^{2}(P) = Q(Q(P)) [=Q(-4, 3) =(-3, -4)] [ Q(P) = (-4, 3)]

Here, H(P) = H(3, 4) = (-3, 4) [ H = half turn, so H (a, b) = (-a, -b)]

∴ Q^{2} (P) = H(P)

a. through 270^{o} anti-clockwise. Write down the coordinates of the image so formed.

Soln:

Rotationof positive 270^{o} about origin is

P(5, 7)→ P'(7, -5) [ P(a, b)→ P'(b, -a)]

b.through 90^{o} clockwise . Write down the coordinates of the image so formed.

Soln:

Rotation of P(5, 7) about through negative 90^{o} is

P(5, 7)→ P'(7, -5)

c. What is the difference between the image of (a) and (b)?

Soln:

The image of a and b are same.

Soln:

In 8(a), rotation rotates A(3, 5) about origin through negative 90^{o} into A'(5, -3) which is shown in the figure along side. From the same figure, the rotation of A(3, 5) about origin through positive 270^{o} is same and gives A'(5, -3). So rotation of a point about origin through negative 90^{o} is equivalent to the rotation about origin through positive 270^{o}

Similarly, in 8(b), when drawing figure, the rotation of a point about origin through positive 90^{o} isequivalent to the rotation about origin through negative270^{o}.

Q^{-1} (P) = Q^{-1} (3 , 4) = (4 , -3)

\(\therefore\) [ q1 is the negative quarter turn ,so q-1 (a , b) = (b , -a)]

Here , Q^{2} (P) = Q^{-1} (Q^{-1} (P))

= Q^{-1} (4 , -3) = (-3 , -4)

Again , H(P) = H (3 , 4) = (-3 , -4)

\(\therefore\) Q^{-2} (P) = H(P) is true .Proved.

Reflection on the line y = -x is P(a , b)→P' (-b , -a) .So ,

P(2 , 3)→ P'(-3 , -2) , Q(-4 , 1)→ Q' (-4 , 1) , R(2 , -5)→ R'(5 , -2)

S(-2 , -4) ,→ S'(4 , 2) and T(0 , 2)→ T' (2 , 0) Ans.

Reflection on the line x = -2 is P(a , b)→ P'(2h - a , b) where , h = -2

So ,

P(2 , 3)→ P'(2 \(\times\) 2 - (-1) , 4) = Q' (-4 + 1 , 4) = Q' (-3 , 4)Ans.

R(2 , -5)→ R' (2 \(\times\) -2 -2 , -2) = R' (-6 , -5) Ans.

S(-2 , -4)**→**S'(2 \(\times\) -2 (-2) , -4) = S' (-4 + 2 , -4) = S'(-2 , -4) Ans.

T(0 , 2)→ T' (2 \(\times\) -2 -0 , -2 ) = T'(-4 , -2) Ans.

Reflection on the line y = 2 is P(a , b)→P' (a , 2k - b) where k = 2

So , P(2 , 3)→ P' (2 , 2 \(\times\) -4) = Q' (-1 , 0) Ans.

R(2 , -5)→ R' (2 , 2 \(\times\) 2 - (-5)) = R'(2 , 9) Ans.

S(-2 , -4)→ S' [ -2 , 2 \(\times\) 2 - (-4)] = S'(-2 , 8) Ans.

T(o , -2)→ T' [o \(\times\) 2 - (-2)] = T' (0 , 6) Ans.

Here , reflection on y -axis gives P(a , b)→ P'(-a , b). So we have , P(1 , 1)→ P'(-1 , 1)

Q(3 , 1)→ Q' (-3 , 1) and R (3 , -1)→ R' (-3 , -1)

The graph is as shown below :

Here , the shaded portion \(\triangle\) P'Q'R' is the image of \(\triangle\) PQR.

a. Positive quarter turn.

Soln:

Rotation of parallelogram ABCD about O through positive 90^{o} is given below:

A(0, -2)→ A'(2, 0)

B(2, 1)→ B'(-1, 2)

C(-1, 2)→ C'(-2, -1) and

D(-3, -1)→ D'(1, -3)

The graph of parallelogram ABCD and its image A'B'C'D' are as shown below:

b.negative quarter turn.

Soln:

Rotation of parallelogram ABCD about O through negative 90^{o} is as given below:

A(0, -2)→ A'(-2, 0)

B(2, 1)→ B'(1, -2)

C(-1, 2)→ C'(2, 1)

and D(-3, -1)→ D'(-1, 3)

The graph of parallelogram ABCD and its image A'B'C'D' are as shown below:

c. half turn

Soln:

Rotation of parallelogram ABCD about O through half turn (180^{o}) is as given below:

A(0, -2)→ A'(0, 2)

B(2, 1)→ B'(-2, -1)

C(-1, 2)→ C'(1, -2)

and D(-3, -1) → D'(3, 1)

The graph of parallelogram ABCD and its image A'B'C'D' are as shown below:

Reflection on x -axis X(P) | Reflection on y - axis Y(P) | Reflection on the line y = x W(P) |

a. X(A) = (4 , -1) | Y (A) = (-4 , 1) | W(A) = (1 , 4) |

b. X(B) = (2 , -1) | Y(B) = (-2 , 1) | W(B) = (1 , 2) |

c. X(C) = (-1 , 2) | Y(C) = (1 , -2) | W(D) = (-3 , 0) |

d. X(D) = (0 , 3) | Y(D) = (0 , -3) | W(D) = (-3 , 0) |

e. X(E) = (3 , 0) | Y(E) = (-3 , 0) | W(E) = (0 , 3) |

f. X(F) = (4 , -4) | Y(F) = (-4 , 4) | W(F) = (4 , 4) |

g. X(G) = (-1 , 5) | Y(G) = (1 , -5) | W(G) = (-5 , -1) |