Solution:

When a point P(x, y) is rotated about origin through = 90°,

P(x,y)→ P(-y, x)

A(3, 4)→ A(-4, 3)

B(-2, 5)→B(-5, -2)

C(-2, -5)→ C(5, -2)

Now, we plot ΔABC and ΔA'B'C' in the same graph as below.

Solution:

When rotated through 90° about the origin in anticlockwise direction. The new positions of the above points are:

(i) The new position of point P (3, 3) will become P' (-3, 3)

(ii) The new position of point Q (-3, -4) will become Q' (3, -4)

(iii) The new position of point R (-5, 9) will become R' (-9, -5)

(iv) The new position of point S (2, -8) will become S' (8, 2)

Solution:

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

(i) The new position of the point P (2, 4) will be P' (-2, -4)

(ii) The new position of the point Q (-2, 7) will be Q' (2, -7)

(iii) The new position of the point R (-5, -8) will be R' (5, 8)

(iv) The new position of the point S (9, -4) will be S' (-9, 4)

Solution:

On plotting the points M (-2, 3) and N (1, 4) on the graph paper to get the line segment MN.

Now, rotating MN through 180° about the origin O in anticlockwise direction, the new position of points M and N is:

M (-2, 3) → M' (2, -3)

N (1, 4) → N' (-1, -4)

Thus, the new position of line segment MN is M'N'.

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 are:

P (-3, 1) → P' (3, -1)

Q (2, 3) → Q' (-2, -3)

Thus, the new position of line segment PQ is P'Q'.

Solution:

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)

Solution:

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)

Solution:

P(x,y)\(\rightarrow\)P'(-y,x)

P(2,1)\(\rightarrow\)P'(-1,2)

Q(5,2)\(\rightarrow\)Q'(-2,5)

R(3,3)\(\rightarrow\)R'(-3,3)

Solution:

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)

Solution:

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)

Solution:

When a point P( x,y) is rotated about the origin through 270^{o},

P(x,y) \(\rightarrow\)P'(y, -x)

So, A(2,1) \(\rightarrow\)A'(1,-2)

B(5,1)\(\rightarrow\)B'(1,-5)

C(4,4)\(\rightarrow\)C'(4,-4)

D(1,4)\(\rightarrow\)D'(4,-1)

Solution:

When a point P(x, y) is rotated about the origin through 180°,

P(x, y) → P'(-x, -y)

O(0, 0) → O'(0, 0)

A(1, 0) → A'(-1, 0)

B(1, 1) → B'(-1, -1) and

C(0, 1) → C' (0, -1)

Now, we plot a unit square and its image in the same graph.

Solution:

On plotting the points M (-2, 3) and N (1, 4) on the graph paper to get the line segment MN.

Now, rotating MN through 180° about the origin O in anticlockwise direction, the new position of points M and N is:

M (-2, 3) → M' (2, -3)

N (1, 4) → N' (-1, -4)

Thus, the new position of line segment MN is M'N'.

Solution:

We get triangle PQR by plotting the point P (1, 4), Q (3, 1), R (2, -1) on the graph paper when rotated through 180° about the origin. The new position of the point is:

P (1, 4) → P' (-1, -4)

Q (3, 1) → Q' (-3, -1)

R (2, -1) → R' (-2, 1)

Thus, the new position of \(\triangle\)PQR is \(\triangle\)P'Q'R'.