# 270 DEGREE ROTATION ON 2D SHAPES

Rotating the shape means moving them around a fixed point. There are two directions

i) Clockwise

ii) Counter clockwise (or) Anti clockwise

The shape itself stays exactly the same, but its position in the space will change.

 270° clockwise 270° counter clockwise90° clockwise 90° counter clockwise (x, y) ==> (-y, x)(x, y) ==> (y, -x)(x, y) ==> (y, -x)(x, y) ==> (-y, x)

Note :

90° clockwise and 270° counter clockwise both are same.

Rotate each shape as described.

Problem 1 :

The shape above has the following coordinates:

A (-2, 0), B (-7, 3) and C(-9, 7)

Rotate the shape 270° counter-clockwise.

Solution:

Here, triangle is rotated 270° counterclockwise. So, the rule that we have to apply here is

(x, y) ---> (y, -x)

A(-2, 0) = A'(0, 2)

B(-7, 3) = B'(3, 7)

C(-9, 7) = C'(7, 9)

Vertices of the rotated figure are

A'(0, 2), B'(3, 7) and C'(7, 9)

Problem 2 :

The shape above has the following coordinates:

A(4, 0), B(10, -1), C(8, -6) and D(3, -9)

Rotate the shape 270° counter-clockwise.

Solution:

Here, triangle is rotated 270° counterclockwise. So, the rule that we have to apply here is

(x, y) ---> (y, -x)

A(4, 0) = A'(0, -4)

B(10, -1) = B'(-1, -10)

C(8, -6) = C'(-6, -8)

D(3, -9) = D'(-9, -3)

Vertices of the rotated figure are

A'(0, -4), B'(-1, -10), C'(-6, -8) and D'(-9, -3)

Problem 3 :

The shape above has the following coordinates:

A (0, -1), B (-8, -4) and C (-10, -10)

Rotate the shape 90° counter-clockwise.

Solution:

Here, triangle is rotated 90° counterclockwise. So, the rule that we have to apply here is

(x, y) ---> (-y, x)

A(0, -1) = A'(1, 0)

B(-8, -4) = B'(4, -8)

C(-10, -10) = C'(10, -10)

Vertices of the rotated figure are

A'(1, 0), B'(4, -8) and C'(10, -10)

Problem 4 :

The shape above has the following coordinates:

A (0, 0), B (-9, -3), C(-8, -10) and D(-2, -6)

Rotate the shape 90° clockwise.

Solution:

Here, triangle is rotated 90° clockwise. So, the rule that we have to apply here is

(x, y) ---> (y, -x)

A(0, 0) = A'(0, 0)

B(-9, -3) = B'(-3, 9)

C(-8, -10) = C'(-10, 8)

D(-2, -6) = D'(-6, 2)

Vertices of the rotated figure are

A'(0, 0), B'(-3, 9), C'(-10, 8) and D'(-6, 2)

Rotate each shape as described.

Problem 5 :

The shape above has the following coordinates:

A (-3, 3), B (-7, 2) and C (-8, 6)

Rotate the shape 270° clockwise.

Solution:

Here, triangle is rotated 270° clockwise. So, the rule that we have to apply here is

(x, y) ---> (-y, x)

A(-3, 3) = A'(-3, -3)

B(-7, 2) = B'(-2, -7)

C(-8, 6) = C'(-6, -8)

Vertices of the rotated figure are

A'(-3, -3), B'(-2, -7) and C'(-6, -8)

Problem 6 :

The shape above has the following coordinates:

A (-4, -2), B (-9, 0), C (-6, -7) and D (-5, -8)

Rotate the shape 90° counter-clockwise.

Solution:

Here, triangle is rotated 90° counterclockwise. So, the rule that we have to apply here is

(x, y) ---> (-y, x)

A(-4, -2) = A'(2, -4)

B(-9, 0) = B'(0, -9)

C(-6, -7) = C'(7, -6)

D(-5, -8) = D'(8, -5)

Vertices of the rotated figure are

A'(2, -4), B'(0, -9), C'(7, -6) and D'(8, -5)

Problem 7 :

The shape above has the following coordinates:

A (-5, 3), B (-6, 1) and C (-9, 9)

Rotate the shape 270° clockwise.

Solution:

Here, triangle is rotated 270° clockwise. So, the rule that we have to apply here is

(x, y) ---> (-y, x)

A(-5, 3) = A'(-3, -5)

B(-6, 1) = B'(-1, -6)

C(-9, 9) = C'(-9, -9)

Vertices of the rotated figure are

A'(-3, -5), B'(-1, -6) and C'(-9, -9)

Problem 8 :

The shape above has the following coordinates:

A (-2, -1), B(-6, -3), C(-7, -6) and D(-1, -9)

Rotate the shape 270° clockwise.

Solution:

Here, triangle is rotated 270° clockwise. So, the rule that we have to apply here is

(x, y) ---> (-y, x)

A(-2, -1) = A'(1, -2)

B(-6, -3) = B'(3, -6)

C(-7, -6) = C'(6, -7)

D(-1, -9) = D'(9, -1)

Vertices of the rotated figure are

A'(1, -2), B'(3, -6), C'(6, -7) and D'(9, -1)

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