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.
90° clockwise 90° counter clockwise 270° clockwise 270° 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.
Find the coordinates of the vertices of each figure after the given transformation.
Problem 1 :
rotation 90° counterclockwise about the origin
Solution:
Given,
B(-5, 0), E(-2, 1) and G(-2, -3)
Here, triangle is rotated 90° counterclockwise. So, the rule that we have to apply here is
(x, y) ---> (-y, x)
B(-5, 0) = B'(0, -5)
E(-2, 1) = E'(-1, -2)
G(-2, -3) = G'(3, -2)
Vertices of the rotated figure are
B'(0, -5), E'(-1, -2) and G'(3, -2)
Problem 2 :
rotation 90° counterclockwise about the origin
Solution:
Given,
J(3, 3), K(2, 2) and F(3, 2)
Here, triangle is rotated 90° counterclockwise. So, the rule that we have to apply here is
(x, y) ---> (-y, x)
J(3, 3) = J'(-3, 3)
K(2, 2) = K'(-2, 2)
F(3, 2) = F'(-2, 3)
Vertices of the rotated figure are
J'(-3, 3), K'(-2, 2) and F'(-2, 3)
Problem 3 :
rotation 90° clockwise about the origin
Solution:
Given, M(-5, -1), L(-5, -3) and I(-4, -3)
Here, triangle is rotated 90° clockwise. So, the rule that we have to apply here is
(x, y) ---> (y, -x)
M(-5, -1) = M'(-1, 5)
L(-5, -3) = L'(-3, 5)
I(-4, -3) = I'(-3, 4)
Vertices of the rotated figure are
M'(-1, 5), L'(-3, 5), I'(-3, 4)
Problem 4 :
rotation 90° clockwise about the origin
Solution:
Given, U(3, 3), T(5, 0) and K(2, -2)
Here, triangle is rotated 90° clockwise. So, the rule that we have to apply here is
(x, y) ---> (y, -x)
U(3, 3) = U'(3, -3)
T(5, 0) = T'(0, -5)
K(2, -2) = K'(-2, -2)
Vertices of the rotated figure are
U'(3, -3), T'(0, -5) and K'(-2, -2)
Rotate each shape as described.
Problem 5 :
The shape above has the following coordinates:
A (-4, 5), B (-10, 3) and C (-6, 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(-4, 5) = A'(-5, -4)
B(-10, 3) = B'(-3, -10)
C(-6, 6) = C'(-6, -6)
Vertices of the rotated figure are
A'(-5, -4), B'(-3, -10) and C'(-6, -6)
Problem 6 :
The shape above has the following coordinates:
A (4, 3), B (7, 2), C (9, 6) and D (2, 7)
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(4, 3) = A'(-3, 4)
B(7, 2) = B'(-2, 7)
C(9, 6) = C'(-6, 9)
D(2, 7) = D'(-7, 2)
Vertices of the rotated figure are
A'(-3, 4), B'(-2, 7), C'(-6, 9) and D'(-7, 2)
Problem 7 :
The shape above has the following coordinates:
A (-5, -3), B (-8, 0) and C (-6, -8)
Rotate the shape 270° counter-clockwise.
Solution:
Here, triangle is rotated 270° counter-clockwise. So, the rule that we have to apply here is
(x, y) ---> (y, -x)
A(-5, -3) = A'(-3, 5)
B(-8, 0) = B'(0, 8)
C(-6, -8) = C'(-8, 6)
Vertices of the rotated figure are
A'(-3, 5), B'(0, 8) and C'(-8, 6)
Problem 8 :
The shape above has the following coordinates:
A(-3, 0), B (-6, 1), C (-9, 8) and D (-1, 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(-3, 0) = A'(0, 3)
B(-6, 1) = B'(1, 6)
C(-9, 8) = C'(8, 9)
D(-1, 6) = D'(6, 1)
Vertices of the rotated figure are
A'(0, 3), B'(1, 6), C'(8, 9) and D'(6, 1)
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