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Rotation means a complete turn around a central point. The distance from the center to any point on the shape stays the same. Every point makes a circle around the center.The movement of a geometric figure about a certain point is the rotation. The amount of rotation is described in terms of degrees. If the degrees are positive, the rotation is performed counterclockwise, if they are negative, the rotation is clockwise. The figure will not change size or shape, but, unlike a translation, will change direction.
There are three conditions required to rotate a figure. They are
In the given figure, a centre is an origin, the angle is 90° and direction is anti-clockwise.
In the figure, the centre is an origin, the angle is -90° and direction is clockwise.
When a point P(3, 2) is rotated about origin through +90°,
the image of the point becomes P(-2, 3).
Look at the graph given below.
A general rule:
R_{90°} (x, y) = (-y, x)
When a point P(3, 2) is rotated about origin through 180°,
the image of the point becomes P(-3, -2).
Look at the graph below:
A general rule:
R_{180°}(x, y) → (-x, -y)
When a point P(2, -3) is rotated about origin through 270°,
the image of the point becomes P(2, -3)
Look at the graph given below:
A general rule:
R_{270°}(x, y)→(y, -x)
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:
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)
ASK ANY QUESTION ON Rotation
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