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Consider a point O in a plane.The rotation of the plane about the point O through a given angle \(\theta\) is a transformation of the plane into itself under which to every point P there is a point P' such that OP = OP' and \(\angle\)POP' = \(\theta\)
The point O is called the centre of rotation. It is an invariant point under a rotation.
Rotation: Transformation that turns figure about a fixed point.
Centre of rotation: The point that the rotation happens around.
Angle of rotation: How many degrees clockwise or counterclockwise a shape is turned.
An object can be rotated either in clockwise or in the anticlockwise direction. It is called the direction of rotation. If any object is rotated in the anticlockwise direction, then it is termed as a positive rotation and if it is rotated in the clockwise direction, it is termed as negative rotation.
So, a rotation is defined when following three conditions are given:
Let us consider a triangle ABC and a point O outside the triangle.
Then \(\triangle\)A'B'C' is the image of \(\triangle\)ABC under the rotation about O through an angle of \(\theta\)° in the anticlockwise direction.
Again,
Take a line segment AB and a point O.
Above result leads us to develop a process for finding the centre, angle and direction of rotation.
Let \(\triangle\)A'B'C' be the image of \(\triangle\)ABC. Join A with its image A' and B with its image B'. Draw perpendicular bisectors of AA' and BB'. Let these perpendicular bisectors LM and PQ intersect each other at the point O. Then O is the centre of rotation.Above result leads us to develop a process for finding the centre, angle and direction of rotation.
Join OA and OA'. Let \(\angle\)AOA' = \(\theta\).
Then \(\theta\) is the angle of rotation. Look at the direction of the arrow in the figure. The direction of the arrow is the direction of rotation.
The point of intersection of two lines is called the centre of rotation. If the angle between the intersecting lines is \(\frac{\theta}2\), then \(\theta\) is called the angle of rotation.
The angle of rotation can be taken of any magnitude. But we generally take the following angles of rotation:
For the clockwise or negative rotation, we will take the following angles of rotation:
The rotation of geometric shapes with an angle of 90° is termed as the quarter turn and is denoted by Q. If the direction of rotation is anticlockwise direction, then it is termed as the positive quarter turn and if the direction of rotation is clockwise direction, it is termed as the negative quarter turn.
The positive or the negative turn with an angle 180° is called the half turn. Similarly, the positive or the negative turn with an angle of 360° is called the full turn.
Note: A rotation of 360° maps the object onto itself. This is called identity transformation under the rotation. In a half turn, a line and its image are parallel.
Coordinates can be used for finding images of geometric shapes after rotating them through some specific angles such as 90°, 180°, 270°, 360°, -90°, - 180°, -270° and -360°.
Case I: When the centre of rotation is origin
From (i) and (ii)
x^{2} + y^{2} = a^{2} + (-axyaxy)^{2}
or, x^{2}+ y^{2} = a^{2} + a2x2y2a2x2y2
or, y^{2} (x^{2} +y^{2}) = a^{2}y^{2} + a^{2}x^{2}
or, y^{2} (x^{2} +y^{2}) = a^{2}y^{2} +a^{2}x^{2}
or, y^{2} (x^{2} +y^{2}) = a^{2}(x^{2} +y^{2})
∴ a = +- y
Now, when a =y, then from (i), b = -axyaxy =- yxyyxy
When a = -y, then from (i), b = -axyaxy = -(yyyy)x = x
∴ Image of (x,y) under rotation through 90^{o} about the origin is either (y, -x) or (-y, x). If (x, y) is in the first quadrant, (-y, x) will be the second quadrant and (y, -x0 will be the fourth quadrant.
Hence,
(i) Image of the point (x, y) under rotation about origin through 90^{o} (positive quarter turn) is (-y, x)
(ii) Image of the point (x, y) under rotation about origin through 90^{o} (negative quarter turn) is (y, -x).
If Q^{+}is the positive quarter turn about origin, then we write
Q^{+} : P(x, y) → P' (-y, x)
If Q^{-} is the negative quarter turn about origin, then we write
Q^{-} : P(x, y)→ P' (y, x)
Note the image of a point under rotation through - 270^{o }about the origin is the same as the image of the point under rotation through 90^{o} about the origin. Similarly, the image of a point under rotation through 270^{o} about the origin is same as the image of the point under rotation through -90^{o} about the origin.
S.No. | Object | Image |
1 | A(1,1) | A'(1,-1) |
2 | B(3,2) | B'(2,-3) |
3 | C(2,4) | C'(4,-2) |
S.No. | Object | Image |
1 | A(1,1) | A'(-1,1) |
2 | B(3,2) | B'(-3,-2) |
3 | C(2,4) | C'(-2,-4) |
Translation | Object | Image | Notations |
Rotation about origin through 90° | P (x, y) | P' (-y, x) | R : P (x, y)→ P' (-y, x) |
Rotation about origin through -270° | |||
Rotation about origin through -90° | P (x, y) | P' (y, -x) | R : P (x, y)→ P' (y, -x) |
Rotation about origin through 270° | |||
Rotation about origin through 180° | P (x, y) | P' (-x, -y) | R : P (x, y)→ P' (-x, -y) |
Rotation about origin through -180° |
Angle of rotation | Centre of rotation | Object Image |
+ 90° or -270° | (0,0) | P(x,y) → P^{1}(-y,x) |
-90° or +270° | (0,0) | P(x,y) → p^{1}(y,x) |
±180° | (0,0) | P(x,y) → p^{1}(-x,-y) |
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:
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(6,-2), B(10,4) and C(4,6) are the vertices of Δ ABC.Find the co-ordinates of its image under the following rotation about the origin.
90°(-ve)
A'(-7,2),B'(2,-10),C'(6,-4)
A'(-6,2),B'(4,-10),C'(8,-4)
A'(-6,2),B'(4,-10),C'(6,-4)
A'(6,1),B'(4,-11),C'(6,1)
A(6,-2), B(10,4) and C(4,6) are the vertices of Δ ABC.Find the co-ordinates of its image under the following rotation about the origin.
90°(+ve)
A'(2,6),B'(-4,10),C'(6,4)
A'(-2,6),B'(-4,10),C'(-6,4)
A'(2,6),B'(-4,10),C'(-6,4)
A'(1,6),B'(4,10),C'(-6,4)
A(6,-2), B(10,4) and C(4,6) are the vertices of Δ ABC.Find the co-ordinates of its image under the following rotation about the origin.
180°
A'(-6,1),B'(-10,2),C'(-1,-2)
A'(6,8),B'(10,-4),C'(-4,-6)
A'(-6,2),B'(-9,-4),C'(-4,-6)
A'(-6,2),B'(-10,-4),C'(-4,-6)
The image of points A(4,5) and B(6,3) are A'(-5,4) and B'(a,b).Find the values of a and b.
a=-1 and b=8
a=-3 and b=6
a=-8 and b=7
a=-2 and b=5
If the vertices of ( riangle)MNP are M(1,1), N(3,1) and P(2,3), plot the image of ( riangle) MNP under the rotation through 90° in anti-clockwise direction about the origin.Then determine the co-ordinates of the corresponding vertices.
M'(1,1),N'(-8,3),P'(-3,2)
M'(1,1),N'(1,3),P'(6,2)
M'(-1,1),N'(-1,3),P'(-3,2)
M'(1,2),N'(-1,3),P'(3,2)
If the vertices of ( riangle)ABC are A(2,3), B(4,5) and C(6,2), plot the image of ( riangle) ABC under the rotation through 90° in anti-clockwise direction about the origin.Then determine the co-ordinates of the corresponding vertices.
A'(1,2)B'(-6,4),C'(2,6)
A'(-3,2)B'(5,4),C'(-1,6)
A'(-3,1)B'(-5,8),C'(-2,6)
A'(-3,2)B'(-5,4),C'(-2,6)
P(4,-2), Q(2,1) and R(5,2) are the vertices of ( riangle)PQR.Draw the figure of ( riangle)PQR under the rotation of 270° by taking (0,0) as the centre of rotation and write the image.
P'(-3,-1),Q'(1,-2),R'(2,-5)
P'(-3,-1),Q'(1,-2),R'(2,2)
P'(-3,-1),Q'(1,7),R'(2,-5)
P'(5,-1),Q'(1,-2),R'(2,-5)
A(2,5), B(6,4) and C(3,7) are the vertices of ( riangle)ABC.Draw the figure of image of ( riangle)ABC under the rotation of -270° by taking (0,0) as the centre of rotation and write the co-ordinates of the image.
A'(-5,2),B'(-4,6),C'(-7,3)
A'(5,8),B'(-4,6),C'(7,2)
A'(5,1),B'(-8,6),C'(-7,3)
A'(-5,2),B'(4,6),C'(-7,2)
A triangle with vertices A(2,4) B(6,4) and C(4,2) is rotated by an angle of x° about the (a,b) then the images are A'(4,-2) B'(4,-6) and C'(2,-4).Find the values of x, a and b using graph.
x =-90° or, 270°,a=0,b=0
x =-80° or, 290°,a=0,b=1
x =-50° or, 270°,a=8,b=5
x =-60° or, 270°,a=1,b=0
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