Note on Rotation

  • Note
  • Things to remember
  • Videos
  • Exercise
  • Quiz

.

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:

  • the centre of rotation
  • the angle of rotation
  • a direction of the turn

Let us consider a triangle ABC and a point O outside the triangle.

  1. Join OA. Taking O as the centre and OA as the radius, draw a circle. On the circumference, take a point A' such that \(\angle\)AOA' = \(\theta\)° in anticlockwise direction.
  2. Join OB. Taking O as the centre and OB as the radius, draw another circle. Take a point B' on the circumference such that \(\angle\)BOB' = \(\theta\)° in the anticlockwise direction.
  3. Join OC. Draw another circle with radius OC and Centre O. Take a point C' on the circumference such that \(\angle\)COC' = \(\theta\)° in the anticlockwise direction.
  4. Join A', B', C' in order.
    .

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.

  1. Join OA and draw a circle with radius OA and centre O. Take a point A' on the circumference such that \(\angle\)AOA' = \(\theta\) in an anticlockwise direction.
  2. Join OB and draw a circle with radius OB and centre O. Take a point B' on the circumference such that \(\angle\)BOB' = \(\theta\) in an anticlockwise direction.
  3. Join A' and B'. Then the line segment A'B' is the image of the line AB under the rotation about O through an angle of \(\theta\)° in the anticlockwise direction.
  4. Join AA' and BB'. Construct two lines XY and PQ such that XY is perpendicular bisector of AA' and PQ is perpendicular bisector of BB'.
    .


Process for finding centre, angle and direction of rotation

.

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.

Rotation is the combination of two reflections in two intersecting lines.

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.

Some special rotations

The angle of rotation can be taken of any magnitude. But we generally take the following angles of rotation:

  1. 90°
  2. 180°
  3. 270°
  4. 360°

For the clockwise or negative rotation, we will take the following angles of rotation:

  1. -90°
  2. -180°
  3. -270°
  4. -360°

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.

Properties of Rotation

  1. Rotation is defined if the centre, angle and direction are given.
  2. A rotation transforms all the points of the geometric figures on a plane in the same direction and the same angular displacement.
  3. The image figure under the rotation is congruent with object figure.
  4. The perpendicular bisectors of the line joining the image point and object point pass through the centre of rotation.
  5. The distance of object point from the centre of rotation is equal to that of an image point from the centre of rotation.
  6. In the rotation, only the centre of rotation is invariant.
  7. The rotation about a point through an angle \(\theta\) in the anticlockwise direction is equivalent to that rotation about the same point through the angle (360° - \(\theta\)) in the clockwise direction.

Use of coordinates in Rotation

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

  1. Quarter turn about origin
    .



    Let P (x, y) be a point in the lane and P' (a, b) be its image after rotating through 90° about origin O. Then, \(\angle\)POP' = 90°.

    Now,

    Slope of OP = \(\frac {y - 0}{x - 0}\) = \(\frac yx\)
    Slope of OP' = \(\frac {b - 0}{a - 0}\) = \(\frac ba\)

    Here,

    (slope of OP)× (slope of OP') = -1
    i.e. \(\frac yx\)× \(\frac ba\) = -1
    or, b = - \(\frac {ax}{y}\)........................(i)

    Again,

    OP2 = (x - 0)2+ (y - 0)2= x2 + y2
    OP'2 = (a - 0)2 + (b - 0)2 = a2+ b2

    But: OP2 = OP'2

    So,

    x2 + y2= a2 + b2........................... (ii)

    From (i) and (ii)

    x2 + y2 = a2 + (-axyaxy)2

    or, x2+ y2 = a2 + a2x2y2a2x2y2

    or, y2 (x2 +y2) = a2y2 + a2x2

    or, y2 (x2 +y2) = a2y2 +a2x2

    or, y2 (x2 +y2) = a2(x2 +y2)

    ∴ 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 90o 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 90o (positive quarter turn) is (-y, x)

    (ii) Image of the point (x, y) under rotation about origin through 90o (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 - 270o about the origin is the same as the image of the point under rotation through 90o about the origin. Similarly, the image of a point under rotation through 270o about the origin is same as the image of the point under rotation through -90o about the origin.

  2. Quarter turn (negative) about origin (-90° or +270°)

    Let XX' and YY' be two mutually perpendicular straight lines intersecting at the origin. Let A(1,3), B(3,2) and C(2,4) be any three points on the plane. OA, OB and OC and joined and rotated quarter turn i.e -90° in anti-clockwise (positive) direction such that OA', OB' and OC' are obtained in such a way that OA = OA', OB = OB' and OC = OC' and \(\angle\)AOA' = \(\angle\)BOB' = \(\angle\)COC' = 90°. Now under the rotation about origin through -90°,
    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)

    On generalizing, we get,
    P(x,y)→P'(y,-x)
    Hence,
    Under the rotation through -90° about origin,
    P(x,y)→P(y,-x)
  3. Half turn about origin (±180)

    Let XX' and YY' be two mutually perpendicular straight lines intersecting at the origin. Let A(1,3), B(3,2) and C(2,4) be any three points on the plane. OA, OB and OC and joined and rotated quarter turn i.e 180° in anti-clockwise (positive) direction such that OA', OB' and OC' are obtained in such a way that OA = OA', OB = OB' and OC = OC' and \(\angle\)AOA' = \(\angle\)BOB' = \(\angle\)COC' = 180°. Now under the rotation about origin through 180°,
    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)

    On generalizing, we get,
    P(x,y)→P'(-x,-y)
    Hence,
    Under the rotation through -180° about origin,
    P(x,y)→P(-x,-y)

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)           →    P1(-y,x)
  -90° or  +270°       (0,0)     P(x,y)           →    p1(y,x)
  ±180°        (0,0)     P(x,y)           →     p1(-x,-y)
.

Very Short Questions

Soln:

.

Procedure"

(i) Join O and C and rotate the point C with centre at O and radius OC in the positive direction of 90o 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 90o 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 90o 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 90o about O.

Soln:

.


Procedure:
Join the points E, F, D with given point O. Then rotate the points E, F and D taking O as the centre and OE, OF and OD as the radius in the direction of - 90o and then the points are marked as E', F' and D' respectively. Also join the points E', F' and D' so that the triangle E' F' D' is the image of triangle EFD under rotation of -90o about O.

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 180o 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 270o 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 270o about O.

Soln:

Q(P) = Q(3, 4) =(-4, 3) [ Q = Positive quarter turn, so Q(a, b) = (-b, a)]

Again, Q2(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)]

∴ Q2 (P) = H(P)

a. through 270o anti-clockwise. Write down the coordinates of the image so formed.

Soln:

Rotationof positive 270o about origin is

P(5, 7)→ P'(7, -5) [ P(a, b)→ P'(b, -a)]

 

b.through 90o clockwise . Write down the coordinates of the image so formed.

Soln:

Rotation of P(5, 7) about through negative 90o 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 90o 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 270o is same and gives A'(5, -3). So rotation of a point about origin through negative 90o is equivalent to the rotation about origin through positive 270o

Similarly, in 8(b), when drawing figure, the rotation of a point about origin through positive 90o isequivalent to the rotation about origin through negative270o.

Q-1 (P) = Q-1 (3 , 4) = (4 , -3)
\(\therefore\) [ q1 is the negative quarter turn ,so q-1 (a , b) = (b , -a)]
Here , Q2 (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.

0%
  • 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'(-6,2),B'(4,-10),C'(8,-4)


    A'(6,1),B'(4,-11),C'(6,1)


    A'(-6,2),B'(4,-10),C'(6,-4)


    A'(-7,2),B'(2,-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.

    90°(+ve)

    A'(-2,6),B'(-4,10),C'(-6,4)


    A'(1,6),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(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'(-6,2),B'(-9,-4),C'(-4,-6)


    A'(-6,1),B'(-10,2),C'(-1,-2)


    A'(-6,2),B'(-10,-4),C'(-4,-6)


    A'(6,8),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=-3 and b=6


    a=-8 and b=7


    a=-1 and b=8


    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'(1,2),N'(-1,3),P'(3,2)


    M'(1,1),N'(1,3),P'(6,2)


    M'(-1,1),N'(-1,3),P'(-3,2)


    M'(1,1),N'(-8,3),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'(-3,1)B'(-5,8),C'(-2,6)


    A'(-3,2)B'(5,4),C'(-1,6)


    A'(1,2)B'(-6,4),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'(1,7),R'(2,-5)


    P'(5,-1),Q'(1,-2),R'(2,-5)


    P'(-3,-1),Q'(1,-2),R'(2,-5)


    P'(-3,-1),Q'(1,-2),R'(2,2)


  • 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'(-5,2),B'(4,6),C'(-7,2)


    A'(5,1),B'(-8,6),C'(-7,3)


    A'(5,8),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 =-60° or, 270°,a=1,b=0


    x =-80° or, 290°,a=0,b=1


    x =-50° or, 270°,a=8,b=5


    x =-90° or, 270°,a=0,b=0


  • You scored /9


    Take test again

DISCUSSIONS ABOUT THIS NOTE

No discussion on this note yet. Be first to comment on this note