When (x, y) moves under a rotation about O through an angle of θ to a new position P'(x', y'), then OP = OP' and POP' = θ where positive θ is measured anticlockwise.
O is the only point which does not move under the rotation.
R_{θ} means a rotation about O through an angle of θ^{0}
Original point (x, y) (x, y) (x, y) |
Angle rotation 90 degree -90 degree 180 degree |
Image (-y, x) (y, -x) (-x, -y) |
Problem 1 :
Find the image equation of the line 2x - 3y = -6 under clockwise rotation about O through 90°.
Solution :
Given line is 2x - 3y = -6
Angle rotation to be performed = 90°clockwise rotation.
Rule to be applied :
(x, y) ==> (-y, x)
2(-y) - 3x = -6
-2y - 3x = -6
Multiplying by the negative sign through out the equation, we get
3x + 2y = 6
Problem 2 :
Find the image of the equation when 3x - 4y = 7 is rotated clockwise through a quarter turn about O.
Solution :
Given line is 3x - 4y = 7
Angle rotation to be performed = quarter turn clockwise rotation.
That is, 90°clockwise rotation.
Rule to be applied :
(x, y) ==> (-y, x)
3(-y) - 4x = 7
-3y - 4x = 7
Multiplying by the negative sign, we get
4x + 3y = -7
Problem 3 :
Find the image of the equation when y = -3 is rotated anticlockwise through 90° about O.
Solution :
Given line is y = -3
Angle rotation to be performed = 90° counter clockwise rotation.
Rule to be applied :
(x, y) ==> (y, -x)
-x = -3
Ignoring negative sign on both sides, we get
x = 3
Problem 4 :
Find the image of the equation when x = 7 is rotated through 180° about O.
Solution :
Given line is x = 7
Angle rotation to be performed = 180°
Rule to be applied :
(x, y) ==> (-x, -y)
-x = 7
Multiplying by negative on both sides, we get
x = -7
then x + 7 = 0
Problem 5 :
Find the image of the equation when y = x^{2} is rotated through 180° about O.
Solution :
Given curve is y = x^{2}
Angle rotation to be performed = 180°
Rule to be applied :
(x, y) ==> (-x, -y)
-y = (-x)^{2}
-y = x^{2}
Multiplying by negative on both sides, we get
y = -x^{2}
Problem 6 :
Find the image of the equation when 2x + 3y = 12 is rotated clockwise direction through 90° about O.
Solution :
Given line is 2x + 3y = 12
Angle rotation to be performed = 90° clockwise rotation
Rule to be applied :
(x, y) ==> (-y, x)
2(-y) + 3x = 12
3x - 2y = 12
Problem 7 :
Find the image of the line x - y = 8 under M _{y = -x }followed by the translation (4, -1).
Solution :
x - y = 8
Reflection across the line y = -x, to perform this we have to follow the rule,
(x, y) ==> change x as -y and change y as -x
In the given line doing the above changes,
-y - (-x) = 8
x - y = 8 after the reflection.
After the translation :
Moving 4 units right and 1 unit down.
x' = x + 4 and y' = y - 1
x + 4 - (y - 1) = 8
x + 4 - y + 1 = 8
x - y + 5 = 8
x - y = 8 - 5
x - y = 3
Problem 8 :
Find the image of the line x + 2y = -4 under R_{ -90 }followed by the translation (2, -5).
Solution :
x + 2y = -4
Rotation of 90 degree in anticlockwise direction. To perform this, we have to follow the rule
(x, y) ==> (y, -x)
In the given line doing the above changes,
y + 2(-x) = -4
-2x + y = -4
After the translation :
Moving 2 units right and 5 units down.
x' = x + 2 and y' = y - 5
-2(x + 2) + y - 5 = -4
-2x - 4 + y - 5 = -4
-2x + y - 9 = -4
-2x + y = -4 + 9
-2x + y = 5
Multiplying by negative through out the equation, we get
2x - y = -5
Problem 9 :
Find the image of the line x + y = 1 under M_{ 90 }followed by M_{ y = x }followed by the translation (3, 1).
Solution :
90 degree clockwise rotation :
(x, y) --> (-y, x)
-y + x = 1
x - y = 1
Reflection across y = x :
x - y = 1
To perform the reflection across y = x, we have to follow the rule given below.
x should be changed as y and y should be changed as x
y - x = 1
-x + y = 1
After the translation :
Moving 3 units right and 1 unit up
x' = x + 3 and y' = y + 1
-(x + 3 )+ y + 1 = 1
-x + y - 3 + 1 = 1
-x + y - 2 = 1
-x + y = 3
y = x + 3
x + y = 1 - 2
x + y = -1
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM