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TRANSFORMATION IN THE PLANE
Transformation in the plane is a mapping which shifts an object from one position to another within the same plane.
Examples of transformations in the xy plane are
i. Reflection
ii. Rotation
iii. Enlargement
iv. Translation

REFLECTION
-The action or process of sending back light, heat or sound from a surface.

ISOMETRIC MAPPING
-Is a transformation which the object size is maintained.
-Reflection is an example of isometric mapping.

Reflection in the line included an angle (α) passing through the origin.


inclined at B with the coordinates being (X, Y)
is the image of under reflection is
PP is perpendicular to OS ( is the line of reflection)
POS= α β

Δ OPQ is right angled a
t Q

Hence X = OP cos B……(i)
Y = OP sin B
is perpendicular to the axis of X at R

Coordinates of R are ( X1,0)
OR= X 1
RP1=Y1
Δ OP1R is right angled at R
P1OS= POS= α-β
Angle P1OR = α – β + α – β + β = 2α – β
Cos (2α-β)= X1
OP1
X1= OP1cos (2α-β)….. (iii)

Y1 = sin (2α-β)
OP1
Y1=OP1sin (2α-β)……(iv)
Cos (A+B) = cos A cos B – sin A sin B
Sin(A+B) = sin A cos B+ sin B cos A
X1 = OP1 cos 2α cosβ + OP sin 2α cos β…. (iii)
Y1= OP sin 2α cos β – OP1 sin B cos 2α
X1= OP cos 2α cos β+ OP sin 2α sin β
Y1= OPsin2αcos β – Op sinβ cos 2α
X1 = OP cos β cos2α + OP sin β sin 2α
Y1= OP cos βsin 2α – OP sin β cos 2α
X1 = X cos 2α + Y sin 2α …..( i)
Y1 = X sin 2α – Y cos 2 α …..(ii)
=


Exercise
1. Find the image of the point A (1, 2) after a reflection in the Y= X plane.




Solution;
Y= X
= 1
Tan α = = 1
α= 900
=
=
=


(X1, Y1) = (2, 1)

2. Find the image of B (3,4) after a reflection in the line Y= -X followed by another reflection in the line Y= 0

My = -X
My = 0

Y= -X





y = – x
y = – 1
x
α = – 450 for clockwise movement
Or
α = 1350 anticlockwise movement

=

=

=

(X1, Y1) = (-4, -3)

Followed reflection at Y = 0
Tan α = 0
α = 0
=

=

=

(X11, Y11) = (-4, 3)


In questions 3 to 6, write the matrix of reflection in a given line.
3. Y= 0 ( the X axis)
Y= 0
Tan α= 0
α= 0
=

Mx =


4. Y = X
= 1

Tan α= 1
Α= 900

=

=


5. X = 0
Tan α = 0
α= 0
=
=

6. Find the image of the point (1, 2) after a reflection in the line Y= X followed by another reflection in the line Y= -X.
= 1
Tan α = 90 0
=
=
=

= ( 2, 1)


ROTATION
Find the image of the point B (1,2) after a rotation by 900 about the origin.

Solution:

= =
Rθ = =


IMAGE OF A POINT ROTATED AT INCLINED LINE AT ANGLE B

Let be inclined at an angle B
=
PA is perpendicular to the X – axis at A
= X , = Y
ΔOAP is right angled at A with POA = B
Cos B = X
OP
X = OP Cos B …….(i)
Sin B = Y
OP
Y = OP sin B ….. (ii)

is perpendicular to the X – axis at B
Cos (B+ θ) = X1
OP
X 1 = cos ( B + θ)
X1 = cos B cos θ – OP sin B sinθ
X1 = X cos θ – Y sin θ ……(iii)
Sin (B+ θ) = Y 1
OP
Y1= sin (B+ θ)
Y1= sin B cos θ + sin θ cos B
Y1= Y cos θ + X sin θ …..(iv)

In matrix form
=

Question
The matrix of rotation
1. 900 about the origin.
R =

=


2. find the image of (1,2) after a rotation of 900 followed by another rotation of 2700 about the origin.

Solution
=

=
=

=

3. Find the image of 3X + 4Y + 6 = 0 under a rotation of 900 about the origin.

Solution:
3X + 4Y = -6
X intercept, Y = 0
3X = -6
X = -2
Y intercept, X = 0
4Y= -6
Y =
(-2,)

=

=

=



TRANSLATION
Exercise
1. A translation takes every point a distance 1 unit to the left and 2 units downwards. Find where it takes.
a. (0,0)
b. (1,1)
c. (3,7)

Solution
a. (0,0)

(a,b) = (-1,-2)



b. (1,1)
(a,b) = (-1,-2)



C. (3,7)



2. If translation takes the origin to (8,7). Given

U= (-12, 12) , v= (6,-16)

find T(u+v)
= ( u + v)



3. Find the image if the line 3x + 4Y + 6 = 0 under a translation by the vector (-6,-1)
3x + 4Y -6 = 0

Solution

Y = mx + c

4Y = -3/4 x – 3/2
X intercept, y=0
3x+ 0 + 6=0
3X= -6
X = -2
(-2,0)

Y intercept, X= 0
0 + 4Y + 6 = 0
4Y = -6
Y =
(0,)



(-8,-1) and (-6,)


Y- Y1 = M (X-X1)
Y- -1 = M( X- -8)
Y + 1 = (X+8)
Y+1 = X +
Y= X 7

4. Find the image of the line
Y = X under a translation by the vector (5,4)

Solution

Y = X
X
-2
1
Y
-2
1





(3,2) and (6,5)
Slope =

=

=

M =1 (3 ,2)
Y-Y1 = M ( X –X1)
Y-2 = 1(X – 3)
Y-2 = X – 3
Y = X – 3 + 2
Y = X -1

LINEAR TRANSFORMATIONS
Consider transformation T,
Let u and v be two vectors
Let t be the real number
The t is a linear transformation if it obeys the following properties.
i. T (tu) = t T (u)
ii. T( u+v) = T (u) + T(v)


ENLARGEMENT
The transformation which magnifies an object such that its image is proportionally increased or decreased in size by some factor.
General matrix of enlargement is

where k is non zero or real number (Linear scale factor)

EXERCISE
1. Find the image of (1,2) under the enlargement by T=

Solution

= 1 =

(X1, Y 1) = (5, 10)

2. Find the image of ( -1/2 , -1/3 ) under the enlargement by
T=

Solution



(X‘, Y ‘) = (6, 4)

3. Find the enlargement matrix which maps the point (3,-4) into (18, -24).



-24 = 4k

18 = -3k

K= -6

-24 = 4K

K = -6

(-6,-6)




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2 Comments

  • Munyaradzi Chimchenga, August 12, 2024 @ 3:57 pm Reply

    Thanks for sending information

  • ajuna julius, March 22, 2024 @ 5:49 pm Reply

    thanks

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