Transformation Geometry ~ Bed Prasad Dhakal

Table of Contents

Introduction
Reflection
Equation of Reflection
Theorems of Reflection
Rotation
Theorems on Rotation
Equation of Rotation
Translation
Equation
Theorem 1
Composite Transformation
Non-isometric transformation
Dilation
Enlargement
Reduction
Stretches (and shrinks)
Shearing

ज्यामितिमा स्थानान्तरण लामो समयदेखि व्यापक चासोको विषय बनेको थियो। युक्लिडको आधिकारिक वा स्वयंसिद्ध सिद्धान्त नभएपनि सामान्यतया स्थानान्तरणलाई सुपरपोजिसनको सिद्धान्त (उठाउर अर्कोमा राख्नु) मा आधारित गरिएको थियो । जब १९ औं शताब्दीमा (1872 मा) फ्लेक्स क्लेनले - एर्लान्जेन विश्वविद्यालयमा - ज्यामितिमा एक नयाँ परिप्रेक्ष्य प्रस्तावित गरे तबदेखी स्थानान्तरणलाई एर्लान्जेन कार्यक्रमको रूपमा चिनिन थालिएको छ।

Transformation Geometry

Transformation Geometry is a branch of mathematics that utilize analytic geometry and algebraic function to study geometric invariants. It was introduced (developed) by Flex Klein in19th century through Erlangen programme. According to him, theer are five Kinds of Transformations:

Rigid motions: Plane geometry of congruent figures, preserve distances (e.g., Euclidean transformations)
Similarity transformations: The familiar geometry with similar figures.
Similarities preserve angles and ratios between distances (e.g., resizing)
Affine (matrix) transformations: Geometry of computer animation. Rectangles and parallelograms the same
Affine transformations preserve parallelism (e.g., scaling, shear)
Projective Geeometry: Geometry of photo
Projective transformations preserve collinearity
Continuous (topological) transformations: Any loop is a circle. Any path is a segment

In transformation geometry, the idea of flips, slides, and turn are key features, which will be analyzed based on its mapping. This mapping are basically of two kinds.

Isometric mapping (transformation)
Non-isometric mapping (transformation)

Geometric transformation involves both analytic and algebraic geometry because it can be approached using the graphical perceptive, and algebraic computational theory.

Reflection

Reflection is very common topic in physics and mathematics. When an object is place before the mirror, the image is formed behind the mirror. This is called reflection.

Definition: Reflection

Let l be a line and A be a point.
Then, reflection about the line l is denoted by \(T_1 \), which is a transformation on a plane, where

if A ∉l , then T_l (A) = A'
if A ∈l , then T_l(A)=A

such that

the mispoint of AA' lies on l
the segment AA'is perpendiculat to l

We start a short discussion for reflection in mathematics as below.

Let us take a geometrical figures like triangle. Now, we hold the triangle vertices, one-by-one, about a line source \( l\) in a grid. Then look at the image (shadow) of the triangle on the grid. The shadow have the same size as the original triangle has, but in opposite direction.

In the above figure, ∆XYZ is a triangle and its shadow is ∆X'Y'Z'.
By actual measurements it can be seen that
∡X = ∡X', ∡Y = ∡Y', ∡Z = ∡Z
XY=X'Y', YZ=Y'Z', XZ=X'Z'
Thus, the two triangles are congruent
The important note is that

l ⊥ bisector of XX'
l ⊥ bisector of YY'
l ⊥ bisector of ZZ'

Here \( l\) is said to be line of reflection

In summary,

The image is found opposite to the mirror line \( l\) .
Mirror line \( l\) is the perpendicular bisector of the line joining the object and its image.
X' is called the reflection (image) of X and so on
\( l\) is called the axis of reflection.

Thus, reflection can be seen in water, in a mirror, in glass. In mathematics, an object and its reflection have the same shape and size, but the figures face in opposite directions . Also, a reflection needs

Object
Line of Reflection

Equation of Reflections

Drag Red Points रातो बिन्दु लाई चलाउनुहोस

Theorem 1: X-axis

Reflection on X-axis:Prove that reflection of a point P(x,y) on x-axis is P'(x',y') where x'=x,y'=-y
Given that
P(x,y)
P’(x’,y’)
Line of reflection l: X-axis
We know that, Mid-point of PP’ lies on l
So we have
\(\left( \frac{x+x'}{2},\frac{y+y'}{2} \right)\) lies on x-axis
or \(\left( \frac{x+x'}{2},\frac{y+y'}{2} \right)=(x,0)\)
or \(\frac{x+x'}{2}=x,\frac{y+y'}{2}=0\)
or x'=x,y'=-y

Theorem 2: Y-axis

Reflection on Y-axis:Prove that reflection of a point P(x,y) on y-axis is P'(x',y') where x'=-x,y'=y
Given that
P(x,y)
P’(x’,y’)
Line of reflection l: Y-axis
We know that, Mid-point of PP’ lies on l
So we have
\(\left( \frac{x+x'}{2},\frac{y+y'}{2} \right)\) lies on y-axis
or \(\left( \frac{x+x'}{2},\frac{y+y'}{2} \right)=(0,y)\)
or \(\frac{x+x'}{2}=0,\frac{y+y'}{2}=y\)
or x'=-x,y'=y

Theorem 3: X=a

Reflection on Y=mx:Prove that reflection of a point P(x,y) on x=a is P'(x',y') i.e.
\(x'=x-2a,y=y\)

Theorem 4: Y=b

Reflection on Y=mx:Prove that reflection of a point P(x,y) on y=b is P'(x',y') i.e.
\(x'=x,y'=y-2b\)

Theorem 5: Y=X

Reflection on Y=X axis: Prove that reflection of a point P(x,y) on y=x line is P'(x',y') where x'=y,y'=x
Given that
P(x,y)
P’ (x’,y’)
Line of reflection l: Y=X
We know that
Mid-point of PP’ lies on l
or \(\left( \frac{x+x'}{2},\frac{y+y'}{2} \right)\) lies on y=x
or \( \frac{x+x'}{2}=\frac{y+y'}{2}\)
or x + x'-y - y'=0 (1)
Next
Slope of PP’ x Slope of l = -1
or \(\frac{y'-y}{x'-x}\times 1=-1\)
or y'-y+x'-x=0 (2)
Therefore, Solving (1) and (2), we get
x'=y,y'=x

Theorem 6: Y=-X

Reflection on Y=-X axis: Prove that reflection of a point P(x,y) on y=-x line is P'(x',y') where x'=-y,y'=-x
Given that
P(x,y)
P’ (x’,y’)
Line of reflection l: Y=-X
We know that
Mid-point of PP’ lies on l
or \(\left( \frac{x+x'}{2},\frac{y+y'}{2} \right)\) lies on y=-x
or \( \frac{x+x'}{2}=-\frac{y+y'}{2}\)
or x + x' + y + y' = 0 (1)
Next
Slope of PP’ x Slope of l = -1
or \(\frac{y'-y}{x'-x}\times -1=-1\)
or x-x'-y+y'=0 (2)
Therefore, Solving (1) and (2), we get
x'=-y,y'=-x

Theorem 7:Y=mX

Reflection on Y=mx:Prove that reflection of a point P(x,y) on y=mx is P'(x',y') i.e.
\(x'=x-\frac{2m ( mx-y)}{1+m^2 },y'=y-\frac{-2( mx-y)}{1+m^2 }\)

Proof
We know that, PP' is bisected by y=mx line, so, by mid-point formula,
\(\left( \frac{x+x'}{2},\frac{y+y'}{2} \right)\) lies on y=mx+c
Thus, we have
\(\frac{y+y'}{2}=m\frac{x+x'}{2}\)
or y'+y-mx-mx'=0 (1)
Also, we know that
(Slope of PP') x (Slope of y=mx) =-1
or \(\frac{y'-y}{x'-x} \times m =-1\)
or my'-my+x'-x=0 (2)
Therefore, solving (1) and (2), we get
\(x'=-\frac{-x+m^2 x-2my}{1+m^2 },y'=-\frac{-2mx+y-m^2 y}{1+m^2 }\)
or \( x'=-\frac{-x-m^2 x+2m^2 x-2my}{1+m^2 },y'=-\frac{-2mx+2y-y-m^2 y}{1+m^2 }\)
or \(x'=x-\frac{2m( mx-y)}{1+m^2 },y'=y-\frac{-2( mx-y)}{1+m^2 }\)

Theorem 8:Y=-mX

Reflection on Y=mx:Prove that reflection of a point P(x,y) on y=mx is P'(x',y') i.e.
\(x'=x-\frac{2m ( mx-y)}{1+m^2 },y'=y-\frac{-2( mx-y)}{1+m^2 }\)

Theorem 9: Y=mX+c

Reflection on Y=mx+c:Prove that reflection of a point P(x,y) on y=mx+c is P'(x',y') i.e.
\(x'=x-\frac{2m ( mx+c-y)}{1+m^2 },y'=y-\frac{-2( mx+c-y)}{1+m^2 }\)

Proof
We know that, PP' is bisected by y=mx+c line, so, by mid-point formula,
\(\left( \frac{x+x'}{2},\frac{y+y'}{2} \right)\) lies on y=mx+c
Thus, we have
\(\frac{y+y'}{2}=m\frac{x+x'}{2}+c\)
or y'+y-mx-mx'-2c=0 (1)
Also, we know that
(Slope of PP') x (Slope of y=mx+c) =-1
or \(\frac{y'-y}{x'-x} \times m =-1\)
or my'-my+x'-x=0 (2)
Therefore, solving (1) and (2), we get
\(x'=-\frac{2cm-x+m^2 x-2my}{1+m^2 },y'=-\frac{-2c-2mx+y-m^2 y}{1+m^2 }\)
or \( x'=-\frac{2cm-x-m^2 x+2m^2 x-2my}{1+m^2 },y'=-\frac{-2c-2mx+2y-y-m^2 y}{1+m^2 }\)
or \(x'=x-\frac{2m( mx+c-y)}{1+m^2 },y'=y-\frac{-2( mx+c-y)}{1+m^2 }\)

Theorem 10:ax+by+c=0

Reflection on ax+by+c=0: Prove that reflection of a point P(x,y) on a line ax+by+c=0 is P'(x',y') where
\( x'=x-\frac{2a(ax+by+c)}{a^2+b^2},y'=y-\frac{2b(ax+by+c)}{a^2+b^2}\)

Proof
Given that
P(x,y) is object
P’(x’,y’) is image, and
Line of reflection is l: ax+by+c=0
Now, we know that
Mid-point of PP’ lies on the line l
or \( \left( \frac{x+x'}{2},\frac{y+y'}{2} \right) \) lies on the line ax+by+c=0
or\(a\frac{x+x'}{2}+b\frac{y+y'}{2}+c=0 \)
or ax+ax'+by+by'+2c=0 (1)
Next
Slope of PP’ x slope of l = -1
or \(\frac{y'-y}{x'-x}\times \frac{-a}{b}=-1 \)
or ay'-ay-bx'+bx=0 (2)
Therefore, solving (1) and (2), we get
\(x'=x-\frac{2a(ax+by+c)}{a^2+b^2},y'=y-\frac{2b(ax+by+c)}{a^2+b^2}\)

Theorems on Reflection

Theorem 1

Prove that reflection is distance preserving transformation.
OR

Prove that reflection is isometric transformation.

Proof
Let T_l be a reflection.
Also let, AB be a given distance such that
T_l(A) = A'
T_l(B) = B'
To prove
AB=A'B'
Construction
Join A and A’ which intersects l at M
Join B and B’ which intersects l at N
Join A and N, join A’ and N

Proofs

SN	Statements	Reasons
1	∆1≅∆2	by SAS
2	∆3≅∆4	by SAS
3	AB=A'B'	CSCT

Similarly, we can prove that AB=A'B in other different cases as mentioned below.

Case 1: If A and B both points lies on the line \(l\)
Case 2: If only one point (say A) of A and B lies on the line \(l\)
Case 3: If both points A and B does NOT lie on the line \(l\), but lies on the same side of AB
Case 4: If both points A and B does NOT lie on the line \(l\), but lies on the opposite side of AB

Theorem 2

Prove that reflection preserves collinearity.
Let T_l is a reflection.
Also let A,B,C are three collinear points such that
T_l(A)=A’
T_l(B)=B’
T_l(C)=C’
To prove,
A’,B’,C’ are three collinear points.
Here,
Given A,B,C are collinear points
or AB+BC=AC
By the definition of reflection
A'B'+B'C'=A'C'
Therefore,
A’,B’,C’ are collinear points.

Rotation

Let us take a center O=(a,b), an angle \(\theta\) and given direction
Also, take a geometrical figures like triangle. Now, we move the triangle's vertices about O, one-by-one, by the \(\theta\) with given direction

When working in the coordinate plane: (i) assume the center of rotation to be the origin unless told otherwise and (ii) assume a positive angle (anti-clockwise) of rotation.

Then shadow cast of the triangle is called rotation of the geometrical figures (in this case , triangle) with center O, this shadow is congruent to the original triangle.

In the figure below, ∆XYZ is a triangle and its shadow is ∆X'Y'Z'.

By actual measurements it can be seen that
∡X = ∡X', ∡Y = ∡Y', ∡Z = ∡Z'
XY=X'Y', YZ=Y'Z', XZ=X'Z'
Thus, the two triangles are congruent.
The important note is that
∡XOX'= ∡YOY'=∡ZOZ'=θ , say
Here θ is said to be angle of rotation and the point O is known as the center of rotation

Formally, a rotation is a transformation that turns a figure about (around) a point. The point in which the figure turns around is called the center of rotation. The center of rotation can be on/in or outside the shape.The direction of rotation can be clockwise or anticlockwise.

Definition

Rotation means to spin a shape around a point. Thus, in a rotation, an object and its rotation image are of the same shape and size, but the figures (object and image) may be turned in different directions.

To define a rotation, we need

Center
Direction
Angle

A rotation is a transformation that turns a figure about a fixed point called the center of rotation. In a rotation, an object and its rotation image are of the same shape and size, but the figures (object and image) may be turned in different directions.

If O be a center and θ be an angle, then rotation about O with angle θ is denoted by T_O,θ, is a transformation such that if A is any point, then T_O,θ(A)=A' such that

OA and OA’ are equal and
∡AOA'=θ

Theorems on Rotation

Theorem 1

Prove that rotation is distance preserving transformation.
OR

Prove that rotation is isometric transformation.

Proof
Let T_O,θ be a rotation .
Also let, AB be a given distance such that
T_O,θ(A) = A'
T_O,θ(B) = B'
To prove
AB=A'B'
Construction
Join AB and A'B'
Proofs

SN	Statements	Reasons
1	∆1≅∆2	by SAS
3	AB=A'B'	CSCT

Theorem 2

Prove that rotation preserves collinearity.
Let T_O,θ is a rotation .
Also let A,B,C are three collinear points such that
T_O,θ(A)=A’
T_O,θ(B)=B’
T_O,θ(C)=C’
To prove,
A’,B’,C’ are three collinear points.
Here,
Given A,B,C are collinear points
or AB+BC=AC
By the definition of rotation
A'B'+B'C'=A'C'
Therefore,
A’,B’,C’ are collinear points.

Equation of Rotation

Equation of Rotation_M1

Prove that rotation of P(x,y) about origin O by angle \(\theta\) is P'(x',y') where
\( x'=x\cos \theta -y\sin \theta ,y'=x\sin \theta +y\cos \theta \)

Proof
Let P(x,y) be an arbitrary point, and suppose that it makes an angle \( \alpha\) with x-axis,then
\( x=r \cos \alpha, y=r \sin \alpha \)
Also let the point P(x,y) is rotated about the center O with an angle \(\theta\) to obtain P'(x',y') then OP' makes an angle \( \alpha +\theta\) with x-axis,hence
\( x'=r \cos (\alpha+\theta), y=r \sin (\alpha+\theta) \)
or \( x'=x \cos \theta -y \sin \theta, y'=x\sin \theta +y\cos \theta \)

Theorem on Rotation_M2

Prove that rotation of P(x,y) about origin O by angle \(\theta\) is P'(x',y') where
\( x'=x\cos \theta -y\sin \theta ,y'=x\sin \theta +y\cos \theta \)

Proof
Let m be a line given by
\( y=\tan \frac{\theta }{2}x \)
or \( x\sin \frac{\theta }{2}-y\cos \frac{\theta }{2}=0 \) (1)
Then we suppose P(x,y) be a point on m.
Now,
Reflection of P(x,y) on x-axis is Q(x,-y).
Finally, rotation of P(x,y) by \(\theta\) = Reflection of Q(x,-y) on \( x\sin \frac{\theta }{2}-y\cos \frac{\theta }{2}=0 \)
Thus, reflection of Q(x,-y) on \( x\sin \frac{\theta }{2}-y\cos \frac{\theta }{2}=0\) is
\( x'=x-\frac{2a(ax+by)}{a^2+b^2},y'=y-\frac{2b(ax+by)}{a^2+b^2} \)
Keeping the value, x=x, y=-y we get
\(x'=x-\frac{2a(ax-by)}{a^2+b^2},y'=-y-\frac{2b(ax-by)}{a^2+b^2} \)
Keeping the value, \( a=\sin (\frac{\theta }{2}),b=-\cos (\frac{\theta }{2})\) we get
\( x'=x-\frac{ 2\sin \frac{\theta}{2} (\sin \frac{\theta }{2} x-(-\cos \frac{\theta }{2})y)} {(\sin \frac{\theta }{2})2+(\cos \frac{\theta }{2})^2},y'=-y-\frac{2(-\cos \frac{\theta }{2})(\sin \frac{\theta }{2}x-(-\cos \frac{\theta }{2})y)}{(\sin \frac{\theta }{2})^2+( \cos \frac{\theta }{2})^2}\)
or \(x'=x-2\sin \frac{\theta }{2}(\sin \frac{\theta }{2}x+\cos \frac{\theta }{2}y),y'=-y+2\cos \frac{\theta }{2}(\sin \frac{\theta }{2}x+\cos \frac{\theta }{2}y)\)
or \(x'=x-2{{\sin }^{2}}\frac{\theta }{2}x-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}y,y'=-y+2\cos \frac{\theta }{2}\sin \frac{\theta }{2}x+2{{\cos }^{2}}\frac{\theta }{2}y\)
or \(x'=x(1-2{{\sin }^{2}}\frac{\theta }{2})-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}y,y'=2\cos \frac{\theta }{2}\sin \frac{\theta }{2}x+y(2{{\cos }^{2}}\frac{\theta }{2}-1)\)
or \(x'=x\cos \theta -y\sin \theta ,y'=x\sin \theta +y\cos \theta \)
Hence, Rotation of P(x,y) by \( \theta\) is
\(x'=x\cos \theta -y\sin \theta ,y'=x\sin \theta +y\cos \theta \)

Theorem 2

Prove that rotation of P(x,y) about origin A(h,k) by angle \(\theta\) is P'(x',y') where
\( x'=(x-h)\cos \theta -(y-k)\sin \theta+h \)
\(y'=(x-h)\sin \theta +(y-k)\cos \theta +k\)

Translation

A translation "slides" an object a fixed distance in given direction. In translation, the original object and its translation have the same shape and size, and they face in the same direction. To define a translation, we need
a Vector

Definition

If PQ is a vector, then a translation through vector PQ is denoted by T_PQ, is a transformation of the plane such that if A is any point in the plane then

T_PQ(A) = A’ such that AA’ and PQ are equal vector.
AA’ and PQ are congruent and parallel

Equation

Prove that translation of P(x,y) by a vector T=(a,b) is P'(x',y') where
x'=x+a,y'=y+b

Proof
By the definition of translation, we can write that
\(\vec{PP}'=T\)
or\((x'-x,y'-y)=(a,b)\)
or\( x'=x+a,y'=y+b\)

Theorem 1

Prove that translation is distance preserving transformation.
OR

Prove that translation is isometric transformation.

Proof
Let T_PP' be a translation.
Also let, AB be a given distance such that
T_PP'(A) = A'
T_PP'(B) = B'
To prove
AB=A'B'
Construction
Join A and A'
Join B and B’
Proofs

SN	Statements	Reasons
1	PP'\|\|AA' and PP'=AA'	by translation
2	PP'\|\|BB' and PP'=BB'	by translation
3	AA'=BB"	by 1 and 2

Theorem 2

Prove that translation preserves collinearity.
Let T_PP' is a translation.
Also let A,B,C are three collinear points such that
T_PP'(A)=A’
T_PP'(B)=B’
T_PP'(C)=C’
To prove,
A’,B’,C’ are three collinear points.
Here,
Given A,B,C are collinear points
or AB+BC=AC
By the definition of translation
A'B'+B'C'=A'C'
Therefore,
A’,B’,C’ are collinear points.

Composite Transformation

The composition of any two translations is a translation.
Proof:

Let \(T_1 \begin{bmatrix} a \\ b \end{bmatrix}\) and \(T_2 \begin{bmatrix} c \\ d \end{bmatrix}\) are two translations.
Also, let \(T_2T_1\) be the composition of two translations, then we have to show that \(T_2T_1\) is a translation.
We know that,
\(T_2T_1(x,y)= T_2 \begin{bmatrix} x+a \\ y+b \end{bmatrix}\)
or \(T_2T_1(x,y)= \begin{bmatrix} x+a+c \\ y+b+d \end{bmatrix}\)
It shows that,
\(T_2T_1\begin{bmatrix} a+c\\ b+d \end{bmatrix}\) is a translation.
A composition of reflections over two parallel lines is a translation. (May also be over any even number of parallel lines.).

Let \(R_1(l_1)\) and \(R_2(l_2)\) are two reflections, where \(l_1||l_2\).
Without loss of generality, let us assume that two parallel lines \(l_1\) and \(l_2\) are respectively \(x=a\) and \(x=b\).

Also let \(R_2R_1\) be the composition of two reflections then we have to show that \(R_2R_1\) is a translation.
Let \(A(x,y)\) be any arbitrary point, then we know that,
\(R_2R_1(A)= R_2R_1(x,y)= R_2 A'(2a-x,y)=A''(2b-(2a-x),y)=A''(2(b-a)+x,y)\)
It shows that,
\( R_2R_1A(x,y)= A''[(x,y)+ [2(b-a),0]]\)
Which is also a translation by a vector \(\begin{bmatrix} 2(b-a) \\ 0 \end{bmatrix}\).
Thus, the composition of two reflections in parallel lines is a translation in a direction perpendicular to the lines by a distance twice larger than the distance between the lines.
The composition of reflections over two intersecting lines is a rotation. The center of rotation is the intersection point of the lines.

Proof
Let \(R_1(l_1)\) and \(R_2(l_2)\) are two reflections, where \(l_1 \) and \( l_2\) intersect an O and angle between \(l_1 \) and \( l_2\) is \(x+y\) .
Let P(x,y) be any arbitrary point, then
Without loss of generality, let us assume that two intersecting lines \(l_1\) and \(l_2\) respectively maps point P to P' and then to P''.

We know that reflections preserve distances, so,
OP = OP' = OP''
Since the triangle OPP' is isosceles, the two angles labeled by x in the figure are equal.
Similarly the triangle OP'P'' is isosceles, so the two angles labeled by y in the figure are equal.
Therefore, from the triangle OPP'', the angle between P and P'' through center O is
\( \measuredangle POP''=2x + 2y = 2(x + y)\)
So, the mapping P to P'' is a rotation with center O and angle \(2(x + y)\). (the cases can be thre, point on the line, inside the angle, outside the angle).
Thus, the composition of two reflections in non-parallel lines (or intersecting lines) is a rotation about the intersection point of the lines by the angle equal to the doubled angle between the lines.
The composition of rotations is either a rotation or a translation.

Let \(R_1\) and \(R_2\) are two rotations, then \(R_1R_2\) be the composition of two rotations.
Now, we must show that \(R_1R_2\) is either a rotation or a translation.
Also let, P(x,y) be any arbitrary point.
Without loss of generality, let us suppose that
\(R_1\) = rotation about O(0,0) by angle \(\theta\)
\(R_2\) = rotation about O(h,k) by angle \(\alpha\)
Such that P(x,y) is mapped to P’(x’,y’) and P(x’’,y’’) respectively by \(R_1\) and \(R_2\) .
Now,
P(x,y) is mapped to P’(x’,y’) by \(R_1\)
So,
\( x'=x\cos \theta -y\sin \theta ,y'=x\sin \theta +y\cos \theta \)
Again,
P'(x',y') is mapped to P'’(x’',y’') by \(R_2\)
So
\( x''=(x'-h)\cos \alpha -(y'-k)\sin \alpha +h\)
\(y''=(x'-h)\sin \alpha +(y'-k)\cos \alpha +k\)
Taking the x-part for simpllification, we get
\( x''=(x'-h)\cos \alpha -(y'-k)\sin \alpha +h\)
or \( x''=[(x')-h]\cos \alpha -[(y')-k]\sin \alpha +h\)
or \( x''=[(x\cos \theta -y\sin \theta)-h]\cos \alpha -[(x\sin \theta +y\cos \theta)-k]\sin \alpha +h\)
or \( x''=x\cos (\alpha+\theta)-y\sin (\alpha+\theta)-h \cos \alpha +k \sin \alpha +h\)
or \( x''=x\cos (\alpha+\theta)-y\sin (\alpha+\theta)+[h-h \cos \alpha +k \sin \alpha ]\)

Case 1: if both center are same, then h=k=0,then
The mapping given by (A)gives that
\( x''=x\cos (\alpha+\theta)-y\sin (\alpha+\theta)-h \cos \alpha +k \sin \alpha +h\)
or \( x''=x\cos (\alpha+\theta)-y\sin (\alpha+\theta)\)
Which is a rotation with same center with angle \(\theta+\alpha\)
Case 2: if \(\alpha+\theta =\) 0 mod 360, then
The mapping given by (A)gives that
\( x''=x\cos (\alpha+\theta)-y\sin (\alpha+\theta)-h \cos \alpha +k \sin \alpha +h\)
or \( x''=x+[h-h \cos \alpha +k \sin \alpha ]\)
Which is a translation of P(x,y) by vector \(\begin{bmatrix}h-h \cos \alpha +k \sin \alpha \\ k-k \sin \alpha -k \cos \alpha \end{bmatrix}\)
Case 3:if \(\alpha+\theta \ne \) 0 mod 360, then
The equation shows that the P{x'',y'') is the composition of rotation and a translation.
Now,
The composition of a rotation and a translation is a rotation.
Otherwise
If possible, suppose that the composition is a translation, then
RT=T
R=TT
It shows that, the composition of two translations is a rotation, which is a contradiction.
Therefore, the mapping given by (A) ensures that the composition is a rotation.

Non-isometric transformation

Non-isometric transformation, denoted by T, is a mapping into a plane in which ratio between the object and image are preserved. Therefore, it is called proportional transformation. The common types of such non-isometric transformations are given below.

Enlargement
Reduction

Non-isometric transformation preserves the shape but not the size. Therefore, it is called shape-preserving transformation (movement). In summary, it

Do not preserve size
Preserve the shape
Preserves angles, but not distance

So, isometric are subset of Non-isometric transformation. Some more example of Non-isometric transformation are

Stretch
Shear
Projectice transformation
Topological transformation

Dilation

Let O be a point and k be a real number, then dilation about O is denoted by \( T_{O,k}\) is a transformation, in which if A is any point then

\( T_{O,k}(A)=A’ \) such that O, A and A’ are collinear
d(OA’) = k d(OA)