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

- 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

Join B and B’ which intersects

Join A and N, join A’ and N

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. Proof
- 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.

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 \( x''=x\cos (\alpha+\theta)-y\sin (\alpha+\theta)+[h-h \cos \alpha +k \sin \alpha ]\)

#### 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)

Here O is called center and k is called scale factor.

There are two types of Dilation, they are Enlargement (k <-1 and 1 < k) and Reduction (k >-1 and 1 > k).#### Enlargement

Let O be a point and k (k <-1 and 1 < k) be a real number, then enlargement 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)

Here O is called center and k is called scale factor.

Note- If k < -1 The enlargement in on opposite side
- If 1 < k, The enlargement is on same side
- If k = 1, it is identical transformation

#### Reduction

Let O be a point and k (k >-1 and 1 > k) be a real number, then reduction 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)

Here O is called center and k is called scale factor.

Note- If -1 < k < 0 The reduction in on opposite side
- If 0 < k < 1, The reduction is on same side
- If k = 1, it is identical 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)

Here O is called center and k is called scale factor.

There are two types of Dilation, they are Enlargement (k <-1 and 1 < k) and Reduction (k >-1 and 1 > k).#### Enlargement

Let O be a point and k (k <-1 and 1 < k) be a real number, then enlargement 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)

Here O is called center and k is called scale factor.

Note- If k < -1 The enlargement in on opposite side
- If 1 < k, The enlargement is on same side
- If k = 1, it is identical transformation

#### Reduction

Let O be a point and k (k >-1 and 1 > k) be a real number, then reduction 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)

Here O is called center and k is called scale factor.

Note- If -1 < k < 0 The reduction in on opposite side
- If 0 < k < 1, The reduction is on same side
- If k = 1, it is identical transformation

#### Theorem

Show that Dilation (or enlargement or reduction) preserves collinearity.

Proof

Let \(T_{O,k}\) be an Enlargement.

Also let A,B,C are three collinear points, such that

\( T_{O,k} (A)=A'\)

\( T_{O,k} (B)=B'\)

\( T_{O,k} (C)=C'\)

To prove,

\( A',B',C'\) are three collinear points.

Here,

Given A,B,C are collinear points.

AB+BC=AC

By the definition of Enlargement

\( \frac{1}{k}A'B'+\frac{1}{k}B'C'=\frac{1}{k}A'C'\)

or A'B'+B'C'=A'C'

Therefore,

A',B',C' are collinear points.

#### Theorem 2

Show that ratio is preserved by Dilation (or enlargement or reduction)

Proof

Let \( T_{O,k}\) be an Enlargement with center at O and scale factor k.

Also let, AB be a given distance.

Then, by definition

\( T_{O,k} (A)=A'\)

\( T_{O,k} (B)=B'\)

Such that \(\frac{OA'}{OA}=k, \frac{OB'}{OB}=k \)

To prove

\( A'B'=k AB\)

Here,

\( \triangle 1 \sim \triangle 1 \) by SAS proportional theorem

\( A'B'= k AB \) (CSST are proportional)

#### Theorem 1

Prove that enlargement of a point P(x,y) about origin (0,0) by scale factor k is P'(x',y') where x'=kx,y'=ky

Proof

Let \( T_{O,k}\) be an Enlargement and P(x,y) be a given point, then

\( T_{O,k} (P)=P'\)

By definition, we write

OP' = k OP

or
(x',y') = k (x,y)

or
x' = k x, y'= ky

#### Theorem 2

Prove that enlargement of a point P(x,y) about center A(a,b) by scale factor k is P'(x',y') where x'= k(x-a)+a, y'= k(y-b)+b

Proof

Let \( T_{A,k} \) be an Enlargement and P(x,y) be a given point, then

\( T_{A,k} (P)=P'\)

Such that

AP' = k AP

or
OP'-OA=k (OP-OA)

or
(x',y')-(a,b) = k {(x,y)-(a,b)}

or
(x',y') = k (x-a,y-b)+ (a,b)

or
x' = (k (x-a)+a, y'=k(y-b)+ b

#### Theorem 2

Prove that enlargement of a point P(x,y) about center A(a,b) by scale factor k is P'(x',y') where x'= k(x-a)+a, y'= k(y-b)+b

Proof

Let \( T_{A,k} \) be an Enlargement and P(x,y) be a given point, then

\( T_{A,k} (P)=P'\)

Such that

AP' = k AP

or
OP'-OA=k (OP-OA)

or
(x',y')-(a,b) = k {(x,y)-(a,b)}

or
(x',y') = k (x-a,y-b)+ (a,b)

or
x' = (k (x-a)+a, y'=k(y-b)+ b

#### Stretches (and shrinks)

Stretches (and shrinks) are transformations that Do change the shape in given direction. This is done by multiplying the inputs or outputs by some number.

- Multiplying the outside of a function results in multiplying the y-values. Multiplying a function by a number greater than 1 makes the graph taller, or stretched vertically. Multiplying the function by a number between 0 and 1 makes the graph shorter, or shrunk vertically.

For example, multiplying a function \(y=x^{1/3}\) by 2 would produce a graph that is twice as tall. Likewise multiplying by 0.5 results in a graph that is half as tall. - Multiplying the inside of a function results in in multiplying the x-values. Multiplying the x by a number greater that 1 makes the graph skinnier, or shrunk horizontally. Multiplying the x by a number between 0 and 1 will make the graph wider, or stretched horizontally.

For example, multiplying in a function \(y=x^2\), the input x by 2 would produce a graph that is half as wide. But multiplying the input x by 0.5 results in a graph that is twice as wide, or stretched horizontally.

#### Vertical stretch/shrink

Let a be a real number, then Stretches (and shrinks) is denoted by \( T_{a,1}\) is a non-isometric transformation, in which if A is any point then

- \( T_{a,1}(A)=A’ \) such that x'=ax, y'=y
- If A,B are two points, then \( |\overrightarrow{A'B'}_x+\overrightarrow{A'B'}_y| = |\overrightarrow{AB}_x+\overrightarrow{AB}_y|\)
- The matrix representation is \(M=\left(\begin{matrix} a & 0 \\ 0 & 1 \end{matrix}\right) \)

- If a > 1, vertical stretch
- if 0 < a < 1, vertical shrink

#### Horizental stretch/shrink

Let b be a real number, then Stretches (and shrinks) is denoted by \( T_{1,b}\) is a non-isometric transformation, in which if A is any point then

- \( T_{1,b}(A)=A’ \) such that x'=x, y'=by
- If A,B are two points, then \( |\overrightarrow{A'B'}_x+\overrightarrow{A'B'}_y| = |\overrightarrow{AB}_x+\overrightarrow{AB}_y|\)
- The matrix representation is \(M=\left(\begin{matrix} 1 & 0 \\ 0 & b \end{matrix}\right) \)

- If b > 1, horizontal shrink
- if 0 < b < 1, horizontal stretch

#### Stretch/Shrink

Let a and b are two a real number, then Stretches (and shrinks) is denoted by \( T_{a,b}\) is a non-isometric transformation, in which if A is any point then

- \( T_{a,b}(A)=A’ \) such that x'=ax, y'=by
- If A,B are two points, then \( |\overrightarrow{A'B'}_x+\overrightarrow{A'B'}_y| = |\overrightarrow{AB}_x+\overrightarrow{AB}_y|\)
- The matrix representation is \(M=\left(\begin{matrix} a & 0 \\ 0 & b \end{matrix}\right) \)

#### Summary

For functions written as \(y = af(b(x − h)) + k\), the details are given below- a is vertical stretch/shrink/reflection
- If a > 1, then vertical stretch by factor of a.
- If 0 < a < 1, then vertical shrink by factor of a.
- If a is −1, then vertical reflection over the x-axis.

- b is horizontal stretch/shrink/reflection
- If b > 1, then horizontal shrink by factor of \(\frac{1}{b}\)
- If 0 <b < 1, then horizontal stretch by factor of \(\frac{1}{b}\)
- If b is −1, then horizontal reflection over the y-axis.

- h is horizontal translation
- If h > 0, then translates right.
- If h < 0, then translate left.

- k is vertical translation
- If k > 0, then translates up.
- If k < 0, then translate down.

#### Shearing

Shearing deals with changing the shape and size of the 2D object along x-axis and y-axis. It is similar to sliding the layers in one direction to change the shape of the 2D object.It is an ideal technique to change the shape of an existing object in a two dimensional plane. In a two dimensional plane, the object size can be changed along X direction as well as Y direction.

#### A shear parallel to the x axis

Transformation with horizental shear, x-shear, the y co-ordinates remain the same but the x co-ordinates changes. If P(x, y) is the point then the new points will be P’(x’, y’) given as –

x'=x+ay, y'=y

which is a maping where the vector [1,0] pointing in the x-direction is fixed, but
the vector [0,1] pointing in the y-direction is taken to the vector [a,1], where a is some
real number. Notice that if a > 0, then horizontal shear pushes the top of the green
parallelogram to the right, resulting in the red parallelogram shown. If a < 0, then horizontal shear pushes
the top of the green parallelogram to the left. It is easy to see from the definition of this
horizontal shear that the corresponding matrix is

\( M = \left( \begin{matrix}
1&a\\0&1\end{matrix}\right)\)

Let a be a real number, then shear is denoted by \( T_{a,X}\) is a non-isometric transformation, in which if A is any point then

- \( T_{a,X}(A)=A’ \) such that x'=x+ay, y'=y
- If A is a point, then \( \overrightarrow{A'A'}=(ay,0)\)
- The matrix representation is \( M = \left( \begin{matrix} 1&a\\0&1\end{matrix}\right)\)
- The determinant will always be 1

#### A shear parallel to the y axis

Transformation with vertical shear, y-shear, the x co-ordinates remain the same but the y co-ordinates changes. If P(x, y) is the point then the new points will be P’(x’, y’) given as –

x'=x, y'=y+bx

which is a maping where the vector [0,1] pointing in the y-direction is fixed, but
the vector [1,0] pointing in the x-direction is taken to the vector [1,b], where b is some
real number. It is easy to see from the definition of this
vertical shear that the corresponding matrix is

\( M = \left( \begin{matrix}
1&0\\b &1\end{matrix}\right)\)

Let b be a real number, then shear is denoted by \( T_{Y,b}\) is a non-isometric transformation, in which if A is any point then

- \( T_{Y,b}(A)=A’ \) such that x'=x, y'=y+bx
- If A is a point, then \( \overrightarrow{A'A'}=(0,bx)\)
- The matrix representation is \( M = \left( \begin{matrix} 1&0\\b&1\end{matrix}\right)\)
- The determinant will always be 1

#### Shear

Let and b be a two real number, then shear is denoted by \( T_{a,b}\) is a non-isometric transformation, in which if A is any point then

- \( T_{a,b}(A)=A’ \) such that x'=x+ay, y'=y+bx
- If A is a point, then \( \overrightarrow{A'A'}=(ay,bx)\)
- The matrix representation is \( M = \left( \begin{matrix} 1&a\\b&1\end{matrix}\right)\)

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