# Transformation

Transformation

Transformation is the movement of an object in plane to bring a change. The original shape of an object is called pre- image and the final image after transformation is called image.There are basic four methods for changing geometrical figures they are:

a. Reflection

Reflection is a mirror image of an object or basically it is called flip of an object.

P(x,y) →P’ (x ,-y)

P(2,3)→P’ (2,-3)

Transformation matrix =$\left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)$

Reflection in y-axis

Reflection in y= x

P(x,y) →P’ ( y ,x)

P(-3,2 )→P’ (2,-3)

Transformation matrix =$\left( {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right)$

Reflection in y= -x

P(x,y) →P’ ( -y ,-x)

P(3,2 )→P’ ( -2,-3)

Transformation matrix =$\left( {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right)$

b. Translation

Translation is a term used in geometry to describe a function that moves an object a certain distance. The object is not altered in any other way. It is not rotated, reflected or re-sized. In a translation, every point of the object must be moved in the same direction and for the same distance.

If there co-ordinates in plane is translated by the vector$\left( {\begin{array}{*{20}{c}}a\\b\end{array}} \right)$then

P(x,y)→ P’ (x+ a ,y+b)

If the translation T1 =$\left( {\begin{array}{*{20}{c}}a\\b\end{array}} \right)\:$is followed by T2 =$\left( {\begin{array}{*{20}{c}}c\\d\end{array}} \right)\:$the combined translation T2OT1=$\left( {\begin{array}{*{20}{c}}{a + c}\\{b + d}\end{array}} \right)\:$

P(x,y)→ P’ (x+ a+c ,y+b+d)

c.Rotation

In this transformation the pre-image is transformed by certain angle like 90°, 180° about the origin or centre (h,k)

Rotation of 90° anti-clockwise about the origin

P(x,y) →P(-y,x)

Transformation matrix =$\left( {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right)$

Rotation of 90° clockwise about the origin

P(x,y) →P(y,-x)

Transformation matrix =$\left( {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right)$

Rotation of 180° clockwise about the origin

P(x,y) →P( -x, -y )

Transformation matrix =$\left( {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right)$

d. Enlargement

Enlargement changes the size of an object .For every enlargement scalar factor kand centre should be defined.

If k>1, then the image is larger than the object.

If k = ± 1, then the image and the object is the same size.

If 0<k<1, then the image is smaller than the object.

If O is an centre of enlargement and k is a scalar factor then , it is [ written as [ 0,K]

P(x,y)→ P(kx,ky)

Transformation matrix =$\left( {\begin{array}{*{20}{c}}K&0\\0&K\end{array}} \right)$

The combination of the enlargement of the enlargement E1 [O, k1] followed by the another enlargement E2 [0,k2 ] is given E2 OE1 =[ O, K1 K2 ]

If the centre is other than origin, say, (a,b) and scalar factork

P(x,y)→ P(k(x-a)+a ,k(y-b) +b)

Example1

A triangle with vertices A(1,2), B(4, -1),C(2,5)is reflected successively in the lines X =5 and y= -2Find by stating the co-ordinates

Soln

A(1,2)$\: \mathop \to \limits^{reflection\: on\: x = 5\: }$A’(2*5 – 1, 2 )= A’(9,2)

B(4, -1)$\: \mathop \to \limits^{reflection\: on\: x = 5\: }$ B’(2*5- 4, -1) =B’(6, -1)

C(2,5)$\: \mathop \to \limits^{reflection\: on\: x = 5\: }$ C ‘(2*5-2, 5) = C’(8,5)

Now,

A’(9,2) $\mathop \to \limits^{reflection\: on\: y = - 2\: }$ A’’(9 ,2 * -2- 2 ) = A’’(9, -6)

B’(6, -1)$\mathop \to \limits^{reflection\: on\: y = - 2\: }$ B’’(6 , 2*-2+1 ) = B’’(6,-3)

C’(8,5) $\mathop \to \limits^{reflection\: on\: y = - 2}$ C’’(8, 2*-2 -5) = C’’(8, -9)

Example 2

IfX(2,3) Y(4,5) and Z(6,2) are vertices of Δ XYZfind the image of Δ XYZunderthe translation vector $X\vec Y$followed by $Y\vec Z\:$

Soln

Translation vector $\overrightarrow {XY}$= $\overrightarrow {OY}$ -$\overrightarrow {OX}$

= (4, 5)- (2,3)

= (2, 2)

||y translation vector$\overrightarrow {YZ}$ = $\overrightarrow {OZ}$- $\overrightarrow {OY}$

= (6, 2) – (4, 5)

= (2,-3)

The translation vector XY ⃗ followed by YZ ⃗ is given by T =(2, 2)+(2,-3)= (4 , -1)

Example 3

Using the matrix method show that a reflection in the line y= x followed by a reflection in y=0 is equivalent to the negative quarter turn about origin.

Soln

The matrix representing the reflection in y = x

T =$\: \left( {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right)$

Let us consider a point P(x,y) as an object.

Pre- multiplying by the given matrix M

$\left( {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right)$

= $\left( {\begin{array}{*{20}{c}}y\\x\end{array}} \right)$

Hence P(x,y) à P’(y , x)

If it is followed by the reflection y = 0 (reflection on x-axis)

Then,

Matrix representing the reflection on x-axis T =$\: \left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)$

$\left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}y\\x\end{array}} \right)$

$= \: \left( {\begin{array}{*{20}{c}}y\\{ - x}\end{array}} \right)$s

∴ P’(y,x)= P’’(y, -x)

P(x, y)= P’’(y,-x)

The matrix representing the negative quarter turn about the origin$\left( {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right)$

If p(x,y) be the object then,

$\left( {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right) = \: \left( {\begin{array}{*{20}{c}}{ - y}\\x\end{array}} \right)$

Hence

P(x,y)à P’(-y , x)

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