Subject: Optional Mathematics
If the point P is transformed by the transformation R1 to the point 'p' is again transformed by the transformation R2 to the point P", then the transformation which maps the point P to p" is said to be the combined transformation of R1 and R2. It is denoted by R2 .R1. The transformation R2.R. is also written as R2. R1
The transformations discussed above are the examples of single transformation. When an object has been transformed, its image can again be transformed to form a new image. Such transformation is called a combination of transformations. After a combination of transformations, the change from the single object to the final image can be described by a single transformation.
Let R1 be a transformation which maps a point P to the point P' and R2 be another transformation which maps P' to the point p''. Then, the transformation which maps a point P to P'' is said to be the combination of R1 and R2 or it is said to be the composite transformation of R1 and R2. It is denoted by R2 or R1.The composite transformation R2 R1 is also called as the transformation R1 followed byR2.
The transformations R1R2 and R2R1 have different meanings. R1R2 gives a combination of transformations R2 followed by R1 whereas, R2R1 gives the combination of transformations R1 followed by R2.
If T and W are two different transformations, then the product TW gives the combination of the transformations W followed by T (this means to work out W first and then apply T), which can be denoted by single transformation say Z and we write it as Z = TW.
The composite transformation of R with itself is denoted by R.R = R2
A translation followed by a translation is equivalent to a single translation.
Let T1 = (\()\frac{a}{b}\) be a translation vector. It describes the translation of units in the X - direction and units in the Y - direction.Let T1 translates a point P to P'.
Let T2 =(\(\frac{c}{d}\)) be another translation vector. It describes the translation ofc units in the X-direction and units in the Y-direction. Let T2 translates P' to the point P''.
Now, we want to find out the translation vector which translates the point P to P'. Then,
x - component of T = PB = PA + AB = PA + P'C = a + c
Y - component of T = BP'' = BC + CP'' = AP' + CP'' = b + d
∴ T = T2oT1 = (\(\frac{a + c}{b + d}\))
Hence, if the translation T1 = (\(\frac{a}{b}\)) is followed by the translation T2 = (\(\frac{c}{d}\)), then the composite translation T2oT1 is described by (\(\frac{a + c}{b + d}\)).
A rotation followed by a rotation is equivalent to a single rotation
A point or an object once rotated about the centre through a given angle can further be rotated about the same centre through another given angle.
Let P (x, y) be any point in the plane. If Q1 be the rotation about origin through 90o, then
Q1: P (x, y)→ P' (-y, x).
Again, if Q2 be the rotation about the origin through 90o, then
Q2: P'(-y, x)→ P'' (-x, -y).
Here, Q2.Q1: P (x, y)→ P''(-x, -y).
But (-x, -y) is the image of the point (x, y) under the rotation about the origin through 180o.
Hence, a rotation of xo followed by a rotation of yo about the same centre is equivalent to the rotation of (xo + yo) about the same centre.
A point or an object once rotated about a centre through a given angle can further be rotated about the different centre through another given angle.
Let R1 be the rotation through an angle xo about the centre and R2 be another rotation through an angle y0 about the different centre. Then, the equivalent transformation (R2oR1) is again a new rotation through the angle (xo + yo) about the third centre. This centre is the meeting point of the perpendicular bisectors of the line segments joining corresponding vertices of the object and its final image. In other words, perpendicular bisectors of the line segment joining corresponding vertices of the object and its final image are concurrent and the point of concurrence is the centre of new rotation ( i.e. combined rotation).
Let ABC be the given triangle (object). It is rotated to Δ A'B'C' through 90o about the centre P. The triangle A'B'C' is again rotated to ΔA''B''C'' through 90o about the centre Q. Then ΔA"B"C" is the image of ΔABC under the rotation through 180o about the third centre R. Perpendicular bisectors of AA", BB", CC" intersect each other at the point R.
A point or object once reflected can further be reflected to form a new image. The axes of these reflections may be parallel to each other or they intersect each other at a point. So, there are two cases for the combination of two reflections.
Let Ab and CD be parallel to each other. Let P' be the image of P under the reflection in the line AB. Then,
PM = MP' and so, PP' = 2 MP'.
Let P' be the image of P" under the reflection of the line CD. Then,
P'N = NP" and so, P'P" = 2P'N.
Here, P" is the image of P under the reflection in the line AB followed by the reflection in the line CD.
Now, \(\overrightarrow{P}{P"}\) = \(\overrightarrow{P}{P'}\) + \(\overrightarrow{P'}{P"}\) = \(\overrightarrow{M}{P'}\) + \(\overrightarrow{P'}{N}\) = 2( \(\overrightarrow{M}{P'} + \overrightarrow{P}'{N})\) = 2 \(\overrightarrow{M}{N}\)
Hence, if the axis of reflection is parallel, a reflection followed by another reflection is equivalent to the translation. The distance of the translation is twice the distance between the axes of reflections and the direction is perpendicular to the axis of reflections.
Let OA and OB intersect each other at the point O. Let P" be the image of P' under the reflection in the line OB. Then, OB is the perpendicular bisector of PP'.
∴∠POM =∠POM =θ (say)
Let P" is the image of P' under the reflection in the line OB. Then, OB is the perpendicular bisector of P'P".
∴∠P'ON =∠P"ON = Φ(say)
Here, P" be the image under the reflection in the line OA followed by the reflection in the line OB.
Now,
∠POP" =∠POM +∠P'OM +∠P'ON +∠P"ON = θ+θ +Φ +Φ = 2(θ +Φ)
= 2(∠P'OM + ∠P'ON) = 2∠MON.
Hence, if the axes of reflections intersect at a point O, then a reflection followed by another reflection is equivalent to a rotation about the centre O through the angle twice the angle between the axes of reflection. The direction of rotation is the direction from OA to OB (i.e. the direction from P to P").
A reflection followed by a reflection is equivalent to either a translation or a rotation.
Fig:- combination of two reflections over intersecting lines
In the following figure, O is the point of intersection of two straight lines OP and OQ. ΔA'B'C' is the image of ΔABC under the reflection in the line OP and ΔA"B"C" is the image of ΔA'B'C' under the reflection in the line OQ. In fact, ΔA"B"C" is the image of ΔABC under the rotation about O through an angle 2∠POQ.
Fig:- combination of two reflections over two parallel lines
In the following figure, AB and CD are two parallel straight lines. ΔP'Q'R' is the image of Δpqr under the reflection in the line AB and ΔP"Q"R" is the image of ΔP'Q'R'. In fact ΔP"Q"R" is the image of ΔPQR under a translation whose magnitude is twice the distance between AB and CD.
An object once enlarged can further be enlarged. Similarly, an object once reduced can further be reduced.
There are two cases in the combination of two enlargements:
Let E1(O, k) be the enlargement with centre O and the scale factor k.
Let A'B' be the image of the given line AB under the enlargement E1 with centre o and the scale factor k.
Then, A'B' =kAB.
Let E2[O, k'] be the enlargement with centre O and scale factor k'.
Let A"B" be the image of A'B' under the enlargement E2 with centre O and scale factor k'. Then A"B" = k'A'B'.
Now,
A"B" = k'A'B' = k'k AB.
∴ A"B" is the image of AB under an enlargement E(O, kk') with centre O and scale factor kk'.
Hence, an enlargement with centre O and scale factor k followed by another enlargement with the same centre O and scale factor k' is equivalent to an enlargement with the same centre and scale factor kk'.
Let E1 [O, k] be the enlargement with centre O and scale factor k. Let AB be the image of AB under enlargement E1 with centre O and scale factor k.
Then, A'B' = kAB.
Let E2 [O', k'] be the enlargement with centre O' and scale factor k'.
Let A"B" be the image of A'B' under enlargement E2 with centre O' and scale factor k'.
Then, A"B" = k'A'B'.
Now, A"B" = k'A'B' = k'kAB.
Join A"A' and B"B'. The intersect each other at the point O".
So, A"B" is the image of AB under an enlargement E[O, k'k] with the third centre O'' and scale factor kk'.
Hence, An enlargement with centre O and scale factor K followed by an enlargement with different centre O' and scale factor k' is equivalent to enlargement with the O" and scale factor KKK'. The centre of combined enlargement can be found by drawing.
An enlargement followed by another enlargement is equivalent to a new enlargement.
An object once reflected can further be rotated to get a new image. A reflection followed by a rotation is also a type of combination of two transformations.
Let R0 be the reflection in X-axis.
Then, R0: P' (x, -y)→ P"(y, x)
∴ R0.Re: P(x, y) → P"(y, x) ...............................(i)
Again, let R be the reflection in the line y = x.
Then, R:P (x, y) → P"(y, x) ......................................................(ii)
So, reflection in the X-axis followed by a rotation about origin through 90o is equivalent to the reflection in the line y = x.
Let Re be the reflection in Y-axis.
Then, Re: P(x, y)→ P'(-x, y).
Let R0 be the rotation about origin through 900.
Then, R0: P (x, y) → P" (-y, -x)
∴ R0. Re: p (x, y) = P"(-y, -x) .........(i)
Again, let R be the reflection in the line y = -x.
Then, R: P(x, y) → P"(-y, -x) ............................(ii)
So, reflection in Y-axis followed by a rotation about origin through 90o is equivalent to the reflection in the line y = -x.
Let Re be the reflection in te line y = x.
Then, Re: P (x, y)→ P'(y, x)
Let R0: P' (y, x)→ P" (-x, y) ....................(i)
Again, Let R be the reflection in the Y-axis.
Then, R:P (x, y) → P"(-x, y) .........................(ii)
Hence, reflection in the line y = x followed by a rotation about the origin through 90o is equivalent to the reflection in Y-axis.
Similarly, we ca establish the following results:
4. Reflection in X-axis followed by a rotation through - 90o about the origin is equivalent to the reflection in the line y = -x.
5. Reflection in Y-axis followed by a rotation through -90o about the origin is equivalent to the reflection in the line y = x.
6. Reflection in the line y = x followed by a rotation through -90o about the origin is equivalent to the reflection in X-axis.
A point or an object once rotated can further be reflected to get a new image. A rotation followed by a reflection is also a type of combination of transformation.
Let Ro be the rotation about origin through 90o. Then,
Ro: P(x, y)→ P'(-y, x).
Let Re be the reflection in the x-axis.
Then, Re: P' (-y, x) → P"(-y, -x)
∴ Re.R0: P (x, y) → P" (-y, -x) .....................(i)
Again, Let R be the reflection in the line y = -x.
Then, R: P (x, y) →P"(-y, -x) .............................(ii)
Hence, rotation about origin through 90o followed by a reflection in X-axis is equivalent to the reflection in the line y = -x.
Let Ro be the rotation about origin through 90o.
Then, Ro: P (x, y) → P'(-y, x).
Let Re be the reflection in the y-axis.
Then, Re: P' (-y, x) → P"(y, x).
∴ Re.Ro: P (x, y) → P"(y, x) ........................(i)
Again, let R be the reflection in the line y = x.
Then, R: P(x, y)→ P" (y, x) .................(ii)
So, rotation about origin through 90o followed by the reflection in Y-axis is equivalent to a reflection in the line y = x.
Let Ro be the rotation about origin through 900.
Then, Ro:P (x, y)→ P' (-y, x).
Let Re be the reflection in the line y = x.
Then,Re: P'(-y, x)→ P"(x, -y)
∴ Re.Ro: P(x,y)→ P"(x, -y) ............................................(i)
Again, let R be the reflection in X-axis.
Then, R:P (x, y)→ P"(x, -y) ...............................(ii)
So, rotation about origin through 90o followed by a reflection in the line y = x is equivalent to a reflection in X-axis.
As above, we can establish the following results:
4. A rotation about origin through - 90o followed by a reflection in X-axis is equivalent to a reflection in the line y = x.
5. A rotation about origin through -90o followed by a reflection in Y-axis is equivalent to a reflection in the line y = -x.
6. A rotation about origin through -90o followed by a reflection in the line y = x is equivalent to a reflection in the y-axis.
A point or an object once reflected can further be translated. Similarly, a point or an object once translated can further be reflected.
Let r be the reflection in X-axis and T =(\(\frac{a}{b}\)) be a translation.
Then, R:P (x, y)→ P'(x, -y) and
T: P' (x, -y)→ P"(x + a, -y + b).
ToR: P (x, y)→ P" (x + a, -y + b)
So, P"(x + a, -y + b) is the image of P (x, y) under the reflection in X-axis followed by translation T.
Again,T:P (x, y)→ P'(x + a, y + b) and
R:P' (x + a, y + b)→ P' (x + a, -y - b)
RoT: P (x, y) → P"(x + a, -y -b)
So, P" (x + a, -y - b) is image of P(x, y) under translation T followed by reflection R.
A point or object once rotated can further be translated. Similarly, a point or an object once translated can further be rotated.
Let R be the rotation through 90o about the origin and T = \(\begin{bmatrix}a\\b\\ \end{bmatrix}\) be a translation.
Then, R: P (x, y)→ P' (-y, x).
∴ T: P' (-y, x)→ P" (-y + a, x + b)
ToR: P(x, y) → P" (-y -b, x + a)
So, P' (-y -b, x + a) is image of P(x, y) under reflection R followed by translation T.
Again, T: P (x, y)→P' (x + a, y + b) and
R: P' (x + a, y + b)→ P" (-y -b, x + a)
∴ RoT: P (x, y)→ P"(-y -b, x + a).
So, P" (-y - b, x + a) is the image of P (x, y) under translation T followed by rotation R.
Let R be the reflection in X-axis and E [O, k] be the enlargement with centre O and scale factor k.
Then, R: P (x, y) → P'(x, -y) and
E:P' (x, -y) → P' (kx, -ky).
∴ EoR: P (x, y) → P'' (kx, -ky)
So, P" (kx, -ky) is image of p (x, y) under reflection r followed by enlargement E.
Again, E: P(x, y) → P'(kx, ky) and
R: P'(kx, ky) → P" (kx, -ky)
∴ RoE: P(x, y) → P"(kx, - ky).
So, P" (kx, -ky) is an image of p (x, y) under enlargement E followed by reflection R.
Let R be the rotation through 90o about origin and E[O, k] be the enlargement with centre O and scale factor k.
Then, R:P(x, y) → P'(-y, x) and
E:P'(-y, x)→ P"(-ky, kx)
∴EoR:P(x, y)→ P"(-ky, kx)
So, P" (-ky, kx) is an image of P (x, y) under rotation R followed by enlargement E.
Again, e:P(x, y)→ P' (kx, ky) and
R:P'(kx, ky)→ P"(-ky, kx)
RoE: p(x, y)→ p"(-ky, kx)
So, P"(-ky, kx)is an image of P(x, y) under enlargement E followed by rotation R.
Let T =\(\begin{bmatrix}a\\b\\ \end{bmatrix}\) be a translation and E [o, k] be an enlargement with centre O and scale factor k.
Then, T:P(x, y)→ P'(x + a, y + b)
E:P'(x + a, y + b)→ P"(kx + ka, ky + kb)
∴ EoT: P(x, y)→ P"(kx + ka, ky + kb)
So, P" (kx + ka, Ky + kb) is image of P (x, y) under translation T followed by enlargement E.
Again, E:P(x, y)→ P'(kx, ky) and
T:P'(kx, ky) → P"(kx + a, ky + b)
∴ToE: P(x, y)→ P" (kx + a, ky + b)
So, P"(kx + a, ky + b) is image of P(x, y) under enlargement e followed by translation T.
Transformations | Object point | Image point |
Reflection in X-axis (y = 0) | P(x, y) | P'(x, -y) |
Reflection in Y-axis(x = 0) | P(x, y) | P'(-x, y) |
Reflection in the line y = x | P(x, y) | P'(y, x) |
Reflection in the line y = -x | P(x, y) | P'(-y, -x) |
Reflection in the line x = k(parallel to Y-axis) | P(x, y) | P'(2k - x, y) |
Reflection in line y = k (parallel to X-axis) | P(x, y) | P'(x, 2k - y) |
Rotation about origin through 90o | P(x, y) | P'(-y, x) |
Rotation about origin through -90o | P(x, y) | P'(y, -x) |
Rotation about origin through 180o | P(x, y) | P'(-x, -y) |
TranslationT = \(\begin{bmatrix}a\\b\\ \end{bmatrix}\) | P (x, y) | P'(x + a, Y + b) |
enlargement with centre origin and scale factor k, E [(o, o), k] | P(x, y) | P'(kx, ky) |
Enlargement with centre (a, b) and scale factor k, E [(a, b), k] | P(x, y) | P'(kx - ka + a, ky - kb + b) |
Transformation | formula | Matrix |
Reflection on x-axis | (x , Y)→(x , Y) |
\begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix} |
Reflection on y - axis | (x , y)→(-x ,y) | \begin{pmatrix} -1 & 0 \\ 0 & 1 \\ \end{pmatrix} |
Find the coordinates of image of the point (6, -4) when it is first reflected on the x-axis and then translated by \(\begin{pmatrix}-2\\-1\\ \end{pmatrix}\).
Here,
Again,
Point (4, 5) is rotated about the origin O through +90o and the image so obtained is reflected on the y-axis. Find the coordinates of the image.
Here,
Again,
Find the coordinates of the image of the point (-4, 6) when it is first translated by \(\begin{pmatrix}6\\-4\\ \end{pmatrix}\) and then reflected on the X-axis.
Here,
Again,
If r1 is the reflection in y-axis and r2 is the rotation through -90o about the origin. Find the image of point P(-5, 8) under the combined transformation r1r2.
Here,
Again,
P' is the image of P(-3, 5) after a rotation about the origin through -90o and p'' is the image of p' after a reflection on the x-axis. Write down the coordinates of p' and p''.
Here,
Again,
Point (-9, 15) is rotated about the origin through -180o and the image so formed is reflected on the line y = x. Find the coordinates of the image so obtained.
Here,
Again,
A(2, 3), B(2, 6) and C(3, 4) are the vertices of ΔABC. If T1 =\(\begin{pmatrix}-2\\1\\ \end{pmatrix}\) and T2 =\(\begin{pmatrix}0\\2\\ \end{pmatrix}\) are two translation then find the image of ΔABC under the combined transformation T20T1.
Here,
T1 = \(\begin {pmatrix} -2\\ 1\\ \end{pmatrix}\), T2 = \(\begin {pmatrix} 0\\ 2\\ \end{pmatrix}\)
T2T1 =\(\begin {pmatrix} -2\\ 1\\ \end{pmatrix}\) +\(\begin {pmatrix} 0\\ 2\\ \end{pmatrix}\) =\(\begin {pmatrix} -2\\ 3\\ \end{pmatrix}\)
Under the combined translation T2T1 =\(\begin {pmatrix} -2\\ 3\\ \end{pmatrix}\)
Point (-2, 3)is rotated about the origin through 90o and the image so obtained is reflected on the x - axis. Find the coordinates of the image.
Here,
Again,
Point (3, -2) is rotated about the origin O through -90o and the image so formed is reflected on the x-axis. Find the coordinates of the image formed.
Here,
Again,
Point(3, 2) is reflected on the line y = x. The image so obtained is rotated about O through +90o. Find the coordinates of this image.
Here,
On a graph paper draw a triangle PQR having the vertices P(1, 4), Q(4, 1) and R(7, 5). Find the image of triangle PQR by stating coordinates and graphing them after successive reflection in y-axis followed by the reflection in the line x = y.
The co-ordinates of \(\triangle\)PQR are P(1, 4), Q(4, 1) and R(7, 5) when \(\triangle\)PQR reflection on the x - axis then the vertices of image are as follows:
Again,
When \(\triangle\)P'Q'R' reflection on the line x = y the vertices of the images are as follows:
The co-ordinates of the \(\triangle\)PQR, \(\triangle\)P'Q'R' and \(\triangle\)P"Q"R" are as shown in the graph paper as below:
Draw the figure having vertices O(0, 0), A(2, 0), B(3, 1) and C(1, 1) on a graph paper. It is translated by \(\begin{pmatrix}0\\2\\ \end{pmatrix}\) and plot the figure O' A' B' C' on the same graph paper. Then the figure O' A' B' C' is reflected on the line x = 3 to form the figure O'' A" B" C" on the same graph. Determine the vertices of O", A", B" and C".
Given points are: O(0, 0), A(2, 0), B(3, 1) and C(1, 1)
Again,
Plotting the above points in the graph paper:
A triangle with vertices A(1, 2), B(4, -1) and C(2, 5) is reflected successively in the line x = 5 and y = -2. Find by stating coordinates and graphically represent the images under these transformations. Also state the single transformation given by the combination of these transformations.
The co-ordinates of \(\triangle\)ABC are A(1, 2), B(4, -1) and C(2, 5) when \(\triangle\)ABC reflection in the line x = 5 then the vertices of the image are as follows:
Again,
When \(\triangle\)A'B'C' reflection on the line y = -2, then:
The co-ordinates of the \(\triangle\)ABC, \(\triangle\)A'B'C' and \(\triangle\)A"B"C" are shown in the graph paper as below:
The single transformation as half turn about (5, -2).
Draw the triangle MNP whose vertices are M(4, -2), N(2, 1) and P(5, 2). Rotate triangle MNP through clockwise 180o about (1, 1) and followed by 90o+ about (1, 1) and draw the image points on the graph.
The image of the point P(x, y) rotated half turn about (a, b)
The image of the point P(x, y) rotated 90º+ about (a, b) is:
∴ The final image of \(\triangle\)MNP, \(\triangle\)M'N'P' and \(\triangle\)M"N"P" in graph paper is:
Enlarge the ΔPQR with vertices P(-4, 6), Q(2, 8) and R(6, 2) under enlargement E[(2, 3), -2] and then reflect the image under the line y = x. Also draw the triangle ABC and its image on the same graph paper.
The co-ordinates of \(\triangle\)PQR areP(-4, 6), Q(2, 8) and R(6, 2). When \(\triangle\)PQR is enlarged on E[(2, 3), -2] then the vertices of the images are as follows:
Again,
When \(\triangle\)P'Q'R' reflection on the line y = x then, the vertices of the images are as follows:
The \(\triangle\)PQR, \(\triangle\)P'Q'R' and \(\triangle\)P"Q"R" on graph paper as follows:
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