vedanta Excel In Opt. Mathematics - Book 9 Vector
If two vectors oa and ob are two like vectors.
We can write oa = kob , or ob = moa , where k and m are positive scalars.
Example: Let oa = 4 and ob = 8
6 12
Then, we can write
ob = 8 =2 4 = 2oa
12 6
which is in the form of ob = moa , where, m = 2 > 0
Hence, oa and ob are like vectors.
Let T1 be direction of oa , tan T1 = 6 = 3
4 2
3
? T1 = tan–1 2
and T2 be the direction of CoD,
then, tan T2= 12 = 3
8 2
3
? T2 = tan–1 2
This shows that AoB and CoD are like parallel vectors.
(h) Equal Vectors Q
Let PoQ and oRS be two vectors such that S
PoQ = 2 and oRS = 2
3 3
Then, |PoQ| = 22 + 32 = 4 + 9 = 13 units P
Let direction of PoQ be T1 R
then, tan T1= 3
2
3
? T1 = tan–1 2
Also, |oRS| = 22 + 32 = 4+9= 13 units
Let direction of oRS be T2,
then, tan T2 = 3
2
? T2 = tan–1 3
2
Hence, PoQ and oRS have each magnitudes and directions. They are called equal vectors.
Definition: Two vectors oa and ob are called equal vectors if they have equal magnitude and
same direction.
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Then we write oa = ob .
Note :
If two vectors are equal then their corresponding components are equal.
(i) Unlike Vectors B
In the adjoint figure, two vectors AoB and CoD are two vectors C
which have opposite directions.
They are called unlike vectors or unlike parallel vectors.
Definition: Two vectors are said to be unlike if they have opposite A
direction whatever may be their magnitudes. If oa and ob are D
unlike vectors, we can write oa = kob or ob = moa
where m and k are scalars, k < 0 and m < 0.
Example: Let oa = 1 , ob = –2
2 –4
Then ob = –2 =–2 1 = – 2oa
–4 2
which is in the form of ob = moa , m = – 2 < 0.
Parallel Vectors
Two vectors are said to be parallel if they are either like or unlike vectors. So, two vectors
which have same direction or opposite direction whatever may be their magnitudes are
called parallel vectors.
If two vectors oa and ob are parallel, they can be expressed in the form of
oa = kob or, ob = moa
where k and m are scalars.
Note :
If oa = kob .
(i) Two vectors are called like parallel if k is positive.
(ii) Two vectors are called unlike parallel if k is negative.
Expression of a vector in the form of xoi + yoj . Y
Any vector oa = (x, y) or can be expressed in the form of
oa = xoi + yoj , where oi and oj are unit vectors along X-axis X' B(0, 1) X
and Y-axis respectively. O A(1, 0)
In the adjoint figure, Y'
OoA = unit vector along X-axis = oi
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OoB = unit vector along Y-axis = oj Q
Let oa = (4, 5) R
Then we can write, oa = 4oi + 5oj .
Co-initial Vector
In the adjoint figure, two vectors have the same initial point. They are
called co-initial vectors.
Definition: Two vectors having the same initial point are called co- P
initial vectors.
Example : PoQ and oPR co-initial vectors as they have the same initial point P.
Worked out Examples
Example 1. Show that the following vectors are unit vectors:
Solution:
1 1
Example 2. 0
Solution: (a) (b) 2
1
(a) Here, let oa = 1 2
0
|oa | = (x-component)2+(y-component)2 =
12 + 02 = 1
? oa is a unit vector.
Let ob = 1
(b) 2
1
2
|ob | = (x-component)2+(y-component)2
= 3 2 4 2 = 9 + 16
5 5 25 25
+
= 9 + 16 = 25 =1
25 25
? ob is a unit vector.
If P(4, 5) and Q(7, 9) are two points, then find PoQ and unit vector along PoQ.
Here, (x1, y1) = P(4, 5)
(x2, y2) = Q(7, 9)
Then, PoQ = x2 – x1 = 7–4 = 3
y2 – y1 9–5 4
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|PoQ| = (x-component)2+(y-component)2
= 32 + 42 = 9 + 16 = 25 = 5
Unit vector along PoQ is given by,
P^Q = PoQ = (3, 4) = 3 , 4
|PoQ| 5 4 5
PoQ PoQ xoi
Example 3. If displaces P(4, 6) to Q(8, –2), find and express it in the form of
Solution: + yoj . Also find unit vector along PoQ.
Example 4. Here, for PoQ, (x1, y1) = P(4, 6), (x2, y2) = (8, – 2)
Solution:
PoQ = x2 – x1 = 8–4 = 4
y2 – y1 –2–6 –8
Expressing PoQ in the form of xoi + yoj , we have
PoQ = 4oi + (–8oj ) = 4oi – 8oj
|PoQ| = 42 + (–8)2 = 16 + 64
= 80 = 4 5 units
Unit vector along PoQ is given by,
P^Q = PoQ = (4, –8)
|PoQ| 4 5
= 4 , – 8 = 1 , – 2
45 45 5 5
If oa = x, 4 is a unit vector, find the value of x.
5
Here, oa = 4
x, 5 is a unit vector.
? |oa | = 1
Now, |oa | = x2 + 42
5
or, 1= x2 + 16
25
Squaring on both sides, we get,
1 = x2 + 16
25
16
or, 1 – 25 = x2
or, x2 = 25 – 16
25
9
or, x2 = 25
? x = ± 3
5
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Example 5. If AoB displaces A(3, 3) to B(6, 7), and PoQ displaces P(4, 1) to (7, 5), prove that
Solution: AoB = PoQ.
Example 6. For AoB, A(3, 3) = (x1, y1), B(6, 7) = (x2, y2)
Solution:
AoB = x2 – x1 = 6–3 = 3
y2 – y1 7–3 4
Magnitude of AoB = |AoB| = 32 + 42 =
9 + 16 = 25 = 5 units
Direction of AoB, tanT1 = 4
3
4
? T1 = tan–1 3
For PoQ, P(4, 1) = (x1, y1)
Q(7, 5) = (x2, y2)
PoQ = x2 – x1 = 7–4 = 3
y2 – y1 5–1 4
Magnitude of PoQ = 32 + 42 = 9 + 16 =
25 = 5
Direction of PoQ, tanT2 = 4
3
4
? T2 = tan–1 3
Since |AoB| = |PoQ|
T1 = T2
i.e., the magnitudes and directions of AoB and PoQ are equal, they are equal vectors.
If P(3, 4), Q(–3, 6), R(5, 6) and S(x, y) and oRS = PoQ, find the values of x and y.
Here, for PoQ, P(3, 4) = (x1, y1) and Q(–3, 6) = (x2, y2)
PoQ = x2 – x1 = –3–3 = –6
y2 – y1 6–4 2
For oRS, R(5, 6) = (x1, y1), S(x, y) = (x2, y2)
oRS = x2 – x1 = x–5
y2 – y1 y–6
Now, PoQ = oRS
or, –6 = x–5
2 y–6
Equating the corresponding components of equal vectors,
we get,
–6=x–5 or, x = – 6 + 5 = – 1
and 2 = y – 6 or, y = 2 + 6 = 8
? S(x, y) = S(–1, 8)
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Example 7. If op = 6 and oq = 3 are two parallel vectors, find the value of a.
Solution: a –2
Here, op = 6 , oq = 3 are parallel vectors. It means that their direction is
same. a –2
They can be written as:
op = koq
or, 6 =k 3
a –2
or, 6 = 3k
a – 2k
Equating their corresponding components,
? 6 = 3k or, k = 2
and a = –2k = –2 × 2 = – 4
? a=–4
Alternative Method
For op , let its direction be T1.
tanT1 = y-component = a
x-component 6
For oq , let its direction be T2,
tanT2 = y-component = –2
x-component 3
Since op and oq vectors are parallel, their directions are same:
? tanT1 = tanT2
or, a = – 2
6 3
? a=–4
Exercise 12.2
Very Short Questions:
1. Define the following vectors:
(a) Unit vector (b) Position vector
(c) Like vector (d) Equal vector
(e) Parallel vectors (f) Null vector
2. Show that the following vectors are unit vectors:
(a) oa = 1 (b) ob = 0
0 1
(c) OoA = 1 , 1 (d) op = 4 , 7
2 2 65 65
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3. (a) If P(4, 5) and Q(4, 5), then show that PoQ is a null vector.
(b) If A(3, –5), then show that AoA is a null vector.
Short Questions:
4. (a) If AoB displaces a point A(4, –2) to B(5, 2). Find AoB, magnitude of AoB and a unit
vector along AoB.
(b) If PoQ displaces a point P(7, 8) to Q(–2, –1), find the unit vector along .
5. (a) If PoQ displaces a point P(2, 3) to Q(5, 7), find PoQ and express it in the form of
xoi + yoj .
(b) If MoN displaces a point M(1, 2) to N(6, 7), find MoN and express it in the form of
xoi + yoj .
6. (a) If x, 4 is a unit vector, find the value of x.
9
(b) If –1 , y is a unit vector find the value of y.
11 11
7. In the adjoint figure. AE//BC, BA//CD, BD//CE. Then answer the following questions.
A DE
(a) Write a vector parallel and equal to AoB.
(b) Which vector is equal to BoD?
(c) Write two vectors which are equal to BoC. B C
(d) Write two vectors which are negative of BoC.
(e) Which is negative vector of oCE?
(f) Write two pairs of like parallel vectors.
(g) Write two pairs of unlike parallel vectors.
(h) What type of vectors are AoD and BoC?
8. (a) If AoB displaces A(5, 9) to the point B(4, 6) and CD displaces C(2, 4) to D(x, y) such
that AoB = CoD, find the coordinates of D.
(b) If P, Q, R and S are four points with vertices P(2, –2), Q(6, 4), R(x, y) and S(3, –5),
such that PoQ = oRS, find the coordinates of R.
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9. Show that the following pair of vectors are parallel:
(a) op = 4 and oq = 8
5 10
(b) PoQ = 10 and oRS = –5
4 –2
(c) oa = –3 and ob = 1
9 –3
(d) MoN = –2 and oRS = 8
–3 12
10. Find the values of p when the pair of vectors are parallel:
(a) oa = 1 and ob = 4
p 8
(b) PoQ = –2 and oRS = 10
5 p
(c) oEF = p and GoH = –3
–3 5
(d) AoB = 3p and CoD = –6
4 –8
Project Work
11. Prepare a report on vector including the following points.
(a) Definition of vectors.
(b) Examples of vectors with their SI units.
(c) Application of vectors in daily life.
1 – 1
2
4. (a) 1 , 17, 17 (b) –9 ,9 2, 1
4 4 –9 – 2
17
5. (a) 3 , 3oi + 4oj (b) 5 , 5oi + 5oj
4 5
6. (a) ± 65 (b) ± 10
9 (b) oCE
7. (a) DoC (c) AoD, DoE (d) DoA, EoD
(e) DoB (f) BoA and CoD, BoD and oCE
(g) AoD and CoB, BoD and oEC. (h) like parallel and equal vectors
8. (a) (1, 1) (b) (–1, –11)
10. (a) 2 (b) –25 (c) 9 (d) 1
5
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12.3 Operations on Vectors
Review
Group discuss the following questions:
(a) What is difference between 2oa and oa ?
(b) Draw a figure to show the meaning of op and – op .
(c) If op = 4 and oq = 8 , comment on these vectors regarding their relations.
5 10
Operations on Vectors
(a) Multiplication of a vector by a scalar
A vector may be multiplied by a scalar.
Let oa be a given vector and k be any scalar, then the scalar multiple of k and oa is
koa . koa represents a vector k times as long as oa . Its magnitude is |k| |oa |.
In koa ,
If k > 0, k has the same direction of oa .
If k < 0, k has the opposite direction of oa . a
In the figure (i) oa = 2, 1 a
it means that oa is two times the ob . =2
Since k = 2, the two vectors are in the same direction. b
Fig. (i)
In the figure (ii), P p = -3m Q
Let MoN = om and PoQ = op M
Then, op = – 3om Nm
Here, op and om are parallel to each other and scalar k = –3. Fig. (ii)
Hence, op is 3 times greater than om but opposite in direction.
Examples:
(1) Let oa = 1 , then 3oa = 3 1 = 3×1 = 3
2 2 3×2 6
(2) Let op = 4 , then –2p = – 2 4 = –8
5 5 –10
(b) Addition of vectors
Let oa = x1 and ob = x2 be two vectors. Then their sum is denoted by oa + ob and
y1 y2
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it is given by
oa + ob = x1 + x2 = x1 + x2
y1 y2 y1 + y2
Triangle Law of Vector Addition
Let AoB displace point A to B and BoC displaces point B to C. Then, C
the displacement of A to C is given by AoC = AoB + BoC
This addition of two vectors is called triangle law of vector addition. B
Here, AoC is called resultant vector of AoB and BoC.
Definition: The triangle law of vector addition states that if two A
sides of a triangle are taken in order represent two vectors, then their resultant is given by
the third side of the triangle whose initial point is the initial point of the first vector and the
terminal point is the terminal point of the second vector.
Parallelogram Law of Vector Addition BC
Let OoA = oa and OoB = ob be two coinitial vectors with initial b A
point as O. A parallelogram OACB is completed taking OoA and O a
OoB as adjacent sides.
Then we have OA//BC, OB//AC and OoA = BoC = oa and OoB= AoC = ob .
Now, by using triangle law of vector addition, we have,
OoC = OoA + AoC = oa + ob (From 'OAC)
This law is called parallelogram law of vector addition.
Definition: The parallelogram law of vector addition states that if two adjacent sides of a
parallelogram represent two coinitial vectors, then the diagonal of the parallelogram passing
through same point gives its resultant.
Addition of vectors given in terms of components
Let OoA = oa = (x1, y1) and OoB = ob = (x2, y2) be two vectors. A parallelogram OACB is drawn
taking OA and OB as adjacent sides. From A, B, and C, draw AM, BN, and CL perpendiculars
Y
to OX. Again, draw AP perpendicular to CL. C(x1+x2, y1+y2)
Diagonal OC is drawn.
Then 'OBN # 'CAP, (By AAS axiom of congruency) B(x2, y2)
Also, ON = x2, OM = x1, BN = y2, AM = y1 X' O N P X
Now, BN = CP = y2, ML = AP = ON = x2, A(x1, y1)
AM = PL = y1
260 OL = OM + ML = x1 + x2 ML
CL = CP + PL = y2 + y1 = y1 + y2
Y'
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For OoC,
x-component of OoC = OL = x1 + x2
y-component of OoC = CL = y1 + y2
? OoC = x1 + x2
y1 + y2
By the parallelogram law of vector addition, we get
OoC = OoA + OoB = x1 + x2 = x1 + x2
y1 y2 y1 + y2
Polygon law of vector addition
Polygon law of vector addition is the generalized triangle law of vector addition. In polygon
ABCDEF (hexagon), AoB + BoC + CoD + DoE + oEF = AoF ED
Proof : FC
Let ABCDEF be a given polygon (hexagon).
Join AC, AD and AE. AB
Using triangle law of vector addition in triangles 'ABC, 'ACD, 'ADE, 'AEF, we get
from 'ABC
AoB + BoC = AoC ....................... (i)
from 'ACD
AoC + CoD = AoD ........................ (ii)
from (i) and (ii), we get
AoB + BoC + CoD = AoD ........................... (iii)
Again, from 'ADE,
AoD + DoE = AoE .............................. (iv)
from (iii) and (iv)
AoB + BoC + CoD + DoE = AoE
Again, from 'AEF,
AoE + oEF = AoF .......................... (v)
from (iv) and (v), we get
AoB + BoC + CoD + DoE + oEF = AoF proved.
It is also written as:
AoB + BoC + CoD + DoE + oEF + oFA = 0 ( AoF = – oFA)
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Hence, the polygon law of vector addition states that if a number of vectors are
represented by the sides of a polygon taken in order then the resultant vector is
represented by the closing side of the polygon in the opposite order.
Properties of Vector Addition
Let oa , ob and oc be any three vectors and k be any scalar. Then following properties of
vector addition hold.
(i) oa + ob = ob + oa (Commutative Property)
(ii) oa + ob and koa are vectors (Closure Property)
(iii) oa + (ob + oc ) = (oa + ob ) + oc (Associative Property)
(iv) oa + (– oa ) = O (Existence of additive inverse)
(v) k(oa + ob ) = koa + koa (Distributive Property)
(vi) oa + oO = oO + oa = oa (Existence of additive identify of vector addition)
(c) Subtraction or Difference of vectors.
Let oa = a1 and ob = b1 be two vectors then the subtraction of ob from oa is oa – ob
and iat 2is given by b2
oa – ob = a1 – b1 = a1 – b1
a1 b2 a2 – b2
Subtraction is the reverse operation of addition.
If AoB and BoC are two vectors then their sum is denoted by AoB + BoC, by triangle law of
vector addition:
AoB + BoC = AoC
The difference of AoB and BoC is AoB – BoC.
AoB – BoC = AoB + (–BoC)
In the adjoint figure, CB is produced to D such that CB = BD, BoD = CoB = – BoC.
Join AD, by triangle law of vector addition:
AoD = AoB + BoD = AoD – BoC
To show AoB = OoB – OoA b B
Let OoA = oa and OoB = ob in the figure. O A
By triangle law of vector addition, a
OoA + AoB = OoB
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or, AoB = OoB – OoA
? AoB = OoB – OoA or, AoB = ob – oa
Mid-point Theorem
Let OoA = oa and OoB = ob . Then the position vector of the mid-point of AB is given by
OoM = 1 (oa + ob ) B
2 OM
Proof : We have OoA = oa , OoB = ob
Join AB. M is the mid-point of AB. Then BoM = MoA
From triangle OMB, A
OoM = OoB + BoM .......................... (i)
From triangle OAM,
OoM = OoA + AoM .......................... (ii)
adding (i) and (ii), we get
2OoM = OoB + OoA + BoM + AoM
= ob + oa + BoM + (– MoA)
= oa + ob + (BoM – BoM) = oa + ob
? OoM = 1 (oa + ob ) Proved.
2
Worked out Examples
Example 1. Let oa = 1 and ob = 3 , find the following : (a) oa + ob
Solution: 2 4
(b) oa – ob (c) 3oa – 2ob
Here, oa = 1 and ob = 3 , then
2 4
(a) oa + ob = 1 + 3 = 1+3 = 4
2 4 2+4 6
(b) oa – ob = 1 – 3 = 1–3 = –2
2 4 2–4 –2
(c) 3oa – 2oa = 3 1 – 2 3 = 3 – 6
2 4 6 8
= 3–6 = –3
6–8 –2
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Example 2. If oa = 2 and ob = 4 , find oa + 2ob and express in the form of xoi + yoj .
Solution: 4 3
Example 3. Here, oa = 2 and ob = 4
Solution: 4 3
Example 4.
Solution: oa + 2ob = 2 +2 4 = 2 + 8 = 2+8 = 10
4 3 4 6 4+6 10
Example 5.
Solution: Expressing it in the form of xoi + yoj
oa + 2ob = 10oi + 10oj
From the adjoining figure prove that : PoQ = OoQ – OoP. O Q
In 'OPQ, using triangle law of vector addition, we get
P
OoP + PoQ = OoQ D
? PoQ = OoQ – OoP Proved.
C
In the adjoining figure, ABCDE is a pentagon prove that : B
AoB + BoC + CoD + DoE + oEF = 0 E
ABCDE is a pentagon. Join AC, AD.
Using triangle law of vector addition, we get,
AoB + BoC = AoC ...................... (i) A
AoC + CoD = AoD ............................. (ii)
AoD + DoE = AoE .............................. (iii)
From (i) and (ii), we get
AoB + BoC + CoD = AoD ........................ (iv)
Using (iii) and (iv), we get,
AoB + BoC + CoD + DoE = AoE [ AoE = – EoA]
? AoB + BoC + CoD + DoE + EoA = 0 Proved.
In the given figure, OoA = oa and OoB = ob . If AoC = 3AoB, find OoC.
Here, OoA = oa , OoB = ob O
Then, AoB = OoB – OoA = ob – oa
Also, AoC = OoC – OoA= OoC – oa ab C
Now, AoC = 3AoB AB
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or, OoC – OoA = 3(OoB – OoA)
or, OoC – oa = 3(ob – oa )
or, OoC = 3ob – 3oa + oa
? OoC = 3ob – 2oa
Example 6. In the adjoining figure, ABCD is a parallelogram. The diagonals intersect at
Solution: P and O is any point, then prove that : OoA + OoB + OoC + OoD = 4OoP
Example 7. Here, we use triangle law of vector addition. C
Solution:
From 'AOP, OoP = OoA + AoP ................ (i) D
From 'POD, OoP = OoD + DoP ............... (ii) O P
From 'POC, OoP = OoC + oCP ................ (iii) B
From 'POB, OoP = OoB + oBP ................ (iv)
A
Adding (i), (ii), (iii) and (iv), we get
4OoP = OoA + OoB + OoC + OoD + AoP + oCP + oBP + DoP
But AoP = – oCP, oBP = – DoP
Hence, AoP + oCP = AoP – AoP = 0
oBP + DoP = oBP – oBP = 0
? 4OoP = OoA + OoB + OoC + OoD Proved.
In the given figure, ABCDEF is a regular hexagon. Prove that
AoB + AoC + AoD + EoA + oFA = 4AoB E D
Let us join FC. C
Since ABCDEF is a regular hexagon, F B
EoD = AoB, oFC = 2AoB
In 'EAD, EoA + AoD = EoD A
In 'FAC, oFA + AoC = oFC
Now, LHS = AoB + AoC + AoD + EoA + oFA
= AoB + (EoA + AoD) + (oFA + AoC) = AoB + EoD + oFC
= AoB + AoB + 2AoB = 4AoB = RHS Proved.
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Exercise 12.3
Very Short Questions:
1. If oa = 4 , ob = 8 , find the following vectors.
6 4
1 oa 1 ob (c) 2oa (d) 3ob
(a) 2 (b) 4 (d) 2oa + 4ob
2. If oa = 1 and ob = 3 , find the following vectors.
2 4
(a) oa + ob (b) 2oa + 3ob (c) oa + 2ob
3. If oa = 4 and ob = 2 , find the following vectors:
5 3
(a) oa – ob (b) ob – oa
(c) 2oa – ob (d) 3oa – 2ob
4. If oa = 2 and ob = 4 , then find the following:
1 3
(a) |oa + ob | (b) |2oa – ob |
(c) |2oa – 3ob | (d) |ob | + |oa |
Short Questions:
5. (a) Find oa when ob = 4 and oa – ob = 10 .
6 6
(b) Find ob when oa = 3 and 3oa + ob = 6 .
6 8
6. (a) If oa = 2 and ob = –2 , find the unit vector along 2oa + ob and 2oa – 3ob.
3 3
(b) If OoA = oa = 3oi + 2oj and OoB = ob = 4oi – 3oj , find AoB and its magnitude.
(c) If OoP = 2oi + 3oj and OoQ = 3oi + 2oj , find PoQ and its magnitude.
7. If oa = (2, –1), ob = (4, – 3), oc = (4, 2) and k = 2, prove that :
(a) oa + ob = ob + oa (b) (oa + ob ) + oc = oa + (ob + oc )
(c) k(oa + ob ) = koa + kob
8. In the adjoint figure OoA = oa and OoB = ob are co-inital vectors, draw arrow diagram for
each of the following vectors:
(a) oa + ob
(b) ob – oa
(c) oa – ob
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(d) 2oa + ob
(e) 2oa – ob D C
Q
9. (a) ABCD is a quadrilateral and diagonals AC and BD Q
intersect at Q, find a single vector of
BoQ + QoC – BoA A B
R
(b) From the figure alongside, using vector addition, find
a single vector for,
(i) oPT + ToQ + QoS S
(ii) oPT + ToQ + QoS + oSR = oPR P T
(iii) Show that : PoQ + QoS = oPT + oTS O
Q
10. In the adjoint figure, if oPS= 1 PoQ, show that : S
2
OoP + OoQ = 2OoS
P
Long Questions: S
11. In the given figure PQRST is a pentagon. Show that : R
(i) PoQ + QoR + oRS + oST = oPT T
(ii) PoQ + QoR + oRS + oST + oTP = 0
PQ
12. PQRS is a quadrilateral. If oPR = op , SoQ = oq , QoP = on , oSP = om, oSR = os , QoR = or
express the following vectors as a single vector.
(a) on – om Pm S
(b) or – os
(c) op + oq – os n ps
(d) op – os + om
q R
Q
r
13. In a regular hexagon ABCDEF if AoB = oa , BoC = ob . Carry out the following operations.
(a) Express AoC, AoD and AoE in terms of oa and ob . Also show that : D
E
C
AoB + BoC + CoD + DoE + oEF + oFA = 0. b
B
(b) Express the remaining sides in terms of F a
oa and ob taken in order. A
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14. In the adjoining figure PQRS is a parallelogram, two diagonals oPR and QoS bisect at T
and O is any point then prove that OoP + OoQ + OoR + OoS = 4OoT
R
S
T A
Q P
O
P
15. In the adjoining figure, P, Q and R are the mid-point of R Q
BC, CA and AB respectively. Then prove that C
B
AoP + BoQ + CoR = 0
Project Work
16. Prepare a report on "Vector Addition" with their uses in our daily life.
1. (a) 2 (b) 2 (c) 8 (d) 24
3 1 12 12
2. (a) 4 (b) 11 (c) 7 (d) 14
6 16 10 20
3. (a) 2 (b) –2 (c) 6 (d) 8
2 –2 7 9
4. (a) 52 (b) 1 (c) 113 (d) 5 + 5
5. (a) 14 (b) –3
12 – 10
6. (a) 2 , 9 , 10 , –3 (b) oi – 5oj , 26 (c) oi – oj , 2
85 85 109 109
9. (a) AoC
12. (a) – oq (b) – oq (c) – on (d) 0
13. (a) AoC = oa + ob , AoD = 2ob , AoE = 2ob – oa
(b) CoD = ob – oa , DoE = – oa , EoF = – ob , FoA = oa – ob
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13Transformation
13.0 Review
Group discuss the following questions.
(a) What are the equations of X-axis and Y-axis?
(b) What is the meaning of line of symmetry?
(c) What type of image is formed when 'ABC is reflected on line x = 0?
(d) What type of image is formed when a triangle PQR is rotated through 90° in anticlockwise?
(e) What is the invariant point in transformation?
(f) Are images formed and objects congruent in reflection and rotation?
13.1 Introduction
A process in which an object changes its size, shape, A M A'
orientation, or position is called a transformation. A P B'
transformation forms images which are congruent B
or similar to given objects. Transformation is also R
called special type of mapping which maps each C C'
point of a plane with a definite point in the same
plane. In transformation, each point of a plane has Q
exactly one image point and each image point has N
exactly one pre-image point in the same plane.
Hence, transformation is a one to one mapping from
a plane onto itself.
If T is a transformation and A' is the image of point A under T,
it is denoted by T(A) = A' or T : A o A' or simply A o A'.
If another transformation brings the image A' back to the object (pre-image) point A, such a
transformation is called the inverse of the transformation T and it is denoted by T–1.
Transformation sometimes leaves certain points unchanged. These points are said to be
invariant. For example, in a rotations centre of rotation is invariant point. In a reflection the
P points on other features which are left unchanged by the transformation such as distance,
angle, parallelism, etc. are also described as invariant.
Types of Transformation
Transformation may or may not change the size of the given objects on the basis of it, the
transformation is divided into following two types.
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(a) Isometric Transformation
(b) Non-isometric Transformation
(a) Isometric Transformation
A transformation in which the size of an object does not change but the position of the
object changes is known as isometric transformation. In this type of transformation
the distance between any two points of a body is equal to the distance between their
corresponding images after transformation. Thus an isometric transformation is a
distance preserving transformation.
Reflection, translation, and rotation are examples of isometric transformation. Isometric
transformations are also called congruent transformations.
(b) Non-isometric Transformation
The transformation in which the size of an object changes is known as a non-isometric
transformation. In this type of transformation, the distance between any two points of
an object and the distance between their corresponding images after transformation
are unequal. Enlargement and reduction are examples of non-isometric transformation.
Non-isometric transformations are called similarity transformation.
Four Fundamental Transformations
The following are four fundamental transformations:
(a) Reflection (b) Rotation
(c) Translation (d) Enlargement (or reduction)
13.2 Reflection
In the adjoint figure 'ABC is reflected on the line MN and image 'A'B'C' is formed. The line
MN is called axis of reflection. From the figure, discuss the answer of the following:
(a) Measure the lengths AP, A'P, BQ, B'Q, CR, and C'R. M
Are AP = AP', BQ = B'Q, CR = C'R? A P A'
(b) Measure APM, BQN, CRN. Write the
conclusion.
(c) Are AB = A'B', BC = B'C' and CA = C'A'? B R B'
(d) Are 'ABC and 'A'B'C' congruent? C C'
Conclusions about reflection from the above figure.
Q
(a) The distances of the object point and the image N
point from the axis of reflection are equal.
i.e. AP = A'P, BP = B'P, CP = C'P.
(b) The line joining the same ends of the object and images are perpendicular to the
reflecting axis. Axis of reflection is the perpendicular bisector of the line segment
joining same end of the object and image. APM = BQN = CRN = 90°
(c) The distance between two vertices of object of equal to the distance between the
corresponding two image points.
i.e. AB = A'B', BC = B'C' and CA = C'A'
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(d) The object and image are congruents. 'ABC # 'A'B'C'.
Properties of Reflections
(a) The object and the image are equidistant from the axis of reflection.
(b) The straight line joining a point and its image is perpendicular to the axis of reflection.
(c) The line segments joining the vertices of object and their corresponding images are
parallel.
(d) The position of the figure changes but the shape and the size remains same after
reflection.
(e) The object is congruent to the image.
(f) The points of the axis of reflection are invariant.
To find the image of a geometrical figure. Q P R
A CN
Let PQR be a triangle whose image is to be found reflecting on B R'
the axis of reflection MN. Q'
(a) Draw perpendiculars PA, QB, and RC on the axis of
reflection of MN.
(b) Each perpendicular is produces to their corresponding M
image points such that PA = P'A, QB = Q'B, RC = R'C.
(c) Join P'A', Q'R' and R'P' with straight line segment to get
the image 'P'Q'R'.
Hence, 'P'Q'R' is the image of 'PQR. We write P'
'PQR o 'P'Q'R
'P'Q'R' is the required image of 'PQR. The image P'Q'R' is shaded in the figure.
Reflection Using Coordinates
Coordinates can be used for finding the images of geometrical figures after the reflection in
the lines like X-axis, Y-axis, lines parallel to x-axis, lines parallel to Y-axis, the line y = x, the
line y = –x, etc.
(a) Reflection on X-axis (or y = 0)
(i) Reflection points A(2, 3), B(4, 5), and C(6, 7) on X-axis. What do we get?
Plotting these points on the graph and reflecting them on the graph paper we get,
A(2, 3) Reflection X-axis A'(2, – 3)
B(4, 5) Reflection X-axis B'(4, –5)
C(9, 4) Reflection X-axis C'(9, – 4)
Hence, we conclude that- P'(x, –y)
P(x, y) Reflection X-axis
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Y
5 B(4, 5) C'(9, 4)
4 A(2, 3)
3
2
1 X
X' -4 -3 -2 -1-O1 1 2 3 4 5 6 7 8 9 10 11
-2
-3 A'(2, -3)
-4
C'(9, -4)
-5 B'(4, -5)
Y'
(ii) Let P(x, y) be a point on the plane reflected it on X-axis, the image P'(x', y') is obtained.
We have PM = MP'. Join PP' and let M(x, 0) be the mid-point of PP' then by using mid-
point formula, Y P(x, y)
x-coordinate of M = x + x'
2
x + x'
or, x = 2
M(x, 0) X
or, 2x = x + x' X' O
? x = x'
Again, y-coordinate of M = y + y'
2
y + y' Y' P'(x', y')
or, 0 = 2
or, 0 = y + y'
? y' = – y
Hence P'(x', y') = P'(x, –y) P'(x, –y)
? P(x, y) Reflection X-axis
Example : Find the image of P(4, 5) reflecting on X-axis.
Solution:
Here, P(4, 5)
For reflection on X-axis, P'(x, –y)
P(x, y) Reflection X-axis P'(4, –5)
? P(4, 5) Reflection X-axis
(b) Reflection on Y-axis (or x = 0)
(i) Reflect the points A(4, 5), B(8, 6) and C(–6, 8) on the Y-axis. Find their images.
Plotting there points on the graph paper and reflect them on Y-axis, we get,
A(4, 5) Reflection Y-axis A'(–4, 5)
B(8, 6) Reflection Y-axis B'(–8, 6)
C(–6, 8) Reflection Y-axis C'(6, 8)
? P(x, y) Reflection Y-axis P'(–x, y)
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Y
9
C(-6, 8) 8 C'(6, 8)
7
B(-8, 6) 6 B'(8, 6)
5 A(4, 5)
A'(-4, 5) 4
3
2
1
X' -9 -8 -7 -6 -5 -4 -3 -2 -1O 1 2 3 4 5 6 7 8 9 X
Y'
(ii) Let P(x, y) be a point on the plane. Reflect it on Y-axis, the image P'(x', y') is obtained.
Join PP'. Let M(0, y) be the mid-point of PP'. Y
Then, by using mid-point formula,
x-coordinate of M = x + x' M(0, y)
2 P(x, y)
x + x' P'(x', y')
2
or, 0 =
or, x + x' = 0 X' O X
? x' = –x
Again, y-coordinate of M = y + y'
2
y + y' Y'
2
or, y = or, 2y = y + y'
? y = y'
Hence, P(x', y') = P(–x, y) P'(–x, y)
P(x, y) Reflection Y-axis
Example : Find the image of P(6, –7) when reflected on Y-axis.
Solution: Here, P(6, –7)
For reflection on Y-axis, P'(–x, y)
P(x, y) Reflection Y-axis P'(–6, –7)
? P(6, –7) Reflection Y-axis
(c) Reflection on the line y = x (or x – y = 0) Y y=x
P'(x', y')
The line y = x bisects the angle between the axis
making an angle of 45° with X-axis in positive M
direction.
Let P(x, y) be a point on the plane. X' O P(x, y) X
Reflect it on the line y = x,
The image P'(x', y') is obtained.
Then slope of the line y = x is m1 = tan45° = 1 Y'
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Again, slope of PP'(m2) = y2 – y1 = y' – y
x2 – x1 x' – x
As the line PP' is perpendicular to the line y = x.
we write,
m1 . m2 = –1
or, 1 × y' – y = –1 or, –x' + x = y' – y
x' – x
or, x + y = x' + y' .................. (i)
Again, mid-point of PP' = M x + x' , y + y'
2 2
The mid-point lies on the line,
y=x
or, y + y' = x + x' or, x + x' = y + y'
2 2
or, x – y = –x' + y' ............ (ii)
Adding (i) and (ii), we get
2x = 2y'
? y' = x
putting y' = x in equation (ii), we get
x – y = – x' + x
or, x' = y
Hence, P'(x', y') = P'(y, x)
Example : Find the image of P(6, 4) reflecting it on line y = x.
Solution:
Here, P(6, 4)
For reflection on y = x,
P(x, y) P'(y, x)
? P(6, 4) P'(4, 6)
(d) Reflection on line y = –x (or x + y = 0) y = –x Y
The line y = –x is the line bisects the angle between P(x, y)
the axis making 135° with X-axis in positive direction.
Hence, slope of line y = – x is M 135° X
m1 = tan135° = –1 P'(x', y') O
X'
Let P(x, y) be a point on the plane.
Reflect it on the line y = –x and image P'(x', y') is Y'
obtained.
Join PP'. M is the mid-point of PP'
? M x + x' , y + y'
2 2
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M lies on the line y = – x
y + y' = –(x + x')
2 2
or, y + y' = – x – x'
or, x + y = – x' – y' ............. (i)
Again, slope of PP' (m2) = y' – y
x' – x
Since PP' is perpendicular to the line, y = –x
m1 . m2 = –1
or, (–1) × y' – y = – 1
x' – x
or, y – y' = – x' + x
or, – x + y = – x' + y' ............... (ii)
Adding (i) and (ii),
2y = –2x'
? x' = – y
Putting x' = – y in equation (ii), we get
– x + y = y + y'
or, – x = y'
? y' = – x
Hence P'(x', y') = P'(–y, –x)
Example : Find the image of P(4, 5) under reflection on line y = –x.
Solution: For reflection on y = – x
P(x, y) Reflection y = –x
P'(–y, –x)
? P(4, 5) Reflection y = –x P'(–5, –4)
(e) Reflection on line y = k (or parallel to X-axis) Y
Equation of a line parallel to X-axis is y = k, P'(x', y')
where k is the y-intercept of the line.
Hence, reflection on the line parallel to X-axis is y=k
reflection on y = k. Take P(x, y) a point the plane. M(x, k)
Reflect P(x, y) on the line y = k, the image P(x, y)
P'(x', y') is obtained.
Join PP'. Let M be the mid-point of PP' whose X' O X
coordinates are (x, k).
As M(x, k) is the mid-point of PP' Y'
? x-coordinate of M = x + x'
2
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or, x = x + x'
2
or, 2x = x + x'
? x' = x ................... (i)
Again, y-coordinate of M = y + y'
2
y + y'
or, k = 2
or, 2k = y + y'
? y' = 2k – y P'(x, 2k – y)
Hence P(x, y) Reflection y = k
Example : If P(7, 5) is reflected on line y = 4, find the image of P.
Solution:
Here, axis of reflection is y = 4
As, P(x, y) Reflection y = k P'(x, 2k – y)
Now, P(7, 5) Reflection y = 4 P'(7, 2 × 4 – 5) = P'(7, 3)
(f) Reflection on the line x = h (parallel to Y-axis)
Let equation of a line parallel to Y-axis be x = h. Y x=h
Take point P(x, y) on the plane. Reflect it on the P'(x', y') M(h, y)
line, the image P'(x', y') is obtained. Join PP'. Let
M be the mid-point of PP', the coordinates of M O P(x, y)
are (h, y). Y'
X
As M is the mid-point of PP', M x + x' , y + y'
2 2
x + x' X'
x-coordinate of M = 2
or, h = x + x'
2
or, 2h = x + x'
? x' = 2h – x
y-coordinate of M = y + y'
2
y + y'
or, y = 2 or, 2y = y + y'
? y' = y P'(2h – x, y)
P'(2h – x, y)
Hence P(x, y) Reflection x = h
? P(x, y) Reflection x = h
Example : Find the image of P(3, 5) under reflection on the line x = 2.
Solution:
Under reflection on x = h,
P(x, y) Reflection x = h P'(2h – x, y)
P'(2 × 2 – 3, 5) = P'(1, 5)
Now, P(3, 5) Reflection x = 2
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Worked Out Examples
Example 1. Draw the image of 'ABC on the axis L.
C
A
B C' C
L
Solution: 'ABC is reflected on the line L.
Example 2.
Solution: we get image 'A'B'C'. A A'
Example 3. Here, the shaded triangle A'B'C' is B B'
Solution: the image of triangle ABC. L
Find the image of the following points under given conditions.
(a) A(4, 3) reflected on X-axis (b) B(6, 5) reflected on Y-axis
(c) C(1, 3) reflected on line y = – x (d) D(2, 3) reflected on line x = 1
(a) Here, A(4, 3)
For reflection on X-axis, we have, P'(x, –y)
P(x, y) Reflection X-axis A'(4, –3)
? A(4, 3) Reflection X-axis
(b) For reflection on Y-axis, we have,
P(x, y) Reflection Y-axis P'(–x, y)
? B(6, 5) Reflection Y-axis B'(–6, 5)
(c) For reflection on y = –x, we have,
P(x, y) Reflection y = –x P'(–y, –x)
? C(1, 3) Reflection y = –x C'(–3, –1)
(d) For reflection on x = 1, we have, P'(2h – x, 1)
P(x, y) Reflection x = h D'(2 × 1 – 2, 3) = D'(0, 3)
? D(2, 3) Reflection x = 1
The vertices of 'ABC are A(2, 3), B(4, 2), and C(7, 8). Write down the
coordinates of the image vertices A', B', C' when reflected on X-axis. Also
draw 'ABC and 'A'B'C' on the same graph.
For the reflection on the X-axis.
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P(x, y) Reflection X-axis P'(x, –y)
Now, A(2, 3) Reflection X-axis A'(2, –3)
Reflection X-axis B'(4, –2)
B(4, 2) Reflection X-axis C'(6, –6)
C(6, 6) Reflection X-axis 'A'B'C'
'ABC
'ABC and its image 'A'B'C' are plotted in the graph and image triangle is
shaded.
Y
7
6 C
5
4 A
3
2 B
1
X' -4 -3 -2 -1-O1 1 2 3 4 5 6 7 8 9 10 11 X
-2 B'
-3 A'
-4
-5 C'
-6
-7
Y'
Example 4. Find the axis of reflection when,
Solution:
(a) The image of A(4, 5) is (4, –5)
(b) The image of B(5, 4) is (5, –6)
(a) we have,
A(4, 5) A'(4, – 5)
and P(x, y) Reflection X-axis P'(x, –y)
A'(4, –5)
A(4, 5) Reflection X-axis
Hence, the required axis of reflection is X-axis.
(b) We have,
B(5, 4) B'(5, –6)
By given condition of B and B', x-coordinate of B and B' are same.
Hence it is the reflection on y = k B'(5, 2k – 4)
Now, B(5, 4) Reflection y = k
But, B'(5, – 6)
? (5, 2k – 4) = (5, – 6)
or, 2k – 4 = – 6
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or, 2k = – 2 Image
? k=–1 P'(x, –y)
Hence, the required axis of reflection is y = –1. P'(–x, y)
P'(y, x)
Formulae for reflection: Object P'(–y, –x)
P(x, y) P'(2h – x, y)
SN Axis of Reflection P(x, y) P'(x, 2k – y)
1. Reflection on X-axis or y = 0 P(x, y)
2. Reflection on Y-axis or x = 0 P(x, y)
3. Reflection on line y = x or x – y = 0 P(x, y)
4. Reflection on line y = – x or x + y = 0 P(x, y)
5. Reflection on line x = h
6. Reflection on line y = k
Exercise 13.1
Very Short Questions:
1. Write down the coordinates of image of the given points in the following:
(a) A(4, 7) is reflected on X-axis.
(b) B(4, 2) is reflected on Y-axis.
(c) C(6, 7) is reflected on line x + y = 0.
(d) D(7, 2) is reflected on the line x = 2.
(e) E(4, 5) is reflected on the line x – y = 0.
(f) F(–6, –4) is reflected on the line y = –2.
(g) G(–2, –3) is reflected on the line y = – x.
2. (a) What is the image of P(2, 0) when it is reflected on X-axis?
(b) What is the image of R(0, 5) when it is reflected on Y-axis?
Are above points invariant in reflection?
Short Questions:
3. Reflect the points P(–3, –5) and Q(5, 6) under the following axis of reflections:
(a) y = 3 (b) y = –2
(c) x = 3 (d) y = 5
4. (a) If the image of A is A'(6, –7) under the reflection on Y-axis, find the coordinates of A.
(b) If the image of P is P'(4, –5) under the reflection on X-axis, find the coordinates of B.
(c) If the image of Q is Q'(–6, 2) under the reflection in the line x = 2, find the
coordinates of Q.
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5. Draw the image of the following figures reflecting on the line l. R
(a) l (b) Sl
P P
R
R Q
Q
(c) l (d) l
O
M MN
N
Long Questions:
6. (a) Reflect the points P(4, –3) and Q(7, 2) on the line which passes through the origin
and point (4, 4). State the coordinates of the images of P and Q.
(b) Reflect the points M(2, –2) and N(3, 4) on the line which passes through the origin
and the point (–4, –4). State the coordinates of the image of M and N.
(c) Let l1 be the line passing through the point origin and bisecting the angle between
the axes making an angle of 45° with X-axis in positive direction. Find the images
of P(2, 4) and Q(3, 6) reflecting on line l1.
(d) Let l2 be the line passing through the origin and bisecting the angle between the
axis making an angle of 135° with X-axis in positive direction. Find the image of
points A(–3, –4) and B(3, 5) reflecting them on the line l2.
7. (a) Let P(3, 5), Q(–2, –5) and R(2, 2) be the vertices of 'PQR. The 'PQR is reflected
on X-axis. Find the coordinates of its image after the reflection and show both the
object and the image in the same graph.
(b) A triangle formed by joining the points P(0, –4), Q(–3, 1), and R(2, 6) is reflected
in the line x = 0 (or Y-axis). Find the coordinates of the image and hence present
both the image and the object on the same graph.
(c) Determine the vertices of 'P'Q'R' formed when 'PQR with the vertices P(2, 4),
Q(2, 5) and R(6, 3) is reflected on the line y = x. Also draw both the triangles on
the same graph.
(d) A triangle PQR with the vertices P(–1, 4), Q(1, 2) and R(6, 6) is reflected on the
line y = –x. Determine the coordinates of 'P'Q'R', the image of 'PQR. Draw both
triangles on the same graph.
8. (a) Determine the vertices of image of 'P'Q'R' formed when 'PQR with vertices
P(6, 3), Q(–3, 5) and R(4, –2) is reflected on the line x = 3. Also draw both of the
triangles on the same graph.
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(b) Determine the vertices of image 'A'B'C' formed when 'ABC with the vertices
A(–1, 8), B(7, 6) and C(4, 1) is reflected on the line x + 2 = 0. Also draw both
triangles on the same graph.
(c) Determine the vertices of 'A'B'C' formed when 'ABC with vertices A(–5, 1),
B(–3, 1) and C(–5, 3) is reflected on the line y = 3. Also draw both triangles on the
same graph paper.
(d) Determine the vertices of image 'A'B'C' formed when 'ABC with vertices A(–2, 8),
B(8, 6), and C(3, 4) is reflected on the line y + 2 = 0. Also draw both triangles on
the same graph.
9. (a) The image of P(m, 5 – n) when reflected on the line y = 0 is P'(2m – 3, 2n – 8). Find
the values of m and n.
(b) The image of M(2, 6 – p) when reflected in Y-axis is M'(4p – 6, q + 2). Find the
values of p and q.
(c) The image of P(3m, 4n) when reflected in the line y = x is P'(5n – 2, 4m – 3). Find
the values of m and n.
(d) The image of P(3 – m, 4 – n) when reflected on the line y = –x is P'(2n – 6, 2m – 5).
Find the values of m and n.
10. Find the axis of reflection if-
(a) P(3, 5) is reflected to P'(3, –5). (b) M(2, 4) is reflected to M'(4, 2).
(c) R(2, 1) is reflected to R'(2, 5). (d) A(4, 5) is reflected to A'(2, 5).
(e) T(2, 5) is reflected to T'(–5, –2). (f) N(2, 3) is reflected to N'(2, 7).
Project Work
11. Prepare a report on "use of reflections in our daily life". Present the report in your class.
1. (a) A'(4, –7) (b) B'(–4, 2) (c) C'(–7, –6) (d) D'(–3, 2)
(e) (5, 4) (f) (–6, 0) (g) (3, 2)
2. (a) P'(2, 0) (b) R'(0, 5) 3. (a) P'(–3, 11), Q'(5, 0)
(b) P'(–3, 1), Q'(5, – 10) (c) P'(9, –5), Q'(1, 6) (d) P'(–3, 15), Q'(–3, –2)
4. (a) (–6, –7) (b) P(4, 5) (c) Q(10, 2) 6.(a) P'(–3, 4), Q'(2, 7)
(b) M'(–2, 2), N'(4, 3) (c) P'(4, 2), Q'(6, 3) (d) A'(4, 3), B'(–5, –3)
7. (a) P'(3, –5), Q'(–2, 5), R'(2, –2) (b) P'(0, –4), Q'(3, 1), R'(–2, 6)
(c) P'(4, 2), Q'(5, 2), R'(3, 6) (d) P'(–4, 1), Q'(–2, –1), R'(–6, –6)
8. (a) P'(0, 3), Q'(9, 5), C'(2, –2) (b) A'(–3, 8), B'(–11, 6), C'(–8, 1)
(c) A'(–5, 5), B'(–3, 5), C'(–5, 3) (d) A'(–2, –12), B'(8, –10), C'(3, –8)
9. (a) m = 1, n = 3 (b) p = 1, q = 3 (c) m = 3, n = 2 (d) m = 2, n = 2
10. (a) X-axis (b) y = x (c) y = 3 (d) x = 3
(e) y = –x (f) y = 5
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13.2 Rotation
Let us take two points A(4, 5) and B(6, 8). Y
Rotate them through +90° about the origin,
find their images. 9
8 B
A
After rotation through +90° about the origin, B' 7
we get, 6
5
A' 4
A(4, 5) A'(–5, 4) 3
2
B(6, 8) B'(–8, 6) X' 1 X
Hence, we can write -9 -8 -7 -6 -5 -4 -3 -2 -1O 1 2 3 4 5 6 7 8 9
P(x, y) P'(–y, x) Y'
Here, A' and B' are the images of A and B due to rotation of +90° about the origin.
Definition: Rotation is a rule which shifts each point of the given object through an equal
angular displacement about a point in the given direction. In a rotation, the centre of rotation
is invariant.
The following conditions are required to define a rotation.
(i) The fixed point about which it is to be turned or rotated.
(ii) The angle through which it is to be rotated, i.e., the angle of rotation.
(iii) The direction about which it is to be rotated.
The rotation may be positive or negative. If an object is rotated through an angle in anti-
clockwise direction, it is called positive rotation. If an object is rotated through an angle in
clockwise direction, it is called negative rotation.
Geometrical Rotation
Process of Rotation of an object in a given direction about given centre. C
Let us consider a triangle ABC which is to be rotated
through +90° about the centre O. Then, we go through
the following processes:
(i) Join OA, measure the length OA. Take OA as a B'
radius, rotate the point A is anticlockwise direction A B
O
such that AOA' = 90° about centre O, A' is the
A'
image of A. C'
(ii) Join OB, measure the length OB. Take OB as a radius
and rotate the point B is anti-clockwise direction.
Such that BOB' = 90° about centre O, B' is the
image of B.
(iii) Join OC, measure the length OC. Take OC as a radius and rotate the point C in anti-
clockwise direction such that COC' = 90° about centre O, C' is the image of C.
(iv) Join A', B' and C'
Hence, 'A'B'C' is the image of 'ABC under rotation of +90° about the origin.
Are 'ABC and 'A'B'C' congruent?
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Properties of Rotation
The following are the properties of rotation:
(i) A rotation transforms all the points of the geometric figures on a plane in the same
direction and the same angular displacement.
(ii) The image and the object are congruent.
(iii) The perpendicular bisectors of the line joining the image point and object point pass
through the centre of rotation.
(iv) The distance of object point from the centre of rotation is equal to that of image point
from the centre of rotation.
(v) In rotation, the centre of rotation is invariant.
(vi) The rotation about of a point through an angle T in positive direction is equivalent to
that of rotation through the angle (360° – T) in the negative direction.
(vii) If the rotation is through 0°, then it takes identity transformation.
Process of finding centre and angle of rotation:
If 'A'B'C' is the image of 'ABC under A' B
rotation of an angle T in the given direction,
we have to find the centre of rotation, B'
angle of rotation. The following steps are
followed:
(i) Join AA', BB', and CC'. C' A
C
(ii) Find the perpendicular bisectors of
any two of the line segments AA', BB', O
and CC'.
(iii) The perpendicular bisectors meet at a point: let it be O. This point O is the centre of
rotation.
(iv) Find AOA', BOB', COC' which are the angle of rotation.
(v) If AOA' is in anti-clockwise direction, the direction is positive; otherwise, negative.
In the above diagram, the angle of rotation is positive, i.e. AOA' = +90°.
Use of Coordinates in Rotation
If the vertices of a geometric figure are given in coordinates, they can be the rotated through
some specific angles such as 90°, –90°, 180°, 270°, –270°, etc.
(a) Rotation through 90° about origin in anti-clockwise direction (Positive Quarter Turn):
R[(0, 0), O]
Let us take a point P(4, 5) which is to be rotated through 90° in anticlockwise direction
about O.
Rotate the point P(4, 5) through positive 90° about origin, we get,
image P'(–5, 4). i.e. P(4, 5) R[(0, 0), +90°] P'(–5, 4).
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Table Y
Points on the Corresponding image 7
plane under R[(0, 0), +90°]
P(4, 5) 6
Q(–6, 7) P'(–5, 4)
R(2, 5) Q'(–7, –6) P'(-5, 4) 5 P(4, 5)
S(6, 5) R'(–5, 2) 4
P(x, y) S'(–5, 6)
P'(–y, x) 3
2
1
X' -8 -7 -6 -5 -4 -3 -2 -1O 1 2 3 4 5 6 X
Y'
Hence, P(x, y) Rotation [(0, 0), +90°] P'(–y, x)
Note :
Rotation of +90° about origin is equivalent to rotation of –270° about the same centre O.
It is denoted by R[(0, 0), +90°]
P(x, y) Rotation [(0, 0), –270°] P'(–y, x)
(b) Rotation through 90° about the origin in clockwise direction (Negative Quarter Turn):
R[(0, 0), –90°]
Let us take a point P(4, 5). Rotation the point through 90° in clockwise about the origin
O, we get the image P'(5, –4). It is denoted by R[(0, 0), –90°]
Hence, P(4, 5) Rotation [(0, 0), –90°] P'(5, –4) Y
Table
5 P(4, 5)
Points on the Corresponding image 4
plane under R[(0, 0), –90°] 3
2
P(4, 5) P'(5, –4)
1 X
Q(2, 5) Q'(5, –2) X' -2 -1-O1 1 2 3 4 5 6 7 8
R(6, 5) R'(5, –6) -2
S(7, 2) S'(2, –7) -3
-4 P'(5, -4)
Y'
P(x, y) P'(y, –x)
Hence P(x, y) Rotation [(0, 0), –90°] P'(y, –x)
Note :
Rotation of –90° about origin is equivalent to rotation of 270° about the same centre O. It
is denoted by R[(0, 0), 270°]
P(x, y) Rotation [(0, 0), 270°] P'(y, –x)
(c) Rotation through 180° about origin (Half Turn) : R[(0, 0), 180°]
Let us take a point P(4, 5) on the plane. Rotate the point P through 180° about the origin,
we get the image P'(–4, –5). It is denoted by R[(0, 0), 180°]
P(4, 5) Rotation [(0, 0), 180°] P'(–4, –5)
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Table Corresponding image X' 6 Y X
under R[(0, 0), 180°] 5 P(4, 5)
Points on P'(–x, –y) 4
the plane P'(–4, –5) 3 123456
Q'(–2, –3) 2
P(4, 5) 1 Y'
Q(2, 3) R'(4, 5)
R(–4, –5) S'(5, 6) -6 -5 -4 -3 -2 -1-O1
S(–5, –6) P'(–x, –y) -2
P(x, y) Rotation [(0, 0), 180°] -3
-4
Hence P(x, y)
P'(-4, -5) -5
Formulae for Rotation:
SN Angle of Centre of Object Rotation [(0, 0), T] Image
Rotation Rotation
O(0, 0)
1. +90° or –270° P(x, y) Rotation [(0, 0), +90°] P'(–y, x)
O(0, 0) P(x, y) Rotation [(0, 0), –90°] P'(y, –x)
2. –90° or 270° P(x, y) Rotation [(0, 0), 180°] P'(–x, –y)
O(0, 0) P(x, y) Rotation [(a, b), +90°] P'(–y+a+b, x–a+b)
3. 180° P(x, y) Rotation [(a, b), –90°] P'(y+a–b, –x+a+b)
(a, b) P(x, y) Rotation [(a, b), 180°] P'(2a – x, 2b – y)
4. +90° or –270°
(a, b)
5. –90° or 270°
(a, b)
6. 180°
Worked out Examples
Example 1. Find the image of the point A(5, 6), B(7, 8), C(–5, –6), D(2, –5) under the
Solution: rotation through the following angles about the origin.
(a) 90° (b) –90° (c) 180° (d) 270° (e) –270°
(a) Under the rotation through +90° about origin,
P(x, y) R[(0, 0), +90°] P'(–y, x)
A(5, 6) R[(0, 0), +90°] A'(–6, 5)
B(7, 8) R[(0, 0), +90°] B'(–8, 7)
C(–5, –6) R[(0, 0), +90°] C'(6, –5)
D(2, –5) R[(0, 0), +90°] D'(5, 2)
(b) Under the rotation through –90° about origin,
P(x, y) R[(0, 0), –90°] P'(y, –x)
A(5, 6) R[(0, 0), –90°] A'(6, –5)
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B(7, 8) R[(0, 0), –90°] B'(8, –7)
C(–5, –6) R[(0, 0), –90°] C'(–6, 5)
D(2, –5) R[(0, 0), –90°] D'(–5, –2)
(c) Under the rotation through 180° about origin,
P(x, y) R[(0, 0), 180°] P'(–x, –y)
A(5, 6) R[(0, 0), 180°] A'(–5, –6)
B(7, 8) R[(0, 0), 180°] B'(–7, –8)
C(–5, –6) R[(0, 0), 180°] C'(5, 6)
D(2, –5) R[(0, 0), 180°] D'(–2, 5)
(d) Under the rotation through 270° about origin,
P(x, y) R[(0, 0), 270°] P'(y, –x)
A(5, 6) R[(0, 0), 270°] A'(6, –5)
B(7, 8) R[(0, 0), 270°] B'(8, –7)
C(–5, –6) R[(0, 0), 270°] C'(–6, 5)
D(2, –5) R[(0, 0), 270°] D'(–5, –2)
(e) Under the rotation through –270° about origin,
P(x, y) R[(0, 0), –270°] P'(–y, x)
A(5, 6) R[(0, 0), –270°] A'(–6, 5)
B(7, 8) R[(0, 0), –270°] B'(–8, 7)
C(–5, –6) R[(0, 0), –270°] C'(6, –5)
D(2, –5) R[(0, 0), –270°] D'(5, 2)
Example 2. If P(4, 1), Q(5, –1) and R(6, 4) are the vertices of 'PQR. Find the coordinates
Solution: of its image under the rotation through +90° about origin. Show both object
and image in the same graph.
Here, P(4, 1), Q(5, –1) and R(6, 4) are the vertices of 'PQR.
Under the rotation of +90° about origin,
P(x, y) R[(0, 0), +90°] P'(–y, x)
Now, P(4, 1) R[(0, 0), +90°] P'(–1, 4)
Q(5, –1) R[(0, 0), +90°] Q'(1, 5)
R(6, 4) R[(0, 0), +90°] R(–4, 6)
? 'PQR 'P'Q'R'
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Y
R'(-4, 6) 7
6
Q'(1, 5)
5 R(6, 4)
P'(-1, 4) 4
3
2
1 P(4, 1)
X' X
-7 -6 -5 -4 -3 -2 -1-O1 12345678
-2 Q(5, -1)
Y'
Example 3. In the adjoining figure, the line segment PQ is rotated about a point and
Solution: image P'Q' is formed. Find the centre of rotation and angle of rotation.
Example 4. To find the centre and angle of rotation,
Solution:
(i) Join PP' and QQ' P' C A
Q
(ii) Draw the perpendicular
bisectors AB and CD of PP'
and QQ' respectively. Q'
(iii) The point of intersections P
of AB and CD will be the
centre of rotation. O
BD
(iv) Join OP and OP', OQ and
OQ'.
In above figure, O is the centre of rotation and POP' = +90°, which is angle
of rotation.
Let P(5, 2), Q(3, 1), and R (2, –4) be the vertices of 'PQR. 'PQR is rotated
through +90° about origin and image 'P'Q'R' is formed. Again the 'P'Q'R'
is rotated through 180° about the origin in the positive direction and image
'P"Q"R" is formed. Draw 'PQR, 'P'Q'R', and 'P"Q"R" on the same graph.
Rotating the 'PQR through +90° about origin, we have,
P(x, y) R[(0, 0), +90°] P'(–y, x)
Now, P(5, 2) R[(0, 0), +90°] P'(–2, 5)
Q(3, 1) R[(0, 0), +90°] Q'(–1, 3)
R(2, –4) R[(0, 0), +90°] R'(4, 2)
Again rotating 'P'Q'R' through +180° about origin, we get,
P(x, y) R[(0, 0), 180°] P'(–x, –y)
Now, P'(–2, 5) R[(0, 0), 180°] P"(2, –5)
Q'(–1, 3) R[(0, 0), 180°] Q"(1, –3)
R'(4, 2) R[(0, 0), 180°] R"(–4, –2)
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? 'PQR 'P'Q'R' 'P"Q"R"
'PPQR, 'P'Q'R' and 'P"Q"R" are plotted in the graph.
Y
7
P' 6
5
4
Q' 3 R' P
2 Q
1
X' X
-7 -6 -5 -4 -3 -2 -1-O1 12345678
R" -2 Q"
-3
-4 R
-5 P"
Y'
Exercise 13.2
Very Short Questions:
1. (a) Define rotation.
(b) State any two properties of rotation.
(c) State invariant point in rotation.
(d) Find the image of each of given points rotating about origin through the following
angles: A(2, 3), B(4, 6), C(5, 2)
(i) +90° (ii) –90° (iii) 180°
(iv) +270° (v) –270°
Short Questions:
2. Rotate the given objects about O and angle of rotation indicated below:
(a) P (b) S R
Q R P Q
(c) O (180°) O N
(+90°)
(d) M
O (270°) RP
288 O (180°)
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(e) O
(–270°)
3. (a) Find the image of P(4, 5) when it is rotated through +90° about (1, 2).
(b) Find the image of A(2, 4) when it is rotated through –90° about the origin.
(c) Find the image of B(4, 3) when it is rotated through 180° about centre (1, 3).
(d) Find the image of Q(–6, –2) when it is rotated through +90° about (1, 2).
4. If Q denotes a positive quarter turn and H denotes a half turn about origin. Then find
the image of A(5, 4) due to the following:
(a) Q(A) (b) H(A) (c) Q2(A) (d) H2(A)
Also, verify that H(A) = Q2(A).
5. Find the angle of rotation when centre of rotation is at origin in the following transformation.
(a) A(2, 3) A'(–3, 2) (b) B(3, 4) B'(–3, –4)
(c) C(2, 4) C'(4, –2) (d) D(2, 5) D'(–5, 2)
6. (a) If A(2, –3) is mapped into A'(–2, 3) under a rotation about origin. What is the image
of P(4, 5), under the same rotation about the same centre?
(b) If B(6, 7) is mapped into B'(–6, –7) under a rotation about origin, what is the image
of D(2, 5), under the same rotation about the same centre?
(c) If P(2, 4) is mapped into P'(–4, 2) under a rotation about origin, what is the image
of Q(3, 6), under the same rotation about the same centre?
7. (a) If (p, 4) is mapped into (q, –3) under a rotation through +90° about origin, then
find the values of p and q?
(b) If (4, x) is mapped into (y, 4) under a rotation through 180° about origin, find the
values of x and y?
8. (a) If A(4 + p, 10) is rotated through +90° about origin, its image is A'(q – 5, 2), find
the values of p and q?
(b) If B(6 – p, 3) is rotated through 180° about origin, its image is (–4, q), find the
values of p and q?
Long Questions:
9. (a) A(1, 1), B(4, 1) and C(3, 5) are the vertices of 'ABC. The 'ABC is rotated through
an angle +90° to get image 'A'B'C' about origin. Determine the coordinates of A',
B' and C', and plot 'ABC, 'A'B'C' in the same graph.
(b) M(2, 5), N(–3, 3) and P(1, –4) are the vertices of 'MNP. Find the coordinates of
image vertices of the triangle under the rotation –90° about origin. Also draw the
graphs of this transformation.
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(c) A triangle with vertices A(2, 3), B(1, 5) and C(–2, 5) is rotated through 180° about
origin to get its image 'A'B'C'. Find the coordinates of A', B' and C'. Also draw the
graphs of this transformation.
(d) P(3, 1), Q(4, –2), and R(5, 3) are the vertices of 'ABC. Find the coordinates of its
image under rotation through 90° in anti-clockwise direction about origin. Show
both object and its image in the same graph.
10. (a) A triangle with vertices A(3, 4), B(3, –4), and C(1, 1) is rotated through an angle
T° about (p, q) to get the image, A'(–4, 3), B'(4, 3), and C(–1, 1). Find the centre of
rotation and the angle of rotation using graph.
(b) A triangle with vertices P(2, 6), Q(–2, 3) and R(1, 2) is rotated through T° about
(a, b) to get the image P'(–2, –6), Q'(2, –3) and R'(–1, –2). Find the centre and the
angle of rotation using graph.
11. Rotate the following triangles through 180° about centre (1, 2) and write down the
coordinates of image. Also plot both object and image triangles on the same graph.
(a) A(4, 4), B(5, 3) and C(2, 1) (b) P(4, 5), Q(2, –5) and R(–1, –2)
12. (a) Rotate the triangle ABC with vertices A(2, 3), B(6, 1) and C(4, –3) is rotated through
+90° about the origin to get image 'A'B'C', again 'A'B'C' is rotated through +180°
about the same centre to get 'A"B"C". Find the coordinates of vertices of 'A'B'C'
and 'A"B"C". Plot 'ABC, 'A'B'C' and 'A"B"C" on the same graph.
(b) Rotate the triangle ABC with vertices A(1, 1), B(3, 0) and C(2, 4) through +180°
about origin to get image 'A'B'C' and again 'A'B'C' is rotated through +90° about
the same centre to get 'A"B"C". Draw graph of 'ABC, 'A'B'C' and 'A"B"C" on the
same graph.
Project Work
13. Compare the properties in between reflection and rotation. List their applications in
our daily life. Let any point P(x, y) be reflected on X-axis and the image so obtained is
again reflected on Y-axis. Is the final image obtained due to there combined reflection
is same as it is rotated through 180° about origin. Illustrate it with graphical work.
1. (d) (i) A'(–3, 2), B'(–6, 4), C'(–2, 5) (ii) A'(3, –2), B'(6, –4), C'(2, –5)
(iii) A'(–2, –3), B'(–4, –6), C'(–5, –2) (iv) A'(3, –2), B'(6, –4), C'(2, –5)
(v) A'(–3, 2), B'(–6, 4), C'(–2, 5) 3.(a) (–2, 5) (b) (4, –2)
(c) (–2, 1) (d) (–5, –5) 4.(a) (–4, 5) (b) (–5, –4)
(c) (–5, –4) (d) (5, 4) 5.(a) +90° (b) 180°
(c) –90° (d) +90° 6.(a) (–4, –5) (b) (–2, –5)
(c) (–6, 3) 7.(a) p = –3, q = –4 (b) x = –4, y = -4 8.(a) p = –2, q = –5
(b) p = 2, q = –3 9.(a) A'(–1, 1), B'(–1, 4), C'(–5, 3) (b) M'(5, –2), N'(3, 3), P'(–4, –1)
(c) A'(–2, –3), B'(–1, –5), C'(2, –5) (d) P'(1, –3), Q'(–2, –4), R'(3, –5)
10. (a) T = +90°, (p, q) = (0, 0) (b) T = 180°, (a, b) = (0, 0)
11. (a) A'(–2, 0), B'(–3, 1), C'(0, 3) (b) P'(–2, –1), Q'(0, 9), R'(3, 6)
12. (a) A'(–3, 2), B'(–1, 6), C'(3, 4), A"(3, –2), B"(1, –6), C"(–3, –4)
(b) A'(–1, –1), B'(–3, 0), C'(–2, –4), A"(1, –1), B"(0, –3), C"(4, –2)
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13.3 Translation
Let ABC be a triangle. Let it be shifted to the 'A'B'C' such that AA' = BB' = CC' and
AA'//BB'//CC'. The displacement along the direction of AA', BB' or CC' is called translation
vector. It is denoted by T or ot .
Here, the triangle A'B'C' so obtained is the image of 'ABC due to translation. Each point
of 'ABC has translation an equal distance in the same A'
direction. Corresponding to each point on 'ABC, there A
is a unique point on 'A'B'C'. Hence, translation is a kind
of function.
Measure if AB, BC, AC, A'B', B'C', A'C' are AB = A'B', B' C'
BC = B'C', AC = A'C', is 'ABC congruent to 'A'B'C'? B C
Definition: Translation is a rule which shifts every point of the object through a definite
distance in a definite direction. It is the displacement of a geometrical figure so that only its
position relative to fixed axes is changed.
Properties of Translation
(i) The object and its image are congruent, i.e. 'ABC # 'A'B'C'.
(ii) Lines joining the corresponding points of the object and image are equal and parallel.
(iii) The direction and magnitude of the displacements are given by the direction and
magnitude of translation vector ot = a
b
where, a = horizontal displacement (along X-axis)
b = vertical displacement (along Y-axis)
Use of Coordinates in Translation P'(x', y') P'(x', y')
A' Y
5 b ot = a b
b P(x, y) a
A4 P
ya
ob = 4 X' Ox X
5
Y'
Let, ot = AoA' = 4 . What does it means?
5
Here, AoA' = 4 means that x-component and y-component of the displacement vector
AoA' are 5 units and 5 units.
respectively 4
Let, P(x, y) be the initial position of the object and ot = a be the translation vector.
b
If P(x', y') be the image of P due to the translation such that
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oPP' = ot = a . Join OP'
b
Then by triangle law of vector addition,
OoP' = OoP + oPP'
or, x' = x + a
y' y b
or, x' = x+a
y' y+b
? x = x + a and y' = y + b, are called equations of translation.
P(x, y) P'(x + a, y + b)
Hence, ot = displacement vector = translation vector = a
b
Worked Out Examples
Example 1. Find the image of each of the following points under translation ot = 2 .
Solution: 3
(a) A(4, 5) (b) B(2, –3) (c) C(2, 7)
Example 2.
Here, translation vector ot = 2
3
we have,
T= a
b
P(x, y) P'(x + a, y + b)
T= 2
3
(a) A(4, 5) A'(4, + 2, 5 + 3) = A'(6, 8)
T= 2
3
(b) B(2, –3) B'(2 + 2, – 3 + 3) = B'(4, 0)
T= 2
3
(c) C(2, 7) C'(2 + 2, 7 + 3) = C'(4, 10)
Translate the following geometrical figures in magnitude and direction of
the given vector ot . Find their translated image.
A AD
ot
C BC ot
B
Solution: (a) To translate 'ABC under translation vector ot
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(i) From A, draw parallel and equal to translation vector ot .
i.e. ot = AoA'. A'
(ii) Similarly from B and C draw parallel A
and equal to ot such that
BoB' = CoC' = ot . B' C'
ot
(iii) Join A', B' and C' to get the image C
'A'B'C' as shown in the figure. B
? 'ABC 'A'B'C'
Hence 'A'B'C' is the translated image of 'ABC.
(b) To translate parm ABCD under translation vector ot .
(i) From A, draw AoA' = ot . A' D'
(ii) Similarly from B, C, D draw A D
parallel and equal to ot such B' C'
that :
AoA' = BoB' = CoC' = DoD' = ot C ot
(iii) Join A', B', C' and D' to get B
translated image parm A'B'C'D'.
? Parm ABCD parm A'B'C'D'
Hence, parm A'B'C'D' is the translated image of parm A'B'C'D'.
Example 3. If a point P(5, 6) is translated to P'(11, 8), find the translation vector. Use the
Solution: translation vector to translate Q(2, 3).
Let translation vector be T = a
b
a
T= b
Then, P(x, y) P(x + a, y + b)
T= a
b
Then, (5, 6) P'(5 + a, 6 + b)
? (5 + a, 6 + b) = (11, 8)
Equating the corresponding components, we get,
5 + a = 11 and 6 + b = 8
? a=6 and b = 2
? T= a = 6
b 2
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T= 6
2
Now, Q(2, 3) Q'(2 + 6, 3 + 2) = Q'(8, 5)
Example 4. A(–3, –3), B(–2, 3), and C(1, 1) are the vertices of 'ABC. Translated 'ABC to
'A'B'C' under translation vector AoB. Plot 'ABC and 'A'B'C' on the same graph.
Solution: Here, A(–3, –3), B(–2, 3) and C(1, 1) are the vertices of 'ABC.
AoB = x2 – x1 = –2+3 = 1
y2 – y1 3+3 6
Now, translating 'ABC with translation vector AoB = 1
6
a
T= b
P(x, y) P'(x + a, y + b)
T= 1
6
A(–3, –3) A'(–3 + 1, –3 + 6) = A'(–2, 3)
T= 1
6
B(–2, 3) B'(–2 + 1, 3 + 6) = B'(–1, 9)
T= 1
6
C(1, 1) C'(1 + 1, 1 + 6) = C'(2, 7)
? 'ABC 'A'B'C'
Hence, 'A'B'C' is the image of 'ABC.
'ABC and 'A'B'C' are shown in the graph.
Y
9 C'
B' 8 C
7
6
5
A' 4
3
B2
1
X' -7 -6 -5 -4 -3 -2 -1-O1 1 2 3 4 5 6 7 8 X
-2
A -3
Y'
Exercise 13.3
Very Short Questions:
1. (a) Define translation.
(b) State any two properties of translation.
(c) Find the image of each of the following points under translation T = 2 :
3
(i) A(4, 5) (ii) B(6, 4) (iii) C(2, 5)
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(d) If A(4, 5) and B(7, 8), find the translation vector AoB. S
(e) If P(1, 2) and Q(–3, –5), find PoQ.
Short Questions:
2. Translate the following figure by translation vector ot :
(a) A (b) P
ot ot R
BC Q
(c) M (d)
ot
ot
NP
3. (a) If P(4, 5), Q(–2, 3), and R(–2, –2) are the vertices of 'PQR, translate it by
4
T= 5 . Find the image vertices P', Q' and R'.
(b) If A(1, 2), B(4, 5) and C(–3, 4) are the vertices of 'ABC, find the images of A, B, and
4
C under the translation T = 6 .
4. If A(4, 5), B(3, 7) and C(3, –2), find:
(a) the image of a when it is translated by AoB.
(b) the image of B when it is translated by BoC.
(c ) the image of C when it is translated by CoA.
6. (a) If a point P(4, –6) is translated to P'(6, 8) by T. What is the image of Q(–2, –3) by
the same translation?
(b) If a point P(7, 8) is translated to P'(4, 3). What is the image of R(5, 3) by the same
translation?
(c) If a point M(6, 7) is translated by a translation vector to M'(2, 3), what is the image
of N(5, 2) by the same translation?
Long Questions:
7. (a) A(2, 3), B(–2, –3), and C(1, 1) are the vertices of 'ABC. Find the coordinates of
image of triangle A'B'C' which are obtained after translating it by a translation
3
vector T = 4 . Present 'ABC and 'A'B'C' on the same graph.
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(b) A(2, 3), B(6, 3), C(6, 8), and D(2, 8) are the vertices of a parallelogram ABCD.
b43otthotgheet
Each vertex of ABCD is translated by T = image parallelogram A'B'C'D'.
Find the coordinates of the image and plot object and image on the same
graph.
8. (a) A triangle PQR having vertices P(3, 2), Q(–2, 4) and R(5, –2) is translated by oPR.
State the coordinates of the image vertices. Draw both 'PQR and 'P'Q'R' on the
same graph.
(b) A parallelogram having vertices P(3, 2), Q(11, 2), R(14, 7), and S(6, 7) is translated
by its diagonal oPR in its magnitude and direction. Find the coordinates of image
parallelogram. Draw both parallelogram on the same graph.
9. (a) If the image of P(2, p) under the translation vector 2 is P'(2q, 2p + 1). Find the
values of p and q. 3
(b) If the image of P(p, q) under the translation vector 2 is P'(2p – 4, 2q – 3), find the
values of p and q. 3
Project Work
10. State the properties of translation. List the daily use of translation and discuss in the
classroom.
1. (c) (i) A'(6, 8) (ii) B'(8, 7) (iii) C'(4, 8) (d) 3 (e) –4
3 –7
2. Show to your teacher 3.(a) P'(8, 10), Q'(2, 8), R'(2, 3) (b) A'(5, 8), B'(8, 11), C'(1, 10)
4. (a) A'(3, 7), B'(2, 9), C'(2, 0) (b) A'(4, –4), B'(3, –2), C'(3, –11)
(c) A'(5, 12), B'(4, 14), C'(4, 5) 6.(a) Q'(0, 11) (b) R'(2, –2) (c) N'(1, –2)
7. (a) A'(5, 7), B'(1, 1), C'(4, 5) (b) A'(5, 7), B'(9, 7), C'(9, 12), D'(5, 12)
8. (a) P'(5, –2), Q'(0, 0), R'(7, –6) (b) P'(14, 7), Q'(22, 7), R'(25, 12), S'(17, 12)
9. (a) p = 2, q = 2 (b) p = 6, q = 6
13.4 Enlargement
Let PQRS be a square and O be the origin. From O, the S' 6 Y R'
vertices P, Q, R and S are joined. OP, OQ, OR, and OS
are produced to P', Q', R' and S' respectively such that 5
OP' = 3.OP, OQ' = 3.OQ, OR' = 3.OR, OS' = 3.OS. Joining
P', Q', R', and S', we get a new square P'B'R'S'. 4
3
S 2 R
1
Here, P'Q'R'S' is the image of square PQRS with centre at X' -6 -5 -4 -3 -2 -1-O1 1 2Q3 4 5 6 X
O. O is called centre of enlargement. P -2
Also, OP' = OQ' = OR' = OS' = 3, this fixed ratio is -3
OP OQ OR OS
-4
called scale factor of enlargement. -5
P' -6 Y' Q'
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We say that the square PQRS is enlarged about centre O with scale factor 3. Scale factor is
denoted by k. It is denoted by E[(0, 0), k]
Square PQRS E[(0, 0), 3] Sequare P'Q'R'S'
The nature of image depends on the value of scalar factor k. The size of object and image
are different in size but they are similar. This type of transformation is called non-isometric
transformation.
Definition: Enlargement (or reduction) is a geometrical transformation which transfers an
object into an object of the same form by enlarging (or reducing) its size in a given ratio.
An enlargement requires the following two factors:
(i) the centre of enlargement.
(ii) the scale factor by which the object is enlarged (or reduced)
Scale Factor
The constant ratio of distance of image to the distance of object from the centre of the
enlargement is called a scale factor of the enlargement. It is denoted by k.
k = distance of image point from centre = OP'
where, distance of object point from centre OP
P = object point
P' = image point
If the centre of enlargement is O and the scale factor k, then E[O, k] represents the enlargement
of an object about centre O with scale factor k.
Properties of Enlargement
The following are the important properties of enlargement:
(i) If k > 1, the given object is enlarged. It means the image size is greater than the object
and the object and image lie in the same side of the centre of enlargement.
(ii) If 0 < k < 1, the object is reduced. The object and the image lie in the same side of the
centre of enlargement.
(iii) If k < 0 (i.e. k is negative), the object and the image lie on the opposite side of the centre
of enlargement. In this case, the image is inverted.
(iv) If k = 1, the object and the image are coincident.
(v) If k = –1, the object and the image are congruent. In this case, the image is equivalent
to half turn about the centre of enlargement.
(vi) The centre of enlargement is invariant.
(vii) The object and the image are similar.
(viii)Under an enlargement, the ratio of corresponding sides of image and image object are
proportional.
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Examples:
1. E[0, 2], k = 2 i.e. k > 1
A'
A
O B B'
C
C'
2. E 0, 1 , k = 12, k < 1, 0 < k < 1
2 A
A'
O B' B
C' A
C
3. E[0, –1], k = –1, k < 0, k is negative.
C'
B' O B
C
A' C'
C
How to find centre and scale factor:
Let 'A'B'C' be the image of 'ABC under an enlargement
as shown in the diagram.
Join AA', BB' and CC'. Let these line segments intersect O B
at a point O (produce them if necessary). The point of
intersection O is called centre enlargement. A B'
For scale factor, measure OA' and OA, OB' and OB, OC' A'
,and OC, then the scale factor k is given by
k = OA' = OB' = OC'
OA OB OC
Note :
1. If the image and the object are in the same direction of centre of enlargement, scale
factor is positive.
2. If the image and the object are in the opposite direction of the centre of enlargement,
scale factor in negative.
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Use of Coordinates in Enlargement Y B'
10
(a) Enlargement about centre origin and scale factor. 9
Let A(1, 2), B(4, 5) and C(2, –1) be the vertices of 8
7
'ABC. Enlarge the ABC by scale factor about centre 6
origin. Join OA . OB and OC. Produce OA, OB, 5 A' B
and OC such that AoA' = 2 . OoA', OoB'= 2OoB and 4
3
OoC = 2 . OC. Then find the coordinates of A', B', 2 A
1
and C'. we get from graph that : A'(2, 4), B'(8, 10) and X' -2 -1-1O 123456789 X
C'(4, –2) respectively. -2 C
We can write above coordinates using symbol of C'
Y'
enlargement.
$ E[O, 2] A'(2, 4)
B(4, 5) E[O, 2] B'(8, 10)
C(2, –1) E[O, 2] C'(4, –2)
Here, 'A'B'C' is the image of 'ABC about centre O and scale factor 2.
Hence, P(x, y) E[O, k] P'(kx, ky)
(b) Enlargement about centre other than origin
Let C(a, b) the centre of enlargement and k, the scale scale factor. Let P(x, y) be a point
on the plane. Let P'(x', y') be the image of P under enlargement E[(a, b), k]. Join OC, OP
and OP'. Then, OoC = a , OoP = x , OoP' = x'
b y y'
oCP = OoP – OoC = (x – a, y – b)
CoP' = OoP' – OoC = (x' – a, y' – b)
But, CoP' = koCP
x' – a =k x–a
y' – b y–b
or, x' – a = kx – ka
y' – b ky – kb
Equating the corresponding elements, we get
or, x' – a = kx – ka and y' – b = ky – kb
or, x' = k(x – a) + a y' = k(y – b) + b
P(x', y') = P'[k(x – a) + a k(y – b) + b]
Hence, P(x, y) E[(a, b), k] [k(x – a) + a, k(y – b) + b]
Hence, formulae of enlargements are tabulated below:
Centre of enlargement Scale factor Object Image
0 k
k P(x, y) P(kx, ky)
(a, b)
P(x, y) P'(k(x – a) + a, k(y – b) + b)
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Worked out Examples
Example 1. Find the images of A(2, 4) and B(3, –1) under the following condition:
Solution:
(a) E[0, 2] (b) E[0, –2] (c) E 0, 1
Example 2. Since we have, P(x, y) 2
Solution: E[0, 2]
Example 3. P'(kx, ky)
Solution:
(a) A(2, 4) E[0, 2] A'(2 × 2, 2 × 4) = A'(4, 8)
Example 4.
Solution: B(3, –1) E[0, 2] B'(2 × 3, 2 × (–1)) = B'(6, –2)
(b) A(2, 4) E[0, –2] A'(–2 × 2, 2 × 4) = A'(–4, 8)
B(3, –2) E[0, –2] B'(–2 × 3, –2 × (–2)) = B'(–6, –4)
0, 1 1 1
2 2 2
(c) A(2, 4) A' × 2, × 4 = A'(1, 2)
0, 1 1 1 3
2 2 2 2
B(3, –2) B' × 3, × (–2) = B' , –1
Find the images of P(1, 2) and Q(–1, 2) under enlargement E[(–2, 1), 3].
Since we have, P'[k(x – a) + a, k(y – b) + b]
P(x, y) E[(a, b), k] P'[3(1 + 2) + 2, 3(2 – 1) + 1] = P'(7, 4)
Now, P(1, 2) E[(–2, 1), 3] Q' [3(–1 + 2)+(–2), 3(2 – 1)+1] = P'(1, 4)
Q'(–1, 2) E[(–2, 1), 3]
The vertices of a triangle ABC are A(1, 4), B(5, 3) and C(4, 5). Find the
coordinates of the vertices of image of 'ABC under enlargement with centre
at origin and scale factor 2.
Here, A(1, 4), B(5, 3) and C(4, 5) are the vertices of 'ABC. 'ABC is enlarged
about centre origin and scale factor 2.
We have, P(x, y) E[O, k] P'(kx, ky)
Now, A(1, 4) E[O, 2] A'(2 × 1, 2 × 4) = A'(2, 8)
B(5, 3) E[O, 2] B'(2 × 5, 2 × 3) = B'(10, 6)
C(4, 5) E[O, 2] C'(2 × 4, 2 × 5) = C'(8, 10)
? A'(2, 8), B'(10, 6) and C'(8, 10) are the vertices of image 'A'B'C'.
P(2, 3), Q(4, 5), and R(7, 0) are the vertices of 'PQR. Find the coordinates
of image of 'PQR with centre at (1, 1) and scale factor 2. Present 'PQR and
'P'Q'R' on the same graph.
Here, centre of enlargement = (1, 1)
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