5. If the vector PQ displaces a point P(– 1, – 2) to Q(3, – 5) then,
(a) Express PQ in the column vector. (b) Calculate the length of PQ.
(c) Find the direction of PQ. (d) Find the unit vector along the direction of PQ.
6. Find the column vector MN in each of the following pair of points. Also, find their magnitude,
direction and unit vector along MN.
(a) M(2, 4) and N(7, 2) (b) N(4, – 5) and M(– 3, – 2) (c) N(2, 0) and M(– 4, – 2)
7. (a) If AB displaces A(2, 3) to B(– 2, 7) and PQ displaces P(– 2, 4) to Q(2, 0), prove that AB and
PQ are negative vectors to each other.
(b) If LM displaces L(2, 5) to M(– 2, 7) and KN displaces K(3, – 2) to N(– 3, 1), show that LM
and KN are like vectors.
8. (a) Prove that the vectors AB and PQ are parallel vectors where AB displaces A(– 1, 1) to B(2, 2)
and PQ displaces P(– 1, – 2) to Q(5, 0).
(b) Show that the vectors KL and LM are collinear if KL displaces K(– 4, –3) to L(2, –1) and LM
displaces L(2, –1) to M(8, 1).
9. In the adjoining figure, ABCD and BCED are two parallelograms standing on the same base BC and
between the same parallel lines AE and BC. Give answer of the following questions:
(a) Which vector is equal to BD ? AD E
(b) What are the vectors that are equal to BC ?
(c) Which is the negative vector of AB ? BC
(d) Which vectors are equal but negative to AD ? A
10. In the given figure, ABC is an equilateral triangle and P, Q and PR
R are the midpoints of its sides AB, BC and AC respectively. If
VECTOR
AB = 2 c , BC = 2 a and PQ = b , find QR , CA , CB and BP
in terms of a , b and c .
ANSWERS BQ C
3. (a) 2 (b) 0 (c) 7 (d) 4 (e) – 2
0 –3 –3 2 –5
4. (a) 1 (b) – 2/ 5 (c) 3/5 (d) – 4/5 (e) 12/13
0 1/ 5 – 4/5 – 3/5 – 5/13
5. (a) 4 (b) 5 units (c) 323.13o (d) 4/5
–3 – 3/5
6. (a) 5 , 29 units, 338.20o, 5/ 29 (b) 7 , 58 units, 336.80o, 7/ 58
–2 – 2/ 29 –3 – 3/ 58
(c) 6 , 2 10 units, 18.43o, 3/ 10
2 1/ 10
9. (a) CE (b) AD, DE (c) CD (d) CB or ED 10. c , – 2b , – 2a , – c
Introduction to Vector 251
61.21 OPERATIONS ON VECTORS
6.2 (A) Vector Addition
Learning Objectives
At the end of this topic, the students will be able to:
• add the vectors. • prove the properties of vector addition.
The vectors can be added, subtracted or multiplied with one another. These a + b= c
addition, subtraction and multiplication processes between the vectors are ab
called the operations of vectors. In this section we study the addition of vectors,
subtraction of vectors and multiplication of a vector by a scalar.
Addition of Vectors
If a and b are two vectors, then the addition of a and a+b=c b
b is denoted by a + b, say c . i.e., a + b = c
If a + b = c , then the vector c is called the resultant a
vector of a and b. The process of adding two or more
vectors is called the composition of the vectors. Vectors Oh! addition of a and b
possess the magnitude as well as direction. So, they may or may not be non-
cannot be added by simple law of algebra which are collinear or collinear.
applicable to scalars. The addition of vectors is slightly
different from the algebraic addition. Let us discuss the
addition of the two vectors in the following cases.
Case I: When two vectors act in the same direction and lying in the same line
VECTOR In the adjoining figure, the vectors AB and BC are in the same direction 0R
and lying in the same line. The vector AB displaces the point A to the point
B by 2 units distance and the vector BC displaces the point B to the point 1
C by 3 units distance to the right direction. Here, the point A moves to
the point C by 5 units distance to the right with the same direction. Now, AB C
0Q 23
3 O0 0
P
we conclude that if the two vectors act in the same direction and lying in the same line, we only add the
2 3 5
magnitudes of the vectors. In the given figure AB = 0 and BC = 0 and AC = 0 .
Here, 2 + 3 = 2+3 = 5 . ∴ AB + BC = AC.
0 0 0+0 0
Similarly, if PQ = 0 , QR = 0 and PR = 0 ,
3 1 4
Here, 0 + 0 = 0+0 = 0 .
3 1 3+1 4
∴ PQ + QR = PR.
252 Illustrated Optional Mathematics-9
Case II: When two vectors act in the opposite direction but lying in the same line
In the adjoining figure, the vectors AB and BC are in opposite direction but lying in the same line. The
vector AB displaces 6 units to the right but the vector BC displaces 3 units to the left lying in the same
line. Here, the point A moves to the point C by 3 units in the right.
Now, we conclude that if the two vectors act in the same direction lying in the opposite line, we only
6 –3
subtract the magnitude of the vectors. In the given figure, AB = 0 and BC = 0 .
and AC = 3 .
0
6 –3 6–3 3 0Q
Here, 0 + 0 = 0+0 = 0 ∴ AB = BC + AC –3
Similarly, If PQ = 0 and QR = 0 , and PR = 0 RA C B
6 –3 3
0 6 –3
Here, 0 + 0 = 0+0 = 0 ∴ PQ = QR + PR 6 00
6 –3 6–3 3
P
Case III: When two vectors are inclined to each other:
In the adjoining graph, the vector AB = 4 displaces the point A to the C
point B and the other vector BC = 2 1 B to the point 6
3 displaces the point 42
C. Here, the point A is shifted to the point C and the vector AC directly 3
displaces the point A to the point C with 6 units right and 4 units up. A 4B
1
6 6
Therefore, AC = 4 and AB + BC = 4
Here, 4 + 2 = 4+2 = 6
1 3 1+3 4
We conclude that, AB + BC = AC.
Addition of Two Column Vectors VECTOR
If a = x1 and b = x2 are column then the addition of a and b is defined by a + b = x1 + x2 = x1 + x2 .
y1 y2 y1 y2 y1 + y2
This is called the addition of two vectors.
Properties of Vector Addition
If a , b and c are three vectors then the vectors addition holds the following properties:
(i) Commutative property: a + b = b + a .
, 2 1 2 1 2 + 1 3
If a = 3 b= 2 then a + b = 3 + 2 = 3 + 2 = 5
and b + c = 1 + 2 = 1 + 2 = 3
2 3 2 + 3 5
∴ a+ b= b+ a
Hence, the vector addition of two vectors is commutative.
Operations on Vectors 253
(ii) Associative property: ( a + b) + c = a + ( b + c ).
If a = 1 , b= 3 and c = 2 , then
2 1 4
1 3 4 4 2 6
a+ b= 2 + 1 = 3 ∴ ( a + b) + c = 3 + 4 = 4
b+ c= 3 + 2 = 5 ∴ a +(b + c)= 1 + 5 = 6
1 4 5 2 5 7
∴ ( a + b) + c = a + ( b + c)
Hence, the vector addition of vectors is associativity.
(iii) Additive identity property: a + o = o = a + o
( ) ( ) ( ) 1 0 1 0 1 0 + 1 1
If a = 2 , o= 0 , then a + o = 2 + 0 = 2 = a and o + a = 0 2 = 2 = a
∴ a + o = o = a + o, where o is additive identify vector and is also called null vector.
(vi) Additive inverse property: a + ( – a ) = o.
. 2 2 2 2
If a = 3 , then its negative vector is – 3 ∴a + (– a ) = 3 – 3 = 0.
Hence there exists additive inverse property.
Subtraction of Two Vectors
Let AB and BC be two vectors then the differences of AB and BC is denoted by AB – BC and is defined
as the sum of AB and negative of BC. i.e., AB + (– BC). C
In the adjoining figure, AB + BC = AC
Draw a negative vector BD of BC AB + BC
Then, – BC = BD
AB
Now, AB – BC = AB + (– BC) = AB + BD = AD ∴ AB – BC = AD.
If a = x1 and b = x2 are two column vectors. D
y1 y2
VECTOR x1 x2 x1 – x2
Then, the difference of a and b is defined by a – b = y1 – y2 = y1 – x2 .
Example 1 : If a = 3 , b = –1 and c = 0 , find the magnitude and direction of a + b– c.
4 2 1
Solution: Given, a = 3 , b= –1 and c = 0 If a= x ,
4 2 1 y
Now, a + b – c = 3 + –1 – 0 = 4–1–0 = 3 | a | = x2 + y2 and
4 2 1 3+2–1 4
∴ The magnitude of a + b – c = | a + b – c | θ = tan–1 y
x
= 32 + 42 = 9 + 16 = 25 = 5 units
The direction of a+ b– c , θ = tan– 1 4 = 53.13o
3
254 Illustrated Optional Mathematics-9
Example 2 : If p = –2 and q = –1 , find the unit vector along p + q.
–4 0
Solution: Here, p= –2 and q = –1
–4 0
Now, p + q = –2 + –1 = –2–1 = –3
–4 0 –4+0 –4
∴ | p + q| = ( – 3)2 + (– 4)2 = 9 + 16 = 25 = 5 units.
Hence, the unit vector along p + q is 1 –3 = – 3/5 .
5 –4 – 4/5
Example 3 : If A(3, 1), B(0, 2) and C(2, 5) are the vertices of ∆ABC, find AB , BC and AC .
Is AB + BC = AC true?
Solution: Given, A(3, 1), B(0, 2) and C(2, 5) are the vertices of ∆ABC.
Now, AB = 20 –– 13 = –3 , BC = 2–0 = 2 , AC = 2–3 = –1
1 5–2 3 5–1 4
and AB + BC = –3 + 2 = –1 C (2, 5)
1 3 4
∴ AB + BC = AC is true.
EXERCISE - 6.2 (A) A (3, 1) B (0, 2)
C
1. (a) What is the sum of AB and BC in the given figure?
(b) Write the properties of vector addition. AB
2. Write the sum of the following vectors with illustration figure.
(a) a= 2 and b = 4 (b) AB = 0 and BC = 0 (c) LM = –2 and MN = 4
0 0 5 3 3 5
3. Write the difference of the following vectors and illustrate them with figure.
4 1 8 4 (c) PQ = 4 –1
(a) p= 3 and q = 2 (b) AB = 4 and BC = –1 –1 and QR = –4
4. If a = 3 , b= 7 and c = –2 , find the following vectors as column vector. VECTOR
2 6 3
(a) a + b + c (b) a + b – c (c) a – b – c (d) b + c – a
5. If p = 3 , q = –1 and r = 4 , then find the magnitude and direction of the following
2 3 –1
composite vectors.
(a) p + q (b) p + r (c) q – r (d) p + q + r (e) p – q + r
6. If l = 4 , m= –2 and n = 0 , then express the following in column vector and find their
5 –4 1
respective unit vectors.
(a) l + m (b) n – l (c) m – l + n (d) l + m – n (e) l – m – n
7. If a = x1 , b = x2 and c = x3 then prove the following operations.
y1 y2 y3
(a) a + b = b + a (b) ( a + b) + c = a + ( b + c )
Operations on Vectors 255
(c) a + o = o + a = a (d) a + (– a) = 0
8. If p = 2 , q = –2 and r = 2 , show the following relations.
4 5 –3
(a) p – q = – q + p (b) ( p – q) – r = p – ( q + r )
(c) ( p + q) – r = p + ( q – r ) (d) p + (– p) = 0
9. (a) If AB displaces A(4, 3) to B(2, 1) and PQ displaces P(4, 0) to Q(5, 4), find AB + PQ and
| AB + PQ|.
(b) If MN shifts M(– 2, 4) to N(2, – 1) and ST shifts S(4, 2) to T(– 2, – 5), find TS – MN and | NM + ST|.
10. (a) If P(2, 3), Q(0, 4) and R(– 1, – 2) are three points, find the unit vector along PQ – PR and the
direction of PQ + QR.
(b) If L(3, 4), M(3, 1) and N(5, 2) are the vertices of ∆LMN, find the unit vector along LM + MN
and the direction of LM + MN.
11. (a) If A(2, 5), B(5, 1) and C(3, 4) are the vertices of ∆ABC, find AC and AB + BC. Can you write
AB + BC = AC?
(b) If P(– 2, – 3), Q(– 1, 0) and R(2, – 2) are the vertices of ∆PQR, find PR + RQ and PQ. Is
PR + RQ = PQ true? C
12. (a) In the adjoining figure, AB = 5 and BC = –4 ,
2 3
(i) Express AB + BC in column vector. B
(ii) Is AB + BC = AC ? Give the reason. A
(b) In the adjoining figure, PQ = 3 , QR = 3 , RS = –1 and ST = –4
1 2 2 0
(i) Find PQ + QR + RS + ST T S
VECTOR (ii) Is PQ + QR + RS + ST = PT ? Check the result. R
ANSWERS Q
P
2. (a) 6 (b) 0 (c) 2 3. (a) 3 (b) 4 (c) 5 4. (a) 8 (b) 12
0 88 1 53 11 5
(c) – 2 (d) 2 5. (a) 29 units, 68.20o (b) 5 2 units, 8.13o (c) 41 units, 141.34o\
–7 7
(d) 2 13 units, 33.69o (e) 2 17 units, 345.96o 6. (a) 2 , 2/ 5
(b) – 4 , – 1/ 2 1 1/ 5
– 4 – 1/ 2 (c) –6 , – 3/5 (d) 2 , 2/ 5 (e) 6, 3/5
–8 – 4/5 1 1/ 5 8 4/5
9. (a) –1 , 5units (b) 2 ,2 26units 10. (a) 1/ 37 , 239.04o (b) 1/ 2 , 315o
2 12 6/ 37 – 1/ 2
11. (a) 1 , 1 , yes (b) 1 , 1 , yes 12. (a) 1 , yes (b) 1 , yes
–1 –1 33 5 5
256 Illustrated Optional Mathematics-9
6.2 (B) Laws of Vector Addition
Learning Objectives
At the end of this topic, the students will be able to:
• identify the laws of vector addition in triangle, parallelogram and polygon.
• use the laws of vector addition in the geometric figures.
In this section, we discuss three laws of vector addition.
a. Triangle Law of Vector Addition a + b= c C
aB b
Let a and b be two vectors in different directions i.e., inclined to each other
as shown in the adjoining figure. Then the addition of two vectors a and b is A
denoted by a + b, say c , i.e., a + b = c . Here, the vectors a , b and c form
a triangle. Therefore, the law of vector addition a + b = c is called the triangle law of vector addition.
If two vectors are represented by two sides of a triangle taken in the same order, then their resultant is represented
by third side of the triangle taken in opposite order.
b. Parallelogram Law of Vector Addition
Here, OA ( a ) and OB ( b) are two co-initial position vectors and construct a parallelogram OACB taking
OA and OB as the adjacent sides. BC
Now, in the parallelogram OABC, OA = BC, OA // BC, OB = AC and OB // AC. a+b =c
b
Now, OA + AC = OC
or, OA + OB = OC Oa A
or, a + b = c
It means the diagonal OC represents the resultant of the vectors a and b. This geometrical method of the VECTOR
composition of vectors is called the parallelogram law of vectors.
If two coinitial vectors represent the adjacent sides of a parallelogram then the diagonal of the parallelogram drawn
from the same initial point is the resultant of the vectors.
Geometrical Interpretation of Parallelogram Law of Vector Addition
Here, OA = x1 and OB = x2 be two position vectors of the points A(x1, y1) and B(x2, y2) respectively
y1 y2
and draw a parallelogram OACB taking OA and OB as its adjacent sides and OC is its diagonal. Then by
the law of vector addition, OA + OB = OC i.e. a + b = c (suppose). Draw also AM ⊥ OX, BN ⊥ OX,
CP ⊥ OX and AQ ⊥ CP. YC
Then, ∆OBN ≅ ∆CAQ and OM = x1, AM = PQ = y1, B(x2, y2) A(x1, y1)
ON = AQ = x2, BN = y2 = CQ
∴ OP = OM + MP = OM + AQ = OM + ON = x1 + x2 Q
CP = PQ + CQ = AM + BN = y1 + y2 ON MP X
Thus, the coordinate of C is (x1 + x2, y1 + y2) and OC = x1 + x2 .
y1 + y2
Operations on Vectors 257
Also, OA + OB = x1 + x2 = x1 + x2
y1 y2 y1 + y2
∴ OA + OB = OC, i.e., a + b = c .
Hence, the diagonal of the parallelogram represents the sum of its two co-initial adjacent vectors. This is
called the parallelogram law of vector addition.
c. Polygon Law of Vector Addition E D
C
The expanded form of the triangle law of vector addition is called the polygon F
law of vector addition. In the adjoining figure, ABCDEF is a hexagon, the sum
of the first five sides as vectors in the same order is represented by the last one
in the opposite order. i.e., AB + BC + CD + DE + EF = AF. For this, join the
points C, D and E with A and use the triangle law of vector addition as follows:
AB + BC + CD + DE + EF + AC + CD + DE + EF AB
= AD + DE + EF + AE + EF + AF.
Hence, the sum of vectors of a polygon taken in order except the last one with the same initial point is
represented by the last or remaining vector.
Note: The sum of all sides as the vectors in the same order of a polygon is zero, i.e., it represents a null vector.
Example 1 : If the point M is the mid point of the directed line segment joining the points A(2, -5) and
B(1, 3), then find the position vector of M by using:
(a) Vector addition and (b) Vector subtraction.
Solution:
Here, the position vector of A (2, – 5), OA = 2 B (1, 3)
–5
The position vector of B (1, 3), OB = 1 M
3 A (2, -5)
VECTOR M is the mid-point of AB,
Now, we have
(a) The position vector of M, OM = OA + AM O
= OA + 1 AB = OA + 1 (OB – OA ) = OA + 1 OB – 1 OA
2 2 2 2
= 1 OA + 1 OB = 1 2 + 1 = 1 3 = 3
2 2 2 – 5 3 2 –2 2.
–1
(b) The position vector of M, OM = OB – MB
= OB – 1 AB = OB – 1 (OB – OA ) = OB – 1 OB + – 1 OB
2 2 2 2
3
= 1 OA + 1 OB = 1 2 + 1 = 1 3 = 2.
2 2 2 – 5 3 2 –2
–1
258 Illustrated Optional Mathematics-9
Example 2 : In the adjoining figure, ABCD is a quadrilateral. Dc C
If AB = a , BC = b, CD = c , DA = d , AC = k dl kb
and BD = l then express the following vectors as a
single vector:
(a) a + b (b) k + d A aB
(c) a + b + c - d
Solution: Here, AB = a , BC = b, CD = c , DA = d, AC = k and BD = l
Now,
(a) a + b = AB + BC= AC = k. [ Triangle law of vector addition]
(b) k + d = AC + DA = DC = – CD = – c .
(c) a + b + c – d = AB + BC + CD – DA
= AC + CD – DA [ Triangle law of vector addition]
= AD – DA = AD + AD = 2AD = 2 d
Example 3 : In the figure, OA = a and OB = b . If 3AB = 2BC , find OC . O
Solution: Here, OA = a and OB = b
Now, 3AB = 2BC a b?
or, 3(OB – OA) = 2(OC – OB) ABC
or, 3( b – a) = 2(OC – a)
or, 3 b – 3 a = 2OC – 2 a
or, 2OC = 3 b – a VECTOR
or, OC = 12 (3 b – a).
Example 4 : In the given figure, prove that: AB + BC + CD + DE + EF + FA = 0
Solution: Here, F
LHS = AB + BC + CD + DE + EF + FA
= AC + CE + EA = AE + EA A E
= AA = 0 = RHS. Proved. C
''Alternatively''
LHS = AB + BC + CD + DE + EF + FA BD
= AC + CE + EA = AE - AE
= 0 = RHS. Proved.
Operations on Vectors 259
Example 5 : In the adjoining figure, AC and BD are the diagonals of a parallelogram ABCD, prove
that AC + BD = 2 BC . DC
Solution: Here, AC and BD are the diagonals of a parallelogram ABCD.
To Prove: AC + BD = 2 BC AD = BC ] A B
Now,
LHS = AC + BD = AB + BC + AD – AB [
= BC + BC = 2 BC = RHS. Proved.
EXERCISE - 6.2 (B)
1. (a) State the triangle law of vector addition. PM
(b) Which co-initial vector represents the vectors along adjacent sides of
a parallelogram. N
(c) Which vector is equal to KP + PN + NM + ML in the given figure? KL
2. (a) If M is the mid point of the directed line segment joining the points A(3, 2) and B(–1, 4) then
find the position vector of M by using,
(i) Vector addition (ii) Vector subtraction.
(b) If the position vector of A, OA = 2 and AB = 4 , find the position vector of B.
3 6
D C
3. In the adjoining figure, ABCD is a parallelogram whose diagonals AC
and BD intersect at O. If AB = a , AD = b , then express the following b
vectors in terms of a and b : O
(a) AC (b) DB (c) OC Aa B
(d) BO (e) AC + BD (f) CO + DO
4. (a In the adjoining figure, PQRS is a parallelogram. SR
VECTOR If OP = p, OQ = q and OR = r , find OS . P
C
(b) In the given figure, OA = a and OB = b . A O QR
If 3 AC = AB, find OC. a Q
(c) In the given figure, OP = p , OQ = q . O bB OP
If 3 PQ = 2 PR , find OR. Q. No.4(b) Q. No.4(c)
5. In the adjoining figure, ABCDEFGH is a regular octagon. If AB = a , BC = b , CD = c and DE = d , then
(a) Express the AB + BC + CD as a single vector. F E
(b) Find HE as the sum of vectors. G d
(c) Express GE as the sum of two vectors. D
H
Oc
C
(d) Find GA as the sum of two vectors. (e) Prove that : AF = BE b
Aa B
260 Illustrated Optional Mathematics-9
6. In the adjoining figure, PQRS is quadrilateral. If PQ = p , QR = q , RS = r , PS = s and
QS = t and PR = u , express the following vectors as a single vector. P pQ
(a) u + r (b) u – p (c) p + q + r s uq
(d) p + t – r (e) u – q + t – s t
7. In the figure, ABCD is a quadrilateral. Sr R
(a) Find the sum of the following vectors as a single vector. D C
(i) AB + BC + CD (ii) AB + BD (iii) AC + CD
(b) Prove that: A B
(i) AB + BD + DC = AB + BC DP Q
(ii) AD + DC + CB = AD + DB E C R
T
8. (a) In the adjoining figure, ABCDEF is a regular
hexagon. Prove that:
(i) AB + BC + CD + DE + EF = AF F
(ii) AB + AC + AD + EA + FA = 4 AB AB S
(b) In the given diagram, prove that: Q. No. 8(a) Q. No. 8(b)
(i) PR + SP = SR A
(ii) PQ + QR + RS + ST + TP = 0 D
(c) In the adjoining figure, prove that:
AB + BD = AC – DC BC
Q. No. 8(c)
9. If AC and BD are the diagonals of a parallelogram ABCD, prove the following:
(a) AD + BC = AC + BD (b) AC – DB = 2 AD D C VECTOR
(c) AC – BD = 2AB (d) AC – DC + BC = 2AD
ANSWERS AB
2. (a) (i) 1 (ii) 1 (b) 6
3 3 9
3. (a) a + b (b) a – b (c) 1 (a + b ) (d) 1 ( b – a ) (e) 2 b (f) – b
2 2
4. (a) p – q + r (b) 1 (2 a + b ) (c) 1 (3 q – p )
5. (a) AD 3 2
(b) a + b + c (c) a + b (d) – ( c + d )
6. (a) PS (b) QR (c) PS (d) PR (e) PP
7. (a) (i) AD (ii) AD (iii) AD
Operations on Vectors 261
6.2 (C) Multiplication of Vector by Scalar
Learning Objectives
At the end of this topic, the students will be able to:
• multiply a vector by a scalar.
• identify the parallel vectors by using multiplication of a vector by a vector.
• prove some geometric properties by using vector method.
Scalar Multiplication of Vector
Let a = x then 2 a = a + a = x + x = x+x = 2x . C
y y y y+y 2y B
Hence, if a = x and m is a scalar then m a = m x = mx . a 2a
y y my
In the adjoining figure, if AB = a , then 2 a = 2. AB = AC . A –3 p PQ
Similarly, if PQ = p, then –3 p = –3PQ = 3(– PQ) = 3. QP = PR. R p
Parallel Vectors in Terms of Scalar Multiplication
In the adjoining square grid, AB = a and CD = 2 a then CD = 2AB. Therefore, CD and AB are parallel
vectors in the same direction, but not equal, ie., AB//CD and |AB| ≠ |CD|. So, AB ≠ CD. Similarly, if PQ
= p and RS = – 3 p, then RS = – 3 p = 3(– p) = 3(–PQ) = 3QP. Hence, PQ and RS are parallel in opposite
direction, but not equal. i.e., PQ// RS, but PQ ≠ RS. Why? a B
A 2a
Also, we discuss the following example: D pQ
C R
If a = 3 and the scalar m = 2 and n = -2 then,
1 P
–3p
m a = 2 3 = 2 3 and n a = – 2 3 = – 2 3
12 1 –2 S
VECTOR Now, | a | = 3 2 + 12 = 3 + 1 = 4 = 2. Y
|m a | = 2 3 2 + 22 = 12 + 4 = 16 = 4. (–, +) (+, +)
– 2 3 2 + (– 2)2 = 12 + 4 = 16 = 4. X' 120o aa
30o
|n a | = X
O
Here, the magnitude of m a and n a is twice the magnitude of a . –2a (+, –)
(+, +)
Again, the direction of a , θ = tan– 1 1 = 30o Y'
3
The direction of m a , θ = tan– 1 2 = tan– 1 1 = 30o
23 3
The direction of na, θ = tan-1 –2 = tan– 1 1 = (180o + 30o) = 210o. Why ?
–2 3 3
Here, the vectors a and m a are in the same direction but the vectors a and n a are in the opposite
direction to each other.
262 Illustrated Optional Mathematics-9
Hence, if the scalar m is positive real number, then the direction of a and m a are in same direction and if
the scalar n is negative real number then the direction of a and n a are in opposite direction.
Properties of Scalar Multiplication
If a and b are two vectors and m and n are scalars then the scalar multiplication of a vector satisfies the
following properties:
(i) k. a = 0, when k = 0,
(ii) m(n a ) = (mn) a = n(m a ), (Associative property)
(iii) (m + n) a = m a + n a , (Distributive property over scalar addition)
(iv) m( a + b) = m a + m b. (Distributive property over vector addition)
Application of Vector
We can use the vector in various fields. In this section, we use the vector to establish some properties of
geometric statements such as equilateral triangle, isosceles triangle, right-angled triangle, quadrilateral,
parallelogram, rectangle, square, rhombus, etc. The method of verification of these properties are given below:
SN Name Figure Showing properties
Scalene A Magnitudes of vectors of all sides are not equal i.e., | AB| ≠
Triangle
1. | BC| ≠ | AC|
BC
P
2. Isosceles Magnitudes of vectors of any two sides are equal i.e., | PQ
Triangle | = | PR|.
Q R
L Magnitudes of vectors of all sides are equal
i.e., | LM | = |MN| = | LN|.
3. Equilateral
Triangle N
M Square of magnitude of any side is equal to the sum of squares
of magnitude of other two sides. i.e., |AC|2 = |AB|2 + |BC|2.
A VECTOR
C
4. Right-angled
Triangle Magnitudes of any two sides are equal and satisfy
B phythagorus theorem. i.e., |PQ| = |QR| and |PQ|2
R + |QR|2 = | PR |2.
Right-angled P Magnitudes of opposites sides are equal and
5. isosceles Q diagonals are not equal. i.e. |AB| = |DC|, |AD| =
triangle C
D |BC|, |AC| ≠ |BD| 'OR' a vector represented by
B the diagonal of the parallelogram is equal to
6. Parallelogram
sum of its two co-initial adjacent vectors. i.e.,
A |AC| = |AB| + |AD| and |AC| ≠ |BD|.
Operations on Vectors 263
7. Rectangle P Magnitude of opposite sides are equal and
Q
S diagonals are equal. i.e. |PS| = |QR|, |PQ| =
8. Rhombus N |SR|, |PR| = |QS| 'OR' a vector represented by the diagonal
K of the parallelogram is equal to sum of its two co-initial
9. Square D R adjacent vectors and the magnitude of diagonals are equal
A
i.e., PR = PS + PQ and | PR| = |QS|.
Magnitudes of all sides are equal and the
M diagonals are not equal. i.e. |KL| = |NM| = |KN| =
| LM|, |KM| ≠ |NL| 'OR' a vector represented by the diagonal
of the parallelogram is equal to sum of its two co-initial
L adjacent vectors and the magnitude of adjacent sides are equal.
i.e., KM = KL+ KN and |KL| = | KN|.
Magnitudes of all sides are equal and the
C diagonals are also equal. i.e., |AB| = |BC| = |CD| =
|DA|, |AC| = |BD| 'OR' a vector represented by the
diagonal of the parallelogram is equal to sum of its two
co-initial adjacent vectors, the magnitude of adjacent
B sides and diagonals are equal. i.e., AC = AB + AD,
|AB| = |AD| and |AC| = |BD|.
Example 1 : If a = 2 , b = –1 and c = 3 then find the vector 2 a – 3 b + 4 c .
–3 4 0
Solution: Here, a = 2 , b = –1 and c = 3
–3 4 0
Now, 2 a – 3 b + 4 c
VECTOR = 2 –23 – 3 –1 + 4 3 = 4 – –3 + 12 = 4 + 3 + 12 = 19 .
4 0 –6 – 12 –0 – 6 – 12 + 0 – 18
Example 2 : If p = –2 and q = 1 then find the unit vector along (2 p – q) and find its direction.
1 3
Solution: Here, p= –2 and q = 1
1 3
Now, we have 2p – q =2 –2 – 1 = –4–1 = – 5 .
1 3 2–3 – 1
∴ The magnitude of (2 p– q), |2 p – q| = ( – 5)2 + (– 1)2 = 25 + 1 = 26 units.
∴ The unit vector along (2 p – q) = 2p – q –5 = – 5/ 26 a= a
= –1 – 1/ 26
|2 p – q| 26 |a|
Again, the direction of (2 p – q), θ = tan– 1 –1 θ = tan –1 y
–5 x
= 180o + 11.31o = 191.31o.
264 Illustrated Optional Mathematics-9
Example 3 : If a = –1 and b = 2 , then prove that a and b are parallel to each other.
3 –6
Solution: Here, a= –1 and b = 2
3 –6
Now, b = –26 = 2×1 =–2 –1 = – 2. a
–2×3 3
Hence, a and b are parallel to each other. Proved.
Example 4 : If p = 2 and p – 2 q = –4 , find q.
–3 3
Solution: Here, p= 2 and p – 2 q = –4
–3 3
1 6
or, –23 – 2 q = –4 or, 2 –6 =q
3
6/2
or, –23 – –34 = 2 q or, q= – 6/2
or, –66 = 2 q or, q= 3 .
–3
Example 5 : Prove that the points A(4, 3), B(– 2, 3) and C(1, 0) B (–2, 3) A (4, 3)
are the vertices of isosceles right-angled triangle by
using vector method.
Solution: Here, the given three points are A(4, 3), B(– 2, 3) and C(1, 0).
Now, AB = –32––34 –6 C (1, 0)
0
= ∴ | AB | = 36 + 0 = 6 units.
BC = 1 + 2 = 3 ∴ | BC | = 9 + 9 = 18 = 3 2 units.
0 – 3 –3
AC = 1 – 4 = – 3 ∴ | AC | = 9 + 9 = 18 = 3 2 units.
0 – 3 – 3
Here, | BC | = | AC | and (3 2)2 + (3 2)2 = 62 VECTOR
or, (| AC |)2 + (| BC |)2 = (| AB |)2
∴ ∠ACB = 90o.
Hence, the given three points are the vertices of isosceles right-angled triangle. Proved.
Example 6 : The medians of a triangle are concurrent. P
Solution: Here, let RL, PM and QN be the medians of ∆PQR then, LN
RL + PM + QN = RQ + QL + PQ + QM + QR + RN
= – QR + 1 QP + PQ + 1 QR + QR + 1 RP Q M R
2 2 2
= – 1 PQ + PQ + 1 QR + 1 RP = 1 PQ + 1 QR + 1 RP
2 2 2 2 2 2
= 1 (PQ + QR + RP ) = 1 (PR + RP ) = 1 × 0 = 0.
2 2 2
Hence, the medians of a triangle are concurrent.
Operations on Vectors 265
EXERCISE - 6.2 (C)
1. (a) If a is a scalar and p = p , what is the product of a and p?
q
(b) Write any one property of scalar multiplication.
(c) In which condition is ∆ABC an isosceles right-angled triangle?
(d) In which condition is the median of a triangle concurrent?
2. If a = –2 and b = 4 , find the following vectors:
3 2
(a) 3 a + 2 b (b) 2 a – 3 b (c) 4 a – 3 b
3. If p = 2 , q = 4 , and r = –2 , find the following vectors:
3 –1 3
(a) 2 p – 2 q + r (b) 3( p – q ) + 2 r (c) 5 p – 2(2 p + 3 r )
4. (a) If a = 2 and b = –1 , find the unit vector along (3 a + 2 b ) and the direction of (3 a +2
–1 3
b)
(b) If p + q = 2 and q = 1 , find the magnitude and direction of 2p – q . Also, determine
–2 –2
the unit vector along 2 p – q .
5. Prove that the following pairs of vectors are parallel to each other:
–1 – 1 –3 –6
3 2 –4 –8
(a) PQ = and RS = (b) AB = and CD =
1
1 2
6. (a) If AB displaces A(–1, 2) to B(2, 2) and PQ displaces P(5, 0) to Q(–1, – 2), show that the
vector ABand PQ are parallel but in opposite direction.
VECTOR (b) If MN shifts M(– 4, – 3) to N(8, 1) and ST shifts S(– 1, 1) to T(5, 3), prove that the vectors MN
and STare like vectors.
7. (a) If p = 2 and 2 p – q = 4 , find q .
3 5
(b) Find 2 a + 3 b if a = –2 and a + 2 b = 6 .
3 9
8. (a) Prove that the points P(2, 2), Q(– 2, – 2) and R – 2 3, 2 3 are the vertices of an equilateral
triangle.
(b) Show that the points A(2, 1), B(1, 4) and C(– 1, 2) are the vertices of an isosceles triangle.
(c) Show that the points (– 1, – 1), (1, 0) and (– 2, 1) are the vertices of a right-angled isosceles
triangle.
(d) Prove that the points (– 2, 1), (2, 2), (4, 4) and (0, 3) are the vertices of a parallelogram.
(e) Prove that the quadrilateral with vertices (6, 1), (9, 4), (7, 6) and (4, 3) is a rectangle.
266 Illustrated Optional Mathematics-9
(f) Verify that the quadrilateral with vertices (8, 3), (10, 5), (8, 7) and (6, 5) is a square.
(g) Show that the points (– 2, 1), (– 5, 0), (– 8, 1) and (– 5, 2) are the vertices of a rhombus.
P
9. (a) In the adjoining figure, A, B and C are the mid-points of
the sides PQ, QR and RP of ∆PQR respectively, then prove AC
that: RA+ QC+ PB = 0 QB R
(b) In the given figure, E and F are the mid points of A
the sides AD and BC of the quadrilateral ABCD C
respectively, show that: AB + DC= 2EF D
(c) If M and N are the midpoints of AB and AC of MN
∆ABC respectively, prove that BC = 2 MN and EF
BC // MN .
A C
BB Q.No. 9 (c)
Q.No. 9 (b)
(d) In the figure, ABCDEF is a regular hexagon. ED ED
Prove that: AB + AC + AD + EA + FA = 4AB.
(e) In the given figure, O is the centre of a regular F CF O C
hexagon ABCDEF.
AB AB
Prove that: AB + AC + AD + AE + AF = 6
AO Q.No. 9 (d) Q.No. 9 (e)
(f) In the figure, PQRST is a regular pentagon. S SR
Prove that:
TR
PQ + PT + QR + SR + TS = 2 PR .
(g) In the adjoining figure, PR and QS are the PQ PQ VECTOR
diagonals of the parallelogram PQRS. Show
Q.No. 9 (f) Q.No. 9 (g)
that: PR + QS = 2QR and PR – QS = 2 PQ .
ANSWERS (b) – 16 (c) – 20 (b) 8
0 (c) 6 15
2. (a) 2 7. (a) 14
13 (b) – 10
18 – 15
3. (a) – 6
11 (b) 5 units, 63.43o, – 2/ 5 0
1/ 5 1
4. (a) 4/5 , 36.87o
3/5
Project Work
Write the triangle, parallelogram and polygon laws of vector addition with related diagrams.
Operations on Vectors 267
TRANSFORMATION 7UNIT
Estimated Teaching Periods : 18
Competency
To transform the geometric figures by using coordinates.
At the end of this unit, the students will be able to: Learning Outcomes
introduce transformation and tell its types.
transform the geometric shapes under reflection in the x-axis, y-axis, line y = x, line y = – x, line
x = a and line y = b and represent in the graph.
translate the geometric figure in the given magnitude and direction
rotate the geometric shapes and images about the center (a, b) through ± 90°, 180° and 360°.
transform the geometric shapes under enlargement and reduction with center (a, b) and given
scale factor and find the image.
Reflection What do you learn?
� Introduction to reflection
� Reflection in coordinate plane
Translation � Translation in coordinate
� Introduce to translation
Rotation � Rotation in coordinate plane
� Introduction to rotation
Enlargement and Reduction
� Introduction to enlargement and reduction � Enlargement and reduction in coordinate plane
TRANSFORMATION Are you ready?
Meaning of transformation
Basic rules of translation, reflection and rotation.
Translation, reflection and rotation of geometric figures.
Translation, reflection and rotation of geometric figures in coordinate axes under specific rule.
Enlarging and reducing the figure or object from a fixed point.
Specification Grid
Unit Chapter Cognitive Knowl- Under- Applica- Higher Total Total
Domain edge standing tion ability Ques- Marks
tions
Topic 1 mark 2 marks 4 marks 5 marks
7 Transformation Reflection & Rotation 1 – 1 1 3 10
Translation & Enlargement
268 Illustrated Optional Mathematics-9
71.1 INTRODUCTION TO TRANSFORMATION
AND REFLECTION
Learning Objectives
At the end of this topic, the students will be able to:
• define reflection with geometric representation.
• reflect an object in the coordinate axes.
7.1 (A) Introduction to Reflection
Introduction to Transformation
In general, the method of transforming (or changing or moving) an object from one place to another place
in a certain rule is called transformation. For example, study the following figures:
M C'
A'
Q O θ C O A
fig. (iv) fig. (v)
A N O
a fig. (ii) fig. (iii)
P
fig. (i)
In the above fig. (i), the point A displace into A' according as the vector PQ = a such that AA' = PQ and TRANSFORMATION
AA' // PQ . Similarly, in the fig. (ii) the object B reflects on the mirror MN, and forms the image B'. In
this condition, OB = OB' and BB' ⊥ MN. In the fig. (iii), the point C rotates into C' in the anti–clockwise
direction, in which OC = OC'. In fig. (iv), the object I is enlarged into I' in which kOI = OI', where k is a
arbitrary constant. Such a way, the transformation brings the image of an object under certain rule. So, it
is one type of function or mapping in which there is one-one correspondence between the points/objects
and their images. In certain condition, the transformation does not change the object, i.e., it remains the
same image as the given object. In the adjoining fig. (v), the object A lies on the transforming surface or
line remains same as A after transformation.
Thus, there are mainly four types of transformations on the basis of the size and condition of the image
with respect to the object. They are:
a. Isometric Transformation b. Non-isometric Transformation
c. Identity Transformation d. Inversion Transformation
In the above mentioned figures, the fig. (i), (ii) and (iii) are isometric transformation in which the objects
and their images are congruent. But in the fig (iv), the object and its image are not congruent. So, it is
a non-isometric transformation. At last, in the fig (v), the object and its image are the same. So, it is an
identity transformation. The point in which the identity transformation is performed, is called an invariant
point.
Introduction to Transformation and Reflection 269
Further, the isometric transformation is classified into three types. They are:
(i) Translation (ii) Reflection (iii) Rotation
Similarly, the non–isometric transformation can be classified into two types. They are:
(i) Enlargement (ii) Reduction
Classification of transformation
Identity Isometric Non–isometric Inverse
transformation transformation transformation transformation
Translation Reflection Rotation Enlargement Reduction
a. Isometric Transformation
Transformation which leaves the dimensions of the object and its image unchanged is called isometric
transformation (simply isometry). The distance between any two points in the pre–image must be
the same as the distance between the images of the two points. So, it is invariant with respect to
distance. In isometry, the object and its image are congruent and the perpendicular distance from the
object and its image to the transformation plane or centre is always equal shown in the figures below.
M Q'
Q
B PO P'
P
N θ Q
A fig. (vii)
fig. (vi) O
fig. (viii)
In the above fig(vi), AB = PQ and AB//PQ, in the fig (vii), OP = OP', PP'⊥ MN and in figure (viii)
OQ = OQ'. So, this transformation is also called the distance persevering transformation.
b. Non-isometric Transformation
Transformation which leaves the dimensions of the object and its k
image changed is called non-isometric transformation (simply
isometry). In the adjoining figure, A' and A'' are the images of the O
object A with centre O such that OA = kOA' and OA = kOA'' where
TRANSFORMATION k is a scale factor (or magnitude). If k > 1, then the image is bigger A'' A'
than the object and it is called an enlargement. If k <1, then the image A
fig. (ix)
is smaller than the object and it is called reduction. If k = 1, then the image is the same as the object
and it is called identity transformation.
c. Identity Transformation
A transformation in which any object is mapped into the image itself is called identity transformation.
d. Inversion Transformation
A transformation which brings the image back and invertible to the object is called inverse
transformation.
270 Illustrated Optional Mathematics-9
Reflection on Axes
Study the following pictures to understand the reflection:
In the adjoining picture (i), a girl is standing in front of mirror
and her image has formed in the mirror as same as her.
Similarly, in the picture (ii), a bear is looking in the well and fig (i) fig (ii)
its image has formed in the water as same as it.
M M
In the both pictures, the images formed in the given surface
as same as the object. Such types of transformations are
called reflections.
Now, we use a line to represent the mirror or the water
surfaces as the reflecting axis, then the above pictures can be
drawn as follows:
In the both pictures, the perpendicular distance of the object
from the mirror line or water level is equal to its image. N N
An isometric transformation involving a fixed point on the line such that the line segment joining the object to its
image is perpendicularly bisected at that point, is called reflection. The line on which the image of the object casts
is called a reflecting axis or an axis of reflection.
In reflection, the object P is reflected in the reflecting line AB in the adjoining figure and its image is P'.
Then we obtain the following information:
(i) OP = OP', i.e., It preservers distance. A
(ii) PP' ⊥ AB, i.e., It preserves perpendicularity.
(iii) The object P and its image P' are of the same shape and size. i.e., It P O P'
preserves congruency.
(iv) The object P, the image P' and the reflecting point O lie on the same B
line i.e., it preserves the collinearity. Reflecting axis
(v) The image of a point on the reflecting axis is invariant.
The image of P is P' under the reflection on the line AB, which is denoted by α(P) = P'. Sometimes it is
denoted by R(P) = P'.
Reflection in Coordinate Plane
In this section, we derive the equation for three categories of reflection on the reflecting axis as the axes, TRANSFORMATION
equi-oblique lines and lines parallel to axes.
(i) Reflection on x-axis or the line y = 0 Y
Consider point P(2, 3). Reflect the point P(2, 3) on the x-axis P(2, 3)
or the line y = 0. For this, draw PC ⊥ XX' and produce PC up
to P' such that PC = CP'. Then, the coordinates of P' are (2, 3). Q(–1, 2)
Symbolically, we can write, P(2, 3) x-axis P'(2, –3).
X' DO C X
Similarly, the image of Q(–1, 2) under reflecting on x-axis is Q'(–1, –2) P'(2, –3)
Q'(–1, –2). Y'
i.e., Q(–1, 2) x-axis Q'(–1, –2).
Introduction to Transformation and Reflection 271
Thus, it is clear that, in general, the coordinates of the image of Y (x, y)
(x, y) under the x-axis or the line y = 0 are (x, –y).
i.e., (x, y) x-axis (x, –y) = (x', y') X' X
O
∴ x' = x and y' = –y
or, x' = 1. x + 0. y and y = 0. x + (–1) y, Y' (x, –y)
which are the required equations for the reflection about the x-axis or the line y = 0.
Alternatively
Let P'(x', y') be the same of a point P(x, y) under reflecting in the x-axis. Then CP = CP' and the point
C lies on the x-axis. So, its coordinates are (x, 0). Again, C is the midpoint of PP'. Then,
(x, 0) = x + x', y + y'
2 2
x + x'
∴ x = 2
or, 2x = x + x' or, x' = x and 0 = y + y' or, y + y' = 0 or, y' = – y.
2
Hence, the coordinates of the image of P(x, y) under reflection in the x-axis are (x, –y).
(ii) Reflection on y-axis or the line x = 0 Y P(2, 3)
Let a point P(2, 3) reflect on the y-axis or the line x = 0. P'(–2, 3)
For this, draw PC ⊥ YY' and produce PC up to P' such that C
PC = CP'. Then the coordinates of P' are (–2, 3).
i.e., P(2, 3) y-axis P'(–2, 3). X' X
O
Similarly, the image of Q(–4, –2) reflecting on the y-axis is D Q'(4, –2)
Q'(4, –2). Q(– 4, –2)
i.e., Q(– 4, –2) y-axis Q'(4, –2). Y'
Thus, it is clear that, in general, the coordinates of the image of (x, y) under the reflection on the
y-axis or the line x = 0 are (– x, y). Y
i.e., (x, y) y-axis (– x, y) = (x', y') (–x, y) (x, y)
∴ x' = – x and y' = y X' O X
or, x' = (–1)x + 0.y and y' = 0.x + 1.y,
Which are the required equations for the reflection about the y-axis or the line x = 0.Y'
TRANSFORMATION Alternatively
Let P'(x', y') be the image of a point P(x, y) under reflection in the y-axis. Then CP = CP' and the
point lies on the y-axis. So, its coordinates are (0, y).
Again, C is the midpoint of AA'. Y P(x, y)
P'(x', y')
Then, (0, y) = x + x', y + y' (–x, y) C(0, y) (x, y)
2 2 X
X' O
∴ 0 = x + x' or, x' + x = 0 or, x' = –x
2
Y'
y = y + y' or, 2y = y + y' or, y' = y.
2
Hence, the coordinates of the image of P(x, y) under reflection in the y-axis are P(–x, y).
272 Illustrated Optional Mathematics-9
AA
Example 1 : Reflect the given figure in the reflecting line l.
Solution: Draw the image of the given triangle ABC B CB C
under the reflection in the line l as shown in l B' C'
alongside: A'
Example 2 : The vertices of ∆PQR are P(– 1, 3), Q(0, 4) and R (1, 2). Reflect it along the x-axis and
write down the coordinates of the vertices of the image of ∆PQR. Draw the reflection
in the graph. Y
Solution: Here, the vertices of ∆PQR are P(– 1, 3), Q(0, 4) and R(1, 2). Q
Now, reflecting the vertices of ∆PQR in the x-axis, we have PR
(x,y) x-axis (x, – y) X' O X
∴ P(–1, 3) P'(– 1, – 3) R'
P'
Q(0, 4) Q'(0, – 4) Q'
R(1, 2) R'(1, – 2) Y'
Therefore, the coordinates of the vertices of the image of ∆PQR are P'(– 1, – 3), Q'(0, – 4)
and R'(1, – 2). Drawing the reflection in the graph alongside.
Example 3 : Find the coordinates of a point P which transforms to the image P'(– 3, – 2) under
reflection in the x-axis.
Solution: Let P(x, y) be the required point. Transforming P(x, y) under reflection in the x-axis, we get
P(x, y) x-axis P'(x, – y).
By question, P(x, y) x-axis P'(– 3, 2). P'(x, – y) = P'(– 3, 2)
By equating corresponding components, we get
x = – 3 and y = – 2.
Hence, the required point is (– 3, – 2).
Example 4 : Find the coordinates of the image of the point P(– 4, 5) reflecting in the line in which TRANSFORMATION
the point P(–4, –5) reflects into P'(4, – 5).
Solution: Since the point P(– 4, – 5) reflects into P'(4, – 5) on the line. So, the mid point of PP' is
–4+ 4, – 5– 5 = (0, – 5). Hence, all the points have the same y-component – 5 and
2 2
x-component 0.
So, the equation of the axis of reflection is x = 0 or y-axis.
Now, reflecting the point A(– 4, 5) under the reflection line x = 0.
We get, A(– 4, 5) x = 0 A'(4, 5).
Hence, the required coordinates of the image of A(– 4, 5) are A'(4, 5).
Introduction to Transformation and Reflection 273
EXERCISE - 7.1 (A)
1. (a) Define transformation. (b) Write the types of transformation.
(c) List the types of isometric transformation.
2. (a) What is reflection? (b) Write any one property of reflection.
(c) What will be the coordinates of the image of a point P(a, b) when it is reflected on the line y = 0?
3. Reflect the following figures in the given reflecting line l:
(a) (b) (c) l
B
A l
l
(d) (e) (f) l
l
l
4. Reflect the following geometric figures in the x and y axes:
(a) Y (b) Y (c) Y (d) Y
O
P X X' F
A Q L N A
X' O R X' O BX
X' O C B X S
X
M
Y' Y' Y' Y'
5. Reflect the following points in the x-axis (or the line y = 0) and find its image:
(a) P(1, 4) (b) Q(–2, 5) (c) R(–3, –1) (d) S(3, – 5) (e) T(0, 3)
6. Write down the coordinates of the following points under reflection along the line x = 0 or
y-axis:
TRANSFORMATION (a) A(4, 2) (b) B(–3, 0) (c) C(– 4, –5) (d) D(2, –3) (e) E(0, 4)
7. (a) The vertices of ∆ABC are A(2, 3), B(5, 1) and C(3, 5). Find the coordinates of the vertices of the
image of ∆ABC under the reflection on the line y = 0. Draw the reflection in the same graph.
(b) Find the coordinates of the vertices of the image of a triangle OPQ with the vertices O(0, 0), P(3,
4) and Q(2, –5) after reflection about the y-axis. Draw this transformation in the same graph.
8. (a) Find the coordinates of the vertices of the image of a quadrilateral ABCD with the vertices A(1, 3),
B(2, 5), C(4, 1) and D(5, 4) under the reflection in the x-axis. Draw this reflection on the graph.
(b) The points P(–2, 3), Q(0, 2), R(2, 3) and S(0, 4) are the vertices of the parallelogram PQRS. Reflect
the vertices of PQRS in the line x = 0 by using graph and write down the coordinates of the
vertices of the image parallelogram.
274 Illustrated Optional Mathematics-9
9. (a) If L'(2, –3) is the image of the point L under reflection in y-axis, find the coordinates of the point L.
(b) Find the coordinates of the point A which reflects into A'(–5, –7) in the line y = 0.
10. (a) If P'(3x, – 4) is the image of the point P(9, y – 3) under the reflection in the line x = 0, find the values
of x and y.
(b) If the image of T(2a – b, 5) is T'(2, a + b) under the reflection in the x-axis, find the values of a and b.
11. (a) Find the axis of reflection when the point A(2, –3) reflects into A'(–2, –3).
(b) Find the coordinates of the image of the point P(2, –3) reflecting under the line in which the point
Q(– 4, 3) reflects into Q'(– 4, –3).
ANSWERS
1. to 4. Show to your teacher
5. (a) P'(1, – 4) (b) Q'(–2, –5) (c) R'(–3, 1) (d) S'(3, 5) (e) (0, –3)
6. (a) A'(– 4, 2) (b) B'(3, 0) (c) C'(4, –5) (d) D'(–2, –3) (e) (0, 4)
7. (a) A'(2, –3), B'(5, –1), C'(3, –5) (b) O'(0, 0), P'(–3, 4), Q'(–2, –5)
8. (a) A'(1, –3), B'(2, –5), C'(4, –1), D'(5, – 4) (b) P'(2, 3), Q'(0, 2), R'(–2, 3), S'(0, 4)
9. (a) L(–2, –3) (b) A(–5, 7) 10. (a) –3, –1 (b) –1, – 4
11. (a) x = 0 or y – axis (b) P'(2, 3).
7.1 (B) Reflection on Equi–obilique Lines
Learning Objectives
At the end of this topic, the students will be able to:
• reflect an object in the lines parallel to the cordinate axes. • reflect an object in the equi–oblique lines.
(i) Reflection on the line y = x or x – y = 0: Y
Consider a point P(1, 3), draw PC ⊥ AB for reflecting the point Q(–3, 4) A
P(1, 3) on the line AB with the equation y = x or x – y = 0. Produce
PC upto P' such that PC = CP'. Then the coordinates of P' is (3, 1). P(1, 3)
i.e., P(1, 3) y = x P'(3, 1). X' P'(3, 1)
X
O
Similarly, the coordinates of the point Q(–3, 4) reflecting on the
same line are Q'(4, –3) B y=x Q'(4, –3)
(x, y)Y'
i.e., Q(–3, 4) y = x Q'(4, –3). (y, x)
TRANSFORMATIONY
Thus, it is clear that, in general, the coordinate
of the image of (x, y) under reflecting on the line
y = x are (y, x).
i.e., (x, y) y = x (y, x) = (x', y') X' O X
∴ x' = y and y' = x
or, x' = 0.x + 1.y and y' = 1.x + 0.y, y=x
which are the equations for the reflection about the line y = x or x – y = 0.
Y'
Introduction to Transformation and Reflection 275
Alternatively
The line l with equation y = x makes angle of 45o with positive x-axis. Then the slope of the line l
is m1 = tan 45o = 1.
Let P(x, y) be a point and P'(x', y') be the image of a point P(x, y) reflection in the line y = x. Then
y – y'
the slope of PP' is m2 = x – x' . Y P(x, y)
Since the lines l and PP' are perpendicular to each other, C
So, m1 × m2 = –1 P'(x', y')
X
or, 1 × y – y' = –1 45o
x – x' O
or, y – y' = –x + x' y=x
or, x' + y' = x + y ............. (i)
Again, x + x', y + y'
2 2
C is the midpoint of PP'. So, the coordinates of C are , which lies on the line y = x. then,
∴ x + x' = y + y'
2 2
or, x' – y' = y – x ................... (ii)
Adding the equations (i) and (ii), we get
2x' = 2y or, x' = y
From equation (i), we get y + y' = x + y and y' = x.
Hence, the coordinates of the image of P(x, y) under reflection in the line y = x are P'(y, x).
(ii) Reflection on the line y = – x or x + y = 0: Y
D
Consider a point P(2, 3). Draw PE ⊥ CD for reflecting the point P(2, 3)
on the line CD with the equation y = –x or x + y = 0. Produce PE up to P' Q'(–1, 5)
such that PE = EP'. Then the coordinates of P' are (–3, –2).
ie., P(2, 3) y = –x P'(–3, –2). F P(2, 3)
Q(–5, 1) E X
X' O
y = –x
TRANSFORMATION
Similarly, the coordinates of the image of the point Q(–5, 1) under P'(–3, –2) C
reflection on the same line are Q'(–1, 5). Y'
i.e., Q(–5, 1) y = –x Q'(–1, 5).
Thus, it is clear that, in general, the coordinates of the image of a point (x, y) under reflection on the
line y = –x are (–y, –x). Y
i.e., (x, y) y = –x (–y, –x) = (x', y') (x, y)
∴ x' = –y and y' = –x X' O X
or, x' = 0.x + (–1)y and y' = (–1)x + 0.y, (–y, –x)
which are the equations for the reflection about the line y = –x or x + y = 0. Y' y = –x
276 Illustrated Optional Mathematics-9
''Alternatively''
The line l with equation y = –x makes angle of 45o with the negative x-axis. Then the slope of the
line l is m1 = tan (–45o) = tan (180o – 45o) = tan 135o = –1.
Let P(x, y) be a point and P'(x', y') be the image of P(x, y) under reflection in the line y = –x. Then
y' – y
the slope of PP' is m2 = x' – x . P(x, y)
Since the lines l and PP' are perpendicular to each other so, lY
m1 × m2 = –1 C
or, –1 × y' – y = –1 P'(x', y') 45o 135o X
x' – x X' O
or, x' – x = y' – y or, x' – y' = x – y ............. (i) Y' y = –x
Again, C is the midpoint of PP'. Then the coordinates of C are x + x', y + y' , which lies on the line
y = –x. then, 2 2
y + y' = – x + x' or, x' + y' = –x – y ................... (ii)
2 2
Adding the equations (i) and (ii), we get
O 2x' = –2y or, x' = –y
From the equation (i), we get
–y – y' = x – y or, y' = –x
Hence, the coordinates of the image of P(x, y) under reflection in the line y = –x are (–y, –x).
Reflection on Lines Parallel to Axes
(i) Reflection on the line x = a or x – a = 0 (Parallel to y-axis):
Consider a point P(1, 3). Draw PC⊥ AB for reflecting Y
the point P(1, 3) on the line AB with the equation x = 2 or A
x – 2 = 0. Produce PC up to P' such that PC = CP'. Then, the C P'(3, 3)
P(1, 3)
coordinates of P' are (3, 3).
i.e., PC (1, 3) x = 2 P'(3, 3) = P'(2 × 2 –1, 3). X' O X
Q(–2, –1) Q'(6, –1)
Similarly, the coordinate of the image of the point Q(–2, –1). x=2
under reflection on the same line are; Y' B TRANSFORMATION
Q'(6, –1) = Q'(2 × 2 – (–2), –1).
Thus, it is clear that, in general, the coordinate of the image of a point (x, y) under reflection on the
line x = a are (2a – x, y). Y
i.e., (x, y) x = a (2a – x, y) = (x', y') (x, y) (2a – x, y)
∴ x' = 2a – x and y' = y, X' O X
or, x' = 2a – x + 0.y and y' = 0.x + 1.y,
which are the equations for the reflection about the line x = a or x – a = 0. Y' x = a
Introduction to Transformation and Reflection 277
''Alternatively''
In the given figure, the coordinates of the image of a point P(x, y) are P'(x', y') under reflection on the
line x = a, then the point M(a, y) which lies on the line x = a, is the mid–point of PP'. Now, we have
M(a, y) = x + x', y + y' Y
2 2 (x, y)
P'(x', y')
∴ a= x + x' and y = y + y' M(a, y)
2 2
X' X
O
or, 2a = x + x' and 2y = y + y'
x=a
or, x' = 2a – x and y' = y Y'
Hence, the image of P(x, y) is P'(2a – x, y) under the reflection about the line x = a.
(ii) Reflection on the line y = b or y – b = 0 (Parallel to x-axis): Y P(2, 5)
Q'(–3, 4)
Consider a point P(2, 5). Draw PR⊥CD for reflecting the point y=2 RD
P(2, 5) on the line CD with the equation y = 2 or, y – 2 = 0. Produce CS X
PR up to P' such that PR = RP'. Then, the coordinate of P' are (2, –1). X'
i.e., P(2, 5) y = 2 P'(2, –1) = P'(2, 2 × 2 – 5). Q(–3, 0) O P'(2, –1)
Similarly, the image of Q(–3, 0) under reflection on the same line Y'
is Q'(–3, 4).
i.e., Q(–3, 0) y = 2 Q'(–3, 4) = Q'(–3, 2 × 2 – 0).
Thus, it is clear that, in general, the coordinates of the image of a point (x, y) reflecting on the line
y = b are (x, 2b – y). Y (x, y)
i.e., (x, y) y = b (x, 2b – y) = (x', y') y=b
∴ x' = x and y' = 2b – y, X' (x, 2b + y)
or, x' = 1.x + 0.y and y' = 0.x + 2b – y, O X
which are the equations for the reflection about the line y = b or y – b = 0. Y'
'' Alternatively''
In the given figure, the coordinates of the image of a point P(x, y) are P'(x', y') reflecting on the line
TRANSFORMATION y = b, then the point M(x, b) which lies on the line y = b, is the mid point of PP'.
Now, we have y=b Y (x, y)
x = x + x' and b = y + y' (x, b)
2 2
(x', y')
or, 2x = x + x' and 2b = y + y' X' X
O
or, x' = 2x – x and y' = 2b – y Y'
or, x' = x and y ' = 2b – y.
Hence, the image of P(x, y) is P'(x, 2b – y) under the reflection on the line y = b.
278 Illustrated Optional Mathematics-9
Example 1 : Find the coordinates of the image of the rectangle ABCD having the vertices
A(2, 1), B(3, 0), C(5, 2) and D(4, 3) under the reflection on the line y + 1 = 0.
Show this transformation in the graph.
Solution: Here, the vertices of the rectangle ABCD are A(2, 1), B(3, 0), C(5, 2) and D(4, 3). Now,
reflecting the vertices of the rectangle ABCD on the line Y
A
y + 1 = 0 or, y = –1, we have D
C
(x, y) y = b (x, 2b – y)
∴ A(2, 1) y = – 1 A'(2, 2 × (–1) –1) = A'(2, –3). X' O B X
B(3, 0) y = – 1 B'(3, 2 × (–1) –0) = B'(3, –2). B'
C(5, 2) y = – 1 D'(5, 2 × (–1) –2) = C'(5, –4).
D(4, 3) y = – 1 C'(4, 2 × (–1) –3) =D'(4, –5). A'
C'
D'
Y'
Hence, the coordinates of the image rectangle A'B'C'D' are A'(2, –3), B'(3, –2),
C'(5, –4) and D'(4, –5). This transformation is shown in the graph.
Example 2 : If P'(3, –1) is the image of the point P after reflection in the line x = 2, find the
coordinates of P.
Solution: Let (x, y) be the coordinates of P, then reflecting P(x, y) in the line x = 2, so the
coordinates of the image of P(x, y) is P'(2 × 2 – x, y) = P'(4 – x, y)
But by question, P'(3, –1) is the image of P(x, y).
So, P'(4 – x, y) = P'(3, –1)
∴ 4 – x = 3 and y = –1 or, x = 1 and y = –1.
Hence, the coordinates of P are (1, –1).
Example 3 : The image of A(a, 5 – b) when reflected in the line x + y = 0 is A'(2a, 2b – 20). Find the
values of a and b.
Solution: Reflecting A(a, 5 – b) in the line x + y = 0 or, y = – x , we get
A(a, 5 – b) y = –x A'(– (5 – b), – a)
But by question,
A(a, 5 – b) y = –x A'(2a, 2b –20)
(– (5 – b), – a) = (2a, 2b – 20) TRANSFORMATION
∴ – (5 – b ) = 2a and – a = 2b – 20
or, –5 + b = 2a and 20 = a + 2b
or, 2a – b = –5 ….. (i) and a + 2b = 20 ……… (ii)
Multiplying eqn (i) by 2 and adding with eqn (ii), we get
4a – 2b = –10
(+) a + 2b = 20 10
5
5a = 10 a = = 2.
Putting the value of a in eqn (i), we get
2 × 2 – b = –5 or, b = 4 + 5 = 9 ∴ a = 2 and b = 9.
Introduction to Transformation and Reflection 279
Example 4 : The vertices of the square PQRS are P(2, 0), Q(4, 2), R(2, 4) and S(0, –2). If the image
of P(2, 0) is P'(–2, 0), find the coordinates of the images of the remaining vertices of
the square PQRS.
Solution: Here, the vertices of the square PQRS are P(2, 0), Q(4, 2), R(2, 4) and S(0, –2) and the
image of P(2, 0) is P'(–2, 0).
The mid–point of PP' (2 – 2, 0 + 0) = (0, 0).
Here, y-coordinate of P = y–coordinate of P'. So, the axis of reflection is x = a, which is passed
through the point (0, 0). Therefore, the equation of the axis of reflection is x = 0 or y-axis.
Now, reflecting the remaining vertices of the square PQRS in the y-axis, we get
Q(4, 2) x = 0 Q'(– 4, 2), R(2, 4) x = 0 R'(–2, 4), S(0, –2) x = 0 S'(0, –2)
Here, the coordinates of the images of the remaining vertices of the square PQRS are
Q'(– 4, 2), R'(–2, 4) and S'(0, –2).
EXERCISE - 7.1 (B)
1. (a) What is equi-oblique line?
(b) Write down the coordinates of the image of a point A(x, y) when it is reflected on the line
x+y=0
(c) What is the image of a point P(p, q) when it is reflected on the line x – a = 0?
2. Reflect the following shapes in the lines l and l':
(a) Y (b) Y (c) Y l (d) Y l
S
P A P
L
Q Rl B D X' O M X X' Q R X
X' O X X' O N O
C X
l
Y' l' Y' l' Y' l' Y' l'
3. Write the images of the following points under reflection about the line y = x or x – y = 0:
(a) L(3, 5) (b) M(– 2, 5) (c) N(– 3, 0) (d) O(– 4, – 4) (e) P(4, – 4)
4. What are the co-ordiantes of images of the following points after reflection about the line
y = –x or x + y = 0?
(a) G(3, 6) (b) H(– 3, 4) (c) I(2, – 2) (d) J(0, 4) (e) K(0, – 4)
TRANSFORMATION 5. Find the images of the following points after reflection about the line y = 4.
(a) A(3, 1) (b) B(4, 4) (c) C(– 4, 0) (d) D(– 2, 4) (e) E(0, 0)
6. Find the coordinates of the images of the following points under reflection in the line x + 2 = 0. Also
identify the invariant points.
(a) P(1, 4) (b) Q(– 5, 0) (c) R(– 3, – 4) (d) S(– 4, 3) (e) T(– 2, 5)
7. (a) The points A(2, 4), B(–3, 5) and C(0, 1) are the vertices of ∆ABC. Reflect the vertices of
∆ABC in the line x = y and write down the coordinates of the vertices of the image ∆A'B'C'.
Also, draw this reflection on the same graph.
(b) Find the coordinates of the vertices of the image of ∆PQR with vertices P(–1, – 4), Q(– 4, 1) and
R(– 2, 3) under the reflection in the line x = 2. Draw this transformation on the graph.
280 Illustrated Optional Mathematics-9
8. (a) If E(0, 1), F(4, 1), G(6, –1) and H(2, –1) are the vertices of the parallelogram. Then, reflect it in
the line x + y = 0 and write down the co–ordinates of the image parallelogram E'F'G'H'. Also,
show this reflection in the same graph.
(b) Draw the rectangle KLMN with vertices K(–2, 0), L(0, –2), M(6, 1) and N(4, 3) on the graph
paper. Reflect the rectangle KLMN in the line y – 1 = 0 and write down the co–ordinates of the
vertices of the image of the rectangle KLMN. Also, draw the image on the same graph paper.
(c) The vertices of a kite ABCD are A(–2, 0), B(0, –1), C(3, 0) and D(0, 1). Find the co–ordinates
of the vertices of the image kite A'B'C'D' under the reflection on the line x = 2. Also, draw this
transformation on the same graph paper.
9. (a) Write down the coordinates of the image of the point (2, –3) under reflecting about the line
joining the points (3, 5) and (3, 7).
(b) Find the coordinates of the image of the line segment joining the points A(2, 3) and B(4, 5)
under reflection in the line joining the points (2, 1) and (– 4, 1).
(c) Find the coordinates of the vertices of the image of the square PQRS having the vertices P(2, 0),
Q(2, –1), R(3, 0) and S(2, 1) under the reflection in the line joining the points (1, 1) and (– 4, – 4).
10. (a) The coordinates of the image of a point P are P'(–3, –1) under the reflection about the line y +
3 = 0. Find the coordinates of the point P.
(b) If A'(3, – 4), B'(– 4, 1) and C'(–1, 0) are the vertices of the image of ∆ABC under reflection
on the line y + x = 0, then find the coordinate of the vertices of ∆ABC and draw ∆ABC and its
image ∆A'B'C' on the same graph paper.
11. (a) The image of P(2, y) when reflected in the line y = –2 is P'(3x – 1, 3). Find the values of x and y.
(b) The image of the point A(x +1, 2y) is A'(4, – 4) under the reflection on the line x = 2. Find the
coordinates of A.
(c) The image of M(2 – x, 3 – y) is M'(2y – 4, 2x – 3) under the reflection in the line y = –x. Find
the coordinates of M and M'.
(d) The image of P(2a, 3 – b) when reflected in the line x – y = 0 is P'(a – 1, 2b – 2). Find the values
of a and b.
12. (a) Find the reflecting axis when the point P(– 4, 3) reflects into P'(–3, 4).
(b) The vertices of ∆RST are R(– 4, 1), S(–3, 4) and T(–2, 2). If T'(10, 2) is the image of T(–2, 2)
under reflection, find the remaining vertices of the image ∆R'S'T'.
ANSWERS
3. (a) L'(5, 3) (b) M'(5, –2) (c) N'(0, –3) (d) O'(–4, –4) (e) P'(–4, 4)
4. (a) G'(–6, –3) (b) H'(–4, 3) (c) I'(2, –2) (d) F'(–4, 0) (e) K'(4, 0)
5. (a) A'(3, 7), (b) B'(4, 4) (c) (–4, 8) (d) D'(–2, 4) (e) E'(0, 8)
6. (a) P'(–5, 4), (b) Q'(1, 0) (c) R'(–1, –4) (d) S'(0, 3) (e) T'(–2, 5) TRANSFORMATION
7. (a) A'(4, 2), B'(5, –3), C'(1, 0) (b) P'(5, –4), Q'(8, 1), R'(6, 3)
8. (a) E'(–1, 0), F'(–1, –4), G'(1, –6), H'(1, –2) (b) K'(–2, 2), L'(0, 4), M'(6, 1), N'(4, –1)
(c) A'(6, 0), B'(4, –1), C'(1, 0), D'(4, 1) 9. (a) (4, –3) (b) A'(2, –1), B'(4, –3)
(c) P'(0, 2), Q'(–1, 2), R'(0, 3), S'(1, 2) 10. (a) P(–3, –5) (b) A(4, –3), B(–1, 4), C(0, 1)
11. (a) 1, –7 (b) A(0, –4) (c) M(1, 2), M'(–2, –1) (d) 1, 3 12. (a) y = –x (b) R'(12, 1), S'(11, 4)
Project Work
Reflect a point P(x, y) in the x-axis (the line y = 0), the y-axis (the line x = 0), the lines x = a, y = b,
x = y and x = – y in the separate coordinate plane.
Introduction to Transformation and Reflection 281
71.21 ROTATION
Learning Objectives
At the end of this topic, the students will be able to:
• introduce rotation with geometric representation.
• rotate the object in coordinate plane by using rule.
Introduction
Study the following pictures to understand the rotation: 11 12 1
10 90o 2
In the adjoining fig. (i) the minute hand of a clock is turned through 90o 93
about its centre when it moves from 12 to 3. 84
In the adjoining fig. (ii) a horse is tied to a nail by a rope. The horse 765
moves through 45o in the anti-clockwise direction always keeping the rope
tight. Similarly, when it covers 180o, it reaches opposite side of its initial fig. (i)
position. If it completes 360o around the nail always keep the rope tight, it
completes one circle or one rotation.
In both figures, the objects move themselves through some angle from 180o 45o
a fixed point/position. So, their final images are also same as the initial
objects. Such types of isometric transformations are called rotations and 360o
the fixed point is called centre of rotation and the angle through which
it turns, is called the angle of rotation. In these cases, the distance of the fig. (ii)
object and its image from the centre is always equal. This constant distance
is called the radius of the rotation.
An isometric transformation in which an object moves about a fixed point through some angle in either clockwise or
anti-clockwise direction, is called a rotation.
In the rotation, the object PQ is rotated about the centre O through the angle θ with anti-clockwise direction
as shown in the adjoining figure and its image is P'Q'. We can obtain the following properties:
(i) OP = OP' and OQ = OQ'. i.e., it preserves distance. Q'
TRANSFORMATION (ii) The object PQ and its image P'Q' are of the same shape and size. P' Q
i.e., it preserves congruency. P
θ
(iii) The perpendicular bisector of the line joining the points P and P', θ
Q and Q' pass through the centre of rotation.
O
(iv) The centre of rotation is only invariant point.
(v) Every point of the object turns through the same angle in the same
direction.
The image of P is P' under the rotation through the angle θ about the centre O, which is denoted by
R(P) = P' or ρ(rho)(P) = P'. Generally, it is also denoted by P R[O; θ] P'.
282 Illustrated Optional Mathematics-9
Types of Rotation
From the study of the above activities, we have the mainly following types of rotations:
(i) Quarter turn or rotation of ± 90o.
(ii) Half turn or rotation of ± 180o. P' O A
(iii) Full turn or rotation of ± 360o. – 90o
(i) Quarter turn or rotation of ± 90o
A rotation of an object with an angle of 90o is
called quarter turn or rotation of ± 90o. If the +90o
rotation is in anti-clockwise direction, then it is O P A'
Negative quarter turn
called the positive quarter turn or the rotation of Positive quarter turn
+ 90o whereas the rotation is in clockwise direction, then it is called the negative quarter turn or the
rotation of – 90o. For example, in the adjoining figure, P' is the image of P under the rotation about
the centre O through 90o in anti-clockwise direction. So, it is the positive quarter turn or the rotation
of + 90o. P'
Similarly, in the adjoining figure, the point A +90o P + 270o A
is rotated through 90o in clockwise direction. O O
So, it is the negative quarter turn or the – 90o
rotation of – 90o. Also, we study the following – 270o
figures carefully:
A'
In the first figure, the rotation through + 90o and the rotation through – 270o about the same centre
both give the same image. Hence, these two rotations are also the same.
i.e., R[O; + 90o] = R[O; – 270o].
Similarly, in the second figure, the rotation through – 90o and the rotation through + 270o
about the same centre both give the same image. Hence, these two rotations are the same.
i.e., R [O; – 90o] = R [O; + 270o].
(ii) Half turn or rotation of ± 180o
A rotation of an object with an angle of 180o is called half turn or P' + 180o P
rotation of 180o. In the adjoining figure, P' is the image of P under the
positive rotation of 180o or the negative rotation of 180o. In both cases, O TRANSFORMATION
the images are the same. So, the both half turns are the same. – 180o
i.e., R[O; + 180o] = R[O; – 180o].
(iii) Full turn or rotation of ± 360o
A rotation of an object with an angle of 360o is called full turn or rotation + 360o P
of 360o. In this rotation, it completes one circle. In this full turn, the P'
object and its image are the same as shown in the adjoining figure. O
So, it is an identity transformation under rotation. The clockwise and – 360o
anti-clockwise direction of the rotation in the full turn are the same.
i.e., R[O; + 360o] = R[O; – 360o].
Rotation 283
Note: P'
(360o – θ)
(i) The rotation about a point through an angle θ in the anti-clockwise direction
is equivalent to the rotation about the same point through (360o – θ) in the θ
clockwise direction. OP
(ii) R denotes the positive rotation about a point through an angle θ and R –1 denotes P'
the negative rotation about the same point through the same angle θ . If R maps
P to P' and R –1 maps P' to P. In this case, the combined vertices of R and R' are θ –θ
invariant. So, R –1 is called an inverse of R. OP
(iii) The angle of rotation can be taken of any measurement (degree, grade, or radian).
Rotation in Coordinate Plane
Let us draw the image of ∆ABC under the rotation about the centre O through A' A
90o as shown in the adjoining figure. There is some difficulty to draw the BC
image of the given object in the case of rotation. So, we can use the coordinate C'
plane for this process which helps us to reduce our complexity.
O
We use the some specific angles to rotate the given object such as 90o, 180o,
270o and 360o. We take the centre of rotation as the origin O(0, 0) in this topic.
Here, we are going to rotate the given point about the centre as origin O(0, 0) B'
in each case of positive quarter turn, negative quarter turn and half turn. Y
i. Centre of Rotation as Origin
A'(–2, 3)
TRANSFORMATION A(3, 2)
The centre of rotation is different as origin or other than origin. X' B(–5, –1)O X
a) Positive quarter turn or rotation of +90o or –270o about origin
In the adjoining coordinate axes, the points A(3, 2) and B'(1, –5)
B(–5, –1) are rotated about the origin O(0, 0) through 90o in the Y'
anti-clockwise direction. We obtain the images A'(–2, 3) and
B'(1, –5) respectively.
i.e., A(3, 2) R[O; +90o] A'(–2, 3) and B(–5, –1) R[O; +90o] B'(1, –5) Y (x, y)
These show that while rotating positive quarter turn about the origin, (–y, x)
the position of coordinates are interchanging and only sign changes in +90o
the x–coordinate.
i.e., (x, y) R[O; +90o] (–y, x) and we also write that, X' O X
–270o
(x, y) R[O; – 270o] (–y, x) Y'
b) Negative quarter turn or rotation of –90o or +270o about origin Y
B'(–1, 5)
In the adjoining graph, the points A(2, 3) and B(–5, –1) are
rotated about the origin O(0, 0) through 90o in clockwise A(3, 2)
direction. We obtain the images A'(3, –2) and B'(–1, 5) for these 90o
points respectively. X' O –90o X
i.e., A(3, 2) R[O; –90o] A'(2, –3) B(–5, –1)
B(–5, 1) R[O; –90o] B'(–1, 5) A'(2, –3)
Y'
284 Illustrated Optional Mathematics-9
These show that while rotating negative quarter turn about the origin, the position of coordinates are
interchanging and only sign changes in y-coordinate.
(x, y) R[O; – 90o] (y, – x) and we also write that
(x, y) R[O; + 270o] (y, – x) Y
c) Half turn or rotation of ± 180° about origin
X' + 180o – 180 o A(3, 2)
In the adjoining graph, the point A(3, 2) is rotated about the origin O O X
through 180o in anti-clockwise or anti clockwise direction. We obtain A'(–3, –2)
the image A'(– 3, – 2)
Y'
i.e., A(3, 2) R[O; + 180o] A'(–3, –2) and A(3, 2) R[O; – 180o] A'(–3, –2).
This shows that while rotating half turn about the origin, the position of coordinates remains the
same but only change is in their sign. The half turn is also called point reflection.
i.e., (x, y) R[O; ± 180o] (–x, –y)
Note: In the case of full turn, the rotation of 360o about the origin maps the object into itself. So, it is
identity under rotation and (x, y) R[O; 360o] (x, y)
ii. Centre of Rotation of (a, b)
Here we are going to rotate the given point about the centre other than origin of (a, b) in each case of
positive and negative quarter and half turn. X
a. Positive quarter turn or rotation of + 90° or –270° about (a, b)
Let us rotate a point A(5, 3) about the centre C(2, 1) through + 90°. A'(0, 4) A(5, 3)
Plot the point A(5, 3) and the centre of rotation C(2, 1) on the same 90°
graph. Draw L-shaped line from A to M and M to C as shown in
the adjoining figure and then rotate the L-shaped line AMC through X' O C(2, 1) MX
90° about the point C as shown in the figure. The coordinates of the
image A' of A will be at (0, 4). Y'
i.e. A (5, 3) R[(2, 1); + 90o] A'(0, 4)
This is done some difficult to draw the object and its image. Now Y Y1 TRANSFORMATION
we formulate the system of rotation about the centre (a, b) through
90° as given below: P'(–y + b + a, x – a + b)
P(x, y)
Let P(x, y) be a point and C(a, b) is the centre of rotation. Draw
new axes as X1CX1' and Y1CY1' with the origin as C(a, b). 90°
Here, the coordinates of P is (x, y) with respect to the origin O(0, X1' C(a, b) X1
0) and the coordinates of P are (x – a, y – b) with respect to the X' O Y1' X
origin C(a, b). Y'
Now rotate P(x – a, y – b) through 90° about the origin as C(a, b), we get
P(x – a, y – b) R[O; + 90o] P'(– y + b + a, x – a + b)
Rotation 285
Again, the coordinates of P' with respect to the actual origin O(0, 0) are (– y + b + a, x – a + b).
Hence, the coordinates of the image of the point P(x, y) under the rotation of + 90° about the centre
C(a, b) are P'(– y + a + b, x – a + b).
i.e. P(x, y) R[(a, b), + 90°/ – 270°] P'(– y + a + b, x – a + b)
YY
P'(– y + b, x – a) P'(–y + a + b, x – a + b)
P(x – a, y – b) P(x, y)
X' O (0, 0) X C(a, b) X
X' O
Y' Y'
b. Negative quarter turn or rotation of – 90° or + 270° about (a, b)
Let P(x, y) be a point and C(a, b) is the centre of rotation. Y
Draw new axes Y1CY1 with the origin C(a, b). Here, the Y1
coordinate of P is (x, y) with respect to the origin O(0, 0)
and the coordinates of P are (x – a, y – b) with respect to P(x, y)
the origin (a, b). Now, rotate P(x – a, y – b) through – 90°
about the origin of C(a, b), we get C(a, b) 90°
X1
P(x – a, y – b) R[0, – 90°] P'(y – b, – x + a) X' O X
Y' P'(y – b + a, – x + a + b)
Again, the coordinates of P' with respect to the actual
origin O(0, 0) are (y – b + a, – x + a + b).
Hence, the coordinates of the image of the point P(x, y) under the rotation of – 90° about the centre
C(a, b) are (y + a – b, – x + a + b).
i.e. P(x, y) R[(a, b), – 90°/ + 270°] P'(y + a – b, – x + a + b)
c. Half turn or rotation of ± 180° about (a, b) Y
TRANSFORMATION In the adjoining figure, P(x, y) is a point with respect to the origin Y1
O(0, 0) and the coordinates of P are (x – a, y – b) with respect to the P(x, y)
origin as C(a, b). Now, rotate P(x – a, y – b) through + 180° about
the origin of C(a, b), we get C(a, b) X1
P(x – a, y – b) R[0, – 180°] P'(– x + a, – y + b)
X' X
Again, the coordinates of P' with respect to the origin O(0, 0) are P' (–x + 2a , – y + 2b))
(– x + a + a, – y + b + b) = (–x + 2a, – y + 2b) Y'
i.e. P(x, y) R[(a, b), ± 180°] P'(2a – x, 2b – y)
286 Illustrated Optional Mathematics-9
A
A' C
Example 1 : Rotate the adjoining figure through 60o about the C1 B D
given point in anti clockwise direction. D'
A
Solution: Rotate the given figure through 60o about the point C B'
in anti clockwise direction.
C
C
BD
Example 2 : Rotate the point A(4, –5) through + 90o about the origin. Write the coordinates of its
image.
Solution: Rotating the point A(4, –5) through +90o about the origin, we have
(x, y) R[O; +90o] (–y, x)
∴ A(4, –5) R[O; 90o] A'(5, 4)
Hence, the coordinates of the images of A(4, –5) are A'(5, 4) under R[O; + 90o].
Example 3 : Find the coordinates of the vertices of the image of ∆ABC with the vertices A(0, 4),
B(4, 1) and C(2, 6) under the negative quarter turn about the origin. Also draw this
transformation in the graph. Y
Solution: Given, the vertices of ∆ABC are A(0, 4), B(4, 1) and C(2, 6). C
Now, rotating the vertices of ∆ABC under the negative quarter A B
O A' X
turn about the origin, we have
C'
(x, y) R[O; – 90o] (y, –x)
∴ A(0, 4) A'(4, 0) X'
B'(1, –4)
B(4, 1)
C(2, 6) C'(6, –2)
Hence, the required coordinates of the vertices of the B'
image ∆A'B'C' are A'(4, 0), B'(1, –4) and C'(6, –2). This
transformation is drawn in the graph alongside. Y'
Example 4 : The points P(3, 1), Q(3, – 4) and R(6, – 4) are the vertices of ∆PQR. Find the coordinates TRANSFORMATION
of the vertices of the image ∆P'Q'R' of ∆PQR under the rotation about the centre (1, 2)
through 180°. Also draw ∆PQR and ∆P'Q'R' on the same graph.
Solution : Given, the points P(3, 1), Q(3, – 4) and R(6, – 4) are the vertices of ∆PQR.
Now, rotating the vertices of ∆PQR about the centre (1, 2) through 180°, we have
(x, y) R[(1, 2), 180o] (– x + 2a, – y + 2b)
∴ P(3, 1) P'(– 3 + 2 × 1, – 1 + 2 × 2) = P'(– 1, 3)
Q(3, – 4) Q'(– 3 + 2 × 1, – (– 4) + 2 × 2) = Q'(– 1, 8)
R(6, – 4) R'(– 6 + 2 × 1, – (– 4) + 2 × 2) = R'(– 4, 8)
Rotation 287
Therefore, the coordinates of the vertices of the image ∆P'Q'R' of ∆PQR are p'(– 1, 3),
Q'(–1, 8) and R'(– 4, 8)
Drawing of ∆PQR and ∆P'Q'R' on the same graph are given below:
X
R' Q'
180° X
P'
(1, 2) P
X' O
Y' Q R
Example 5 : Under a certain rotation about the origin, the images of points P(2, 3) and Q(a, –2) are
P'(– 3, 2) and (b, 1 + 2a). Find the values of a and b.
Solution: Here, the image of P(2, 3) is P'(–3, 2) under a certain rotation about the centre as origin.
Since we have the point (x, y) has the image (–y, x) under the rotation of +90o about the
centre as origin, so the rotation which transfer the point P(2, 3) into P'(–3, 2) through the
angle of rotation + 90o with centre (0, 0).
Now, rotating through 90o about the origin, we get
Q(a, –2) R[O; +90o] Q'[2, a]. But, by question, Q(a, –2) R[O; +90o] Q'(b, 1 + 2a)
∴ (2, a) = (b, 1 + 2a) ∴ b = 2 and a = 1 + 2a or, a = –1.
Hence, the values of a and b are –1 and 2 respectively.
Example 6 : A line segment AB joining the point A(2, 3) and B(3, – 1) is rotated by a certain angle
about the point (a, b) the images A'(– 4, 5) and B'(0, 6) are formed. Find the angle and
centre of rotation by using graph. Y
Solution: Drawing the line segment AB joining the points B'
A(2, 3) and B(3, –1) and its image line segment A'
TRANSFORMATION joining the points A'(–4, 5) and B'(0, 6) on the
same graph. Now, join AA' and BB'. Draw the A
perpendicular bisector of AA' and BB' which 90o
intersect at the point C(–2, 1). This point (–2, 1) is C(–2, 1)
the center of rotation. X' X
O
Again, join the points A and A' with C and measure B
the angle ACA' by using protector, we get Y'
∠ACA' = 90o.
Hence, the angle of the rotation is 90o in anti–clockwise direction.
288 Illustrated Optional Mathematics-9
EXERCISE - 7.2
1. (a) Define:
(i) Rotation (ii) Half-turn (iii) Full Turn
(iv) Positive Quarter Turn (v) Negative Quarter Turn
(b) Write the co-ordinates of the image of a point A(x, y) when it is rotated around the origin
through 90°.
(c) What is the image of a point P(x, y) when it is rotated through 180° about the point (a, b)?
(d) Write down the images of the points A(a, b) and B(c, d) when they are rotated through 270°
about the centre (0, 0).
2. Rotate the following figures about the given point though the following instructions:
A PS AQ
A RP WZ
OB
BC QR BD XOY
R[O; +90o] O
C P O R[O; –90o]
R[O; 180o]
R[C; +270o] R[P; +90o] R[O; –180o]
3. Write down the image of the point P(–3, –4) through the following angles of rotation about the centre as
origin.
(a) 90o (b) – 90 (c) 180o (d) 270o (e) 360o
4. Rotate the line segment AB joining the points A(–1, 2) and B(4, 2) taking (0, 0) as centre of rotation
and the following angles of rotation:
(a) Positive quarter turn (b) Negative quarter turn
(c) Half turn (d) Full turn
5. (a) The vertices of ∆ABC are A(–1, 1) B(–4, 2) and C(–2, 5). Rotate ∆ABC and find its image
under the rotation about the origin through 90o. Draw this rotation in the graph.
(b) Rotate a parallelogram PQRS having the vertices P(3, –7), Q(5, –5), R(6, –1) and S(4, –3)
about the origin through 180o and write down the coordinates of the vertices of the image of
the parallelogram PQRS. Also, draw the above rotation on the same graph.
(c) Draw the square ABCD having the vertices A(1, 2), B(–1, 4), C(–3, 2) and D(–1, 0) in the
graph and rotate it about the centre as O(0, 0) through quarter turn in clockwise direction.
Write down the coordinates of the vertices of the image square A'B'C'D' and also draw the
image in the same graph.
6. (a) P(4, 5), Q(1, 1) and R(5, 1) are the vertices of ∆PQR. Rotate it about the point (0, 0) through TRANSFORMATION
negative quarter turn and write the coordinates of the image ∆P'Q'R. Also, draw the above
transformation in the graph.
(b) Find the coordinates of the vertices of the images of a rectangle GHEF with the vertices of
G(4, –1), H(7, – 4), E(5, – 6) and F(2, – 3) under the rotation of [(0, 0); 90o] and represent this
rotation in the graph.
(c) Draw ∆ABC with the vertices A(–1, –1), B(–5, –2) and C(–3, – 4) on the graph and rotate
∆ABC by [(0, 0), 180o] on the same graph. Write down the coordinates of the images ∆A'B'C'
of ∆ABC.
Rotation 289
7. (a) The image of the point P(1, 1) and Q(3, 4) are P'(–1, 1) and Q'(–4, 3) under the certain rotation.
Find the centre and angle of rotation by using graph.
(b) The images of the points A(3, –b) and B(a, –4) are A'(–3, –3) and B'(–4, 2) under the rotation
through –90o about (0, 0). Find the values of a and b.
(c) If M'(2, 3) and N'(–2, 5) are the image of M(–3, 2) and N(x, y) under the rotation about (0, 0)
through certain angle then find the coordinates of N.
8. Find the image of a point (2, – 2) under the following rotation:
(a) R[(1, 2); + 90°] (b) R[(0, 2); 180°]
(c) R[(–2, 0); – 90°] (d) R[(– 3, – 1); –270°]
9. (a) P(4, 5), Q(1, 1) and R(5, 1) are the vertices of ∆PQR. Rotate it about the point (3, 2) through
negative quarter turn and write the coordinates of the image ∆P'Q'R'. Also draw the above
transformation in the graph.
(b) Find the coordinates of the vertices of the image of a rectangle GHEF with the vertices
G(4, – 1), H(7, – 4), E(5, – 6) and F(2, – 3) under the rotation of [(2, 1); 90°] and represent this
rotation in the graph.
(c) Draw ∆ABC with the vertices A(– 1, – 1), B(– 5, – 2) and C(– 3, – 4) on the graph and rotate
∆ABC by [(– 1, 3), 180°] on the same graph. Write down the coordinates of the image ∆A'B'C'
of ∆ABC.
ANSWERS
1 and 2 Show to your teacher 3. (a) P'(4, –3) (b) P'(–4, 3) (c) R'(3, 4) (d) P'(– 4, 3)
(e) P'(–3, – 4)
4. (a) A'(–2, –1), B'(–2, 4) (b) A'(2, 1), B'(2, – 4) (c) A'(1, –2), B(– 4, –2) (d) A'(–1, 2), B'(4, 2)
5. (a) A'(–1, –1), B'(–2, – 4), C'(–5, –2) (b) P'(–3, 7), Q'(–5, 5), R' (– 6, 1), S'(– 4, 3)
(c) A'(2, –1), B'(4, 1), C'(2, 3), D'(0, 1) 6. (a) P'(5, – 4), Q'(1, –1), R'(1, –5)
(b) G'(1, 4), H'(4, 7), E'(6, 5), F'(3, 2) (c) A'(1, 1), B'(5, 2), C(3, 4)
7. (a) R[O; +90o] (b) – 2, 3 (c) (–5, –2)
8. (a) (5, 3) (b) (–2, 6) (c) (– 4, – 4) (d) (–2, 4)
9. (a) P'(6, 1), Q'(2, 4), R'(2, 0) (b) G'(4, 3), H'(7, 6), E'(9, 4), F'(6, 1)
(c) A'(–1, 7), B'(3, 8), C'(1, 10)
TRANSFORMATION Project Work
Rotate a point P(x, y) through the angles 90°, 180° and 270° in anti-clockwise and clock-wise direction
about the center as the origin (0, 0) and the point (a, b) in the separate coordinate plane.
290 Illustrated Optional Mathematics-9
71.31 TRANSLATION
Introduction to Translation
Study the following activities to clear the translation: Q A
In the adjoining fig. (i), a bus has moved some
distance on the straight smooth road in the B
specific direction PQ . Similarly, in the picture P Bus Straight Road
(ii), the boy has removed the illustrated Opt
Math from a position A to another position B in fig. (i) fig. (ii)
right. In the both activities, the bus and the book
move some distance in specific direction called vector. These types of isometric transformation on the basis
of vector is called translation. It slides every point of a figure with the same distance in the same direction.
An isometric transformation in which an object maps with its image under given vector is called a translation.
In translation, the line segment AB is moved along PQ as shown in the Q
adjoining figure and its image is A'B'. Then we obtain the following properties: A'
(i) AB = A'B', i.e., it preserves distance. P B'
(ii) AB//A'B', i.e., it preserves direction. A
(iii) P and P' are congruent i.e., it preserves congruency. B
(iv) The magnitude of PQ is equal to the distance between the object and its image and PQ is parallel
to AA' or BB'. i.e., | PQ | = AA' = BB'.
(v) The object P moves to P' from one position to another.
(vi) The image of a point with null translation vector is invariant.
The image of P is P' under translation, which is denoted by T(P) = P' or τ(P) = P' [τ is read as Tau].
A vector on which the object moves into its image in the translation is called translation vector, which gives the
magnitude and direction of the way of translating object.
Translation in Coordinates Plane Y TRANSFORMATION
C' (1, 6)
In the adjoining graph, the position vector of P A' (6, 5)
B (8, 2)
is 4 . i.e., OP = 4 . Now, let us translate the C (–3, 4) B' (12, 4)
2 2 O X
points A(2, 3), B(8, 2), C(– 3, 4) and D(–1, –2) A (2, 3) P
under the translation vector OP that means 4
2
each point is moved 4 units right and 2 units X'
D' (3, 0)
upward. Then the images are A'(6, 5), B'(12,
4), C'(1, 6) and D'(3, 0) respectively. D (–1, –2)
Y'
Translation 291
Similarly, in the adjoining graph the column vector of AB Y
is –4 . Now let us translate the points P(8, 1), Q(–1, 1), Q'(– 5, 4) P'(4, 4)
3
R(3, –2) and S(–1, –2) under AB that means each points B –4
Q(–1, 1) 3 P(8, 1) X
are moved 4 units left and 3 units upward. Then, we find S'(– 5, 1)
X' O A
their images P'(4, 4), Q'(–5, 4), R'(–1, 1) and S'(–5, 1) S(–1, –2) R(3, – 2)
respectively. Y'
What do we find from these activities ? We arrange the
rule for these translations.
We obtain the coordinate of the image A' of the point A(2, 3) under the translation vector OP = 4
2
4
i.e, A(2, 3) 2 A' (6, 5) = A' (2 + 4, 3 + 2).
4
Similarly, B(8, 2) 2 B' (12, 4) = B' (8 + 4, 2 + 2).
4
C(–3, 4) 2 C' (1, 6) = C'(–3 + 4, 4 + 2).
4
D(–1, –2) 2 D' (–3, 0) = D'(–1 + 4, –2 + 2).
Also, we obtain the coordinate of the image P' of the point P(8, 1) under the translation vector AB = –4 .
–4 3
i.e., P(8, 1) 3 P' (4, 4) = P' (8 – 4, 1 + 3).
–4
Similarly, Q(–1, 1) 3 Q' (–5, 4) = Q' (–1 – 4, 1 + 3).
–4
R(3, –2) 3 R' (–1, 1) = R' (3 – 4, –2 + 3).
–4
S(–1, –2) 3 S' (–5, 1) = S' (–1 – 4, –2 + 3).
From the above two patterns we found that the components of the given translation vector are added to
respective coordinates of the point that means x–component and y-component of the vector are added to
TRANSFORMATION x-coordinate and y-coordinate of the points respectively. Let us suppose that (x, y) is a point and translate
a
it under the vector b , then the coordinate of its image will be (x + a, y + b)
i.e., x a x+a or, (x, y) a (x + a, y + b),
y b y+b b
where x is the position vector of (x, y) and x+a is the position vector of the image point of
y y+b
a
(x, y) under the translation vector b .
Note: If (x', y') is the image of (x, y) under the translation vector T = a , then T = a = x' – x .
b b y' – y
292 Illustrated Optional Mathematics-9
Example 1 : Translate the point A(4, –3) by the translation vector –2 . Write the coordinates of
the image of A. 4
Solution: Here,
Translating the point A(4, –3) by the vector –2 , we get
4
–2
A (4, –3) 4 A'(4 – 2, –3 + 4) = A'(2, 1)
Hence, the required coordinate of the image are A'(2, 1).
Example 2 : Plot the vertices P(1, 4), Q(3, 2) and R(4, 5) of ∆ABC on the coordinate plane. Find the
3
coordinate of the image of ∆ABC under the translation T = –1 . Also draw the image
on the same plane.
Solution: Here, plotting the vertices P(1, 4), Q(3, 2) and R(4, 5) of ∆ABC on the coordinate plane.
Now, translating these vertices under the vector T = 3 .
–1
We obtain the following images: Y
Q(1, 4) R(4, 5)
3 R'(7, 4)
P(1, 4) – 1 P' (1 + 3, 4 – 1) = P' (4, 3). P'(4, 3)
Q(3, 2) 3 Q' (3 + 3, 2 – 1) = Q' (6, 1). X' Q(3, 2) Q'(6, 1)
R(4, 5) –1 R' (4 + 3, 5 – 1) = R' (7, 4). X
O3
3 –1
–1
Y'
Again, the image ∆P'Q'R' is drawn on the same plane.
Example 3 : If the image of the point A(3, 2) is A'(–5, 5) under the translation, find the image of the
point B(4, –2) under the same translation.
Solution: Let a be a translation vector for the given translation, then the image of the point A(3, 2)
b
a
under b is A' (3 + a, 2 + b).
But by question, the image of A(3, 2) is A'(– 5, 5). TRANSFORMATION
Therefore, (3 + a, 2 + b) = (– 5, 5)
∴ 3 + a = – 5 and 2 + b = 5 or, a = – 8 and b = 3.
Hence, the translation vector is –8 .
3
Again, the image of the point B(4, –2) under the same translation –8 is given by,
–8 3
B(4, –2) 3 B' (4 – 8, –2 + 3) = B' (– 4, 1).
Translation 293
Example 4 : If (x, – 4) is the image of a point (3, y) under the translation vector –3 , then find the
values of x and y. –4
Solution: Here, (x, –4) is the image of a point (3, y) under the translation vector – 3 .
– 4
So, (3 – 3, y – 4) = (x, –4) or, (0, y – 4) = (x, – 4)
∴ x = 0 and y – 4 = –4, or y = 0.
Hence, the value of both x and y is 0.
Example 5 : Find the coordinates of the vertices of the square ABCD having the vertices A(2, 0),
B(5, 2), C(3, 5) and D(0, 3) under the translation vector AC . Draw the above translation
on the graph paper.
Solution: Given,
The vertices of the square ABCD are A(2, 0), B(5, 2), C(3, 5) and D(0, 3).
The column vector of = AC = x2 – x1 = 3–2 = 1 .
y2 – y1 5–0 5
1 Y
Now, translating the vertices of the square ABCD under 5 ,
C'(4, 10)
we get, D'(1, 8) B'(6, 7)
1
A(2, 0) 5 A' (2 + 1, 0 + 5) = A' (3, 5).
1 A' C(3, 5)
5
B(5, 2) B' (5 + 1, 2 + 5) = B' (6, 7). D(0, 3)
C(3, 5) 1 C' (3 + 1, 5 + 5) = C' (4, 10).
D(0, 3) 5 D' (0 + 1, 3 + 5) = D' (1, 8). B(5, 2)
1 X' O A(2, 0) X
5
Y'
The required coordinates of the vertices of the image square
A'B'C'D' are A'(3, 5), B'(6, 7), C'(4, 10) and D'(1, 8).
Drawing of the above translation on the graph paper is given in the adjacent graph.
Example 6 : Find the image of the point (4, 5) under the translation (x, y) → (x + 2, y – 2).
TRANSFORMATION Solution: Since (x, y) → (x + 2, y – 2). So, the translation vector is 2 .
–2
Now, translating the point (4, 5) under 2 , we get
–2
2
(4, 5) – 2 (4 + 2, 5 – 2) = (6, 3).
Hence, the image of (4, 5) is (6, 3).
"Alternatively"
The image of the point (4, 5) under the translation (x, y) → (x + 2, y – 2) is given by
(4, 5) → (4 + 2, 5 – 2) = (6, 3).
294 Illustrated Optional Mathematics-9
EXERCISE - 7.3
1. (a) Write down any one property of translation.
(b) Which type of isometric transformation is called translation?
(c) What is the image of a point T(p, q) under translation by the vector r ?
s
2. Draw the image of the following figures under the given vectors:
(a) B (b) P (c) A D
R a
C
A Q a B
p T
(e) (f)
(d) K s b
P S c
LN
M QR
3. Translate the following figures by the given translation vectors:
(a) (b) (c) (d)
A Q
P
B4 R3 3 –3
1 –2 4 –4
4. Find the translation vector from the following graphs: (d) Y
(a) Y (b) Y (c) Y
A A' A X' X
X' O X A' X
O
X' BC X X' O
O
B' C'
Y' Y' Y' Y'
5. Write the images of the following points by the given translation vector:
(a) A (1, 4) by 3 (b) B(4, –2) by 1 (c) C(– 3, 1) by –1 (d) D (13, 9) by –8 TRANSFORMATION
2 –3 2 0
6. (a) Find the coordinates of the points of the image of a line segment joining the points A(2, 3) and
B(– 3, – 5) after translating by a vector 2 .
–4
(b) Translate a line segment PQ joining the points P(– 2, 0) and Q(3, – 4). Write down the
coordinates of the image translated by the translation vector –1 .
3
7. Plot the following vertices of the geometric shape on the coordinate plane and find the coordinate of
2
the images under the translation vector –5 . Also draw the image on the same coordinate plane.
(a) P(0, 3), Q(–3, 1) and R(–2, 4) of ∆PQR.
(b) A(1, 2), B(5, 1), C(4, 4) and D(–1, 5) of the quadrilateral ABCD.
Translation 295
8. Find the coordinates of the vertices of the image of the figure having the following vertices under the
–3
translation vector 2 and draw this translation on the graph paper:
(a) The parallelogram ABCD having the vertices A(2, –1), B(–1, 2), C(–3, 1) and D(0, –2).
(b) The rhombus PQRS having the vertices P(2, 1), Q(–1, 0), R(0, –3) and S(3, –2).
9. (a) Find the coordinates of the vertices of the image of the kite ABCD having the vertices A(6, 1),
B(4, –1), C(6, –5) and D(8, –1) under the translation vector BD. Draw the above translation in
the graph paper.
(b) The vertices of a square PQRS are P(3, 4), Q(6, 1), R(9, 4) and S(6, 6). Translate the square
PQRS by the vector SQ + QP and then draw the above transformation on the square grid.
10. (a) Find the image of a point (3, –4) under the translation (x, y) → (x – 3, y + 1).
(b) Find the coordinates of the vertices of the image of ∆ABC with vertices A(–4, 2), B(–1, –2)
and C(–2, 1) under the translation (x, y) → (x + 2, y – 1).
11. (a) If the image of the point (3, – 4) is (1, 5) under the translation vector T = a , find T and then
find the image of the point (– 4, –2) by the same translation. b
(b) The vertices of the rectangle OABC are O(0, 0), A(3, 2), B(1, 5) and C(–2, 3). If the image of
O is (–4, 3), find the coordinate of other vertices of the image of the rectangle OABC. Draw
this translation on the same graph.
12. (a) If (2x – 1, 4) is the image of the point (3, 3y + 1) under the translation vector 4 , find the
values of x and y. 6
(b) If the image of the line segment PQ joining the points P(2x – y, 3) and Q(4, 2) are P'Q' joining
the points P'(1, 2x + 1) and Q'(3, 4) under same translation vector, find the values of x and y.
ANSWERS
1. to 3. Show to your teacher.
4. (a) 4 (b) 4 (c) – 1 (d) 1 5. (a) A' (4, 6) (b) B'(5, –5)
2 –1 –4 –3
(c) C'(–4, 3) (d) D'(5, 9) 6. (a) A' (4, –1), B'(–1, –9) (b) P'(–3, 3), Q'(2, –1)
7. (a) P' (2, –2), Q'(–1, – 4), R'(0, –1) (b) A'(3, –3), B'(7, – 4), C'(6, –1), D'(1, 0)
8. (a) A' (–1, 1), B'(– 4, 4), C'(– 6, 3), D'(–3, 0) (b) P' (–1, 3), Q'(– 4, 2), R'(–3, –1), S'(0, 0)
TRANSFORMATION 9. (a) A'(10, 1), B'(8, –1), C'(10, –5), D'(12, –1) (b) P'(0, 2), Q'(3, –1), R'(6, 2), S'(3, 4)
10. (a) (0, –3) (b) A'(–2, 1), B'(1, –3), C'(0, 0) 12. (a) 4, – 1 (b) 2, 2
11. (a) (– 6, 7), –2 (b) A'(–1, 5), B'(–3, 8) c) (–6, 6)
9
Project Work
Translate a point P(x, y) by the vector AB joining the points (x1, y1) and (x2, y2) with suitable diagram.
296 Illustrated Optional Mathematics-9
71.41 ENLARGEMENT AND REDUCTION
Learning Objectives
At the end of this topic, the students will be able to:
• define enlargement and reduction with its geometric representation.
• enlarge in coordinate plane by using rule.
Introduction
Study the following pictures to understand the enlargement:
In the above pictures, the objects and their images are different in size, but same shape. Here, the object
and its image are not congruent. So, such types of transformation are the non–isometric transformation.
In picture (i), the shadow of the boy casts on the screen, which is larger than the boy. So, this type of
transformation is called enlargement.
In picture (ii), the photocopy machine prints the photo of animal smaller than the original photo. So, this
type of transformation is called reduction.
But, in picture (iii), the camera taken a photo of a man, whose image lies on the negative (reel) of the
camera in backside and opposite or inverse. This is also a type of reduction but inversion.
A non-isometric transformation in which the image of an object is larger than the object about a centre, is called an
enlargement. It is also called a dilation.
A non-isometric transformation in which the image of an object is smaller than the object about a centre is called a reduction.
Notation of enlargement and reduction A' TRANSFORMATION
A
In the adjoining figure, the line segment AB is enlarged in reference to the point C to the
image A'B'. Here, AB is parallel to A'B'. So, ∆ABC and ∆A'C'B' are similar triangles and
the ratios of their sides are equal.
CA CB AB CB B'
i.e., CA' = CB' = AB' = k (say)
This constant ratio k is called a scale factor and the point C is called the centre of the enlargement.
The constant ratio of the length of image to object from the centre of the enlargement is called a scale factor of the
enlargement. It is generally denoted by k.
A point of intersection of the producing lines joining the point of the object and its corresponding image is called a
centre of enlargement. It is generally denoted by O or C. Thus, the enlargement is denoted by E[O/C; k].
Enlargement and Reduction 297
The size of image of the object depends upon the scale factor k. We discuss the following examples:
P' M X
P M' Y'
C L' L OY
O Q Q' N' X'
N
fig. (i) E[O; 2] –3
1 fig. (iii) E[O; 2 ]
fig. (ii) E[C; 2 ]
In the fig. (i), OP' = 2OP and OQ' = 2OQ. So, P'Q' = 2PQ. Here, P'Q' is larger than PQ.
Similarly, in the fig. (ii), CL' = 1 CL, CM' = 1 CM and CN' = 1 CN. So, ∆L'M'N' = 1 ∆LMN. Here,
2 2 2 2
∆L'M'N' is similar to ∆LMN.
In the fig. (iii), OX' = –3 OX and OY' = –3 OY. So, X'Y' = –3 XY. That means the image X'Y' forms
2 2 2
backside of the centre of the enlargement and is the inverse of XY.
The image of an object depends upon the nature of the scale factor k of the enlargement as follows:
(a) If k > 1 then the object is enlarged or the size of the image gets bigger than the object.
(b) If 0 < k < 1 then the object is reduced or the size of the image gets smaller than the object.
(c) If k < 0 then the object is inverse or the image casts behind the centre of enlargement and inverse.
(d) If k = 1 then the object and its image are the same. It is called an identity enlargement. In this
case, the object and its image overlap to each other or the object and image lie in different
position, the centre of the enlargement is at infinity. i.e., the line joining the object points and the
image points are parallel. For example; A A' A A'
Here, in fig. (iv) AB and A'B' are overlapped and in O
fig. (v) AB and A'B' are in different places so, the
centre of enlargement does not appear and AA'//BB'. B B' B B'
Thus, the centre of enlargement may be in the fig. (iv) E[O; 1] fig. (v) E[O; 1]
following conditions. For example:
A D' A' P' L L'
A' DA P N N'
O C' C O O M M'
CB d. At infinity
TRANSFORMATION B' C' B' Q
R
B b. Inside the object
a. Outside the object Q' R'
c. At the boundary of the object
We obtain the following conclusions from the above discussion:
(i) The object and its image under enlargement are not congruent but the same in shape.
So, ∆ABC ~ ∆A'B'C'. So, it preserves the similarity.
(ii) The size of the image of the object is changed.
(iii) The ratios of the length of sides of the object and the length of the corresponding side of its
image are constant, which is called scale factor of the enlargement.
(iv) If the scale factor is greater than 1 then the image of an object will be bigger.
298 Illustrated Optional Mathematics-9
(v) If the scale factor is greater than 0 and less than 1 then the image of an object will be smaller.
(vi) If the scale factor is less than 0, then the image will be inversion. i.e., the object and its image
will be either direction of centre of enlargement.
(vii) If the object and its image lie on the same side of the centre then the scale factor is positive,
otherwise it is negative.
Enlargement in Coordinate Plane
We can use the coordinate plane to identify the image of the given object in the enlargement. In the
coordinate plane, the centre of the enlargement is either the origin or other than origin as any point (a, b).
So, we have two cases for the enlargement of the object according as its centre.
(i) Centre of enlargement as origin (0, 0) and scale factor k. Y
(ii) Centre of enlargement as any point (a, b) and scale factor k. A'(4, 6)
(i) Centre of enlargement as origin (0, 0) and scale factor k A(2, 3)
Let us enlarge a line segment AB joining the points A(2, 3) and B(2, –1) X' OX
about the centre O(0, 0) by scale factor 2. Then the image line segment A'B' B(2, –1) A(4, – 2)
has the points A'(4, 6) and B'(4, –2) as shown in the adjoining graph.
Y'
i.e., A(2, 3) E[O; 2] A'(4, 6) = 2 times of A(2, 3)
B(2, –1) E[O; 2] B'(4, –2) = 2 times of B(2, –1)
Here, the coordinates of the image are 2 times of the respective coordinates of the object. Hence, if the
centre of enlargement is taken as origin O(0, 0) and scale factor k, then the coordinates of the image of a
point (x, y) are (kx, ky).
i.e., (x, y) E[O; k] (kx, ky).
(ii) Centre of enlargement at any point (a, b) and scale factor k:
Let us enlarge a line segment PQ joining the point P(4, 3) and Q(4, 0) about the centres C(2, 1) by scale factor
2, then how does the image line segment P'Q' get the coordinate point change? For this, we use the vector.
CP = OP – OC = 4 – 2 = 2 Y
3 1 2 P'(6, 5)
Since the scale factor is 2 so,
CP' = 2 CP = 2 2 = 4 , P(4, 3)
2 4
OP' = OC + CP' = 2 + 4 = 6 TRANSFORMATION
1 4 5
X' C
Hence, the coordinates of P' is (6, 5). O Q' X
Similarly, CQ = OQ – OC = 4 – 2 = 2 . (4, 0) Q'(6, –1)
0 1 –1
Y'
= 2( –21) = 4
CQ' = 2CQ –2 .
OQ' = OC + CQ' = 2 + 4 = 6 .
1 –2 –1
Hence, the coordinates of Q are (6, –1).
In general, we compute the rule of the enlargement about (a, b) with scaler factor k.
Enlargement and Reduction 299
In the adjoining figure, P'(x', y') is the image of P(x, y) under E[(a, b); k], then
OP' = OC + CP' Y
or, x' = OC + k CP = OC + k (OP – OC )
y'
a x a a kx ka (a, b) P (x, y) P'(x', y')
= b +k y – b = b + ky – kb C
= kx + a – ka = kx + a(1 – b) X' X
ky + b – kb ky + b(1 – k) O
Y'
∴ x' = kx – a(k – 1), y' = ky – b(k –1).
Hence, the coordinates of P' are (kx – a(k –1), ky – b(k – 1).
i.e., (x, y) E[(a, b); k] (kx – a(k – 1), ky – b(k – 1))
or, (k(x – a) + a, k(y – b) + b)). Note: These both rules are satisfied
for reduction and inversion.
or, (kx + a(1 – k), ky + b(1 –k))
or, (a – k (a –x), b – k(b – y)).
Example 1 : Find the coordinates of the image of a point P(3, – 4) under the enlargement with
centre (0, 0) and scale factor 3.
Solution: Enlarging the point P(3, 4) about the centre (0 , 0) with scalar factor 3,
(x, y) E[O; k] (kx, ky)
∴ P(3, –4) E[O; 3] P'(3 × 3, 3 × (–4)) = P' (9, –12).
Hence, the coordinate of the image point P' is (9, –12).
Example 2 : Enlarge the line segment AB joining the points A(4, – 6) and B(– 8, – 4) by E (0, 0); 1
and E[(2, –1); 2] separately. 2
Solution: Enlarging the line segment AB joining the points A(4, – 6) and B(– 8, – 4) by E (0, 0); 1 ,
2
we have
(x, y) E[(0, 0); k] (kx, ky)
∴ A(4, – 6) E[(0, 0); 1 ] 11
2 A' 2 × 4, 2 × (– 6) = A'(2, – 3).
TRANSFORMATION B(– 8, – 4) E[(0, 0); 1 ] 11
2 B' 2 × (–8), 2 × (– 4) = B'(– 4, – 2).
Again, enlarging the same line segment AB by E[(2, –1); 2], we have
(x, y) E[(a, b); k] [kx – a(k – 1), ky – b(k – 1)]
∴ A(4, – 6) E[(2, – 1); 2] A'[2 × 4 – 2(2 – 1), 2 × (–6) – (–1)(2 – 1)]
= A' (8 – 2, – 12 + 1) = A'(6, – 11).
∴ B(–8, –4) E[(2, –1); 2] B'[2 × (– 8) – 2(2 – 1), 2 × (– 4) – (–1)(2 – 1)]
= B' (–16 – 2, –8 + 1) = B'(–18, –7).
300 Illustrated Optional Mathematics-9
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