|PQ | = |QR| then PQ and QR are called equal vectors.
i.e. PQ = QR. P R
Q
Also, two vectors →a = x1 and →b = x2 are equal if x1 = x2 and y1 = y2
y1 y2
(x) Parallel (collinear) vectors:
Two vectors are said to be parallel (collinear) if they are like or unlike. Let two
vectors →a and →b are said to be parallel if →a = k→b or →b = m→a where k and m are
scalars (positive or negative.)
WORKED OUT EXAMPLES
1. If AB = –3 find magnitude of AB.
–4
Solution: Here,
AB = –3
–4
Here, x-component of AB (x) = –3
y-component of AB (y) = – 4
By formula,
|AB | = x2 + y2 = (–3)2 + (–4)2 = 9 + 16 = 25 = 5 units
∴ The magnitude of AB = |AB | = 5 units.
2. If PQ displaces the point P(–7, –3) to the point Q(–3, –8), find the vector
PQ , |PQ |, the direction of PQ and a unit vector along PQ .
Solution: Here
Let, P(–7, –3) = (x1, y1)
Q(–3, –8) = (x2, y2)
Now,
(i) PQ = x2 – x1 = –3 + 7 = 4
y2 – y1 –8 + 3 –5
Infinity Optional Mathematics Book - 9 251
(ii) Here, x-component of PQ (x) = 4
y-component of PQ (y) = –5
∴ |PQ | = x2 + y2 = 42 + (–5)2 = 16 + 25 = 41 units
(iii) For the direction of PQ .
If θ be the angle between PQ and positive x-axis, then,
tanθ = y = –5
x 4
or, θ = tan-1 –5 = –51.34
4
Least positive angle = 51.34°
Since, x-component of PQ is positive and y-component of PQ is negative so
the angle lies in 4th quadrant.
\ Required direction = 360° – 51.34° = 308.66°
(iv) Again, unit vector of PQ
Since, x-component of PQ = P^Q = PQ = (4, –5)
|PQ | 41
∴ P^Q = 1 (4, –5)
41
3. If AB displaces A(5, 2) to B(9, 5). Find AB in form of x→i + y→j . Also find
its unit vector along AB .
Solution: Here,
Let, A(5, 2) = (x1, y1) B(9, 5) = (x2, y2)
(i) Now, AB = x2 – x1 = 9–5 = 4
y2 – y1 5–2 3
Again, AB in form of x→i + y→j is
AB = 4 = 4 + 0 = 4 1 +3 0 = 4→i + 3→j
3 0 3 0 1
∴ AB = 4→i + 3→j in form of x→i + y→j .
252 Infinity Optional Mathematics Book - 9
(ii) Since,
x-component of AB (x) = 4
y-component of AB (y) = 3
∴ |AB |= x2 + y2 = 42 + 32 = 16 + 9 = 25 = 5 units.
The unit vector of AB = A^B = AB
|AB |
4→i + 3→j
= 5
∴ A^B = 4 →i + 53→j .
5
A^B 4 35→j .
Hence, unit vector along AB is = 5 →i +
4. If vector AB displaces A(2, –1) to B(3, 3) and vector PQ displaces
P(5, 5) to Q(4, 1) then show that AB = –PQ .
Solution: Here, AB displaces A(2, –1) to B(3, 3)
PQ displaces P(5, 5) to Q(4, 1)
For AB :
Let, A(2, –1) = (x1, y1)
B(3, 3) = (x2, y2)
Now, AB = x2 – x1 = 3–2 = 1
y2 – y1 3+1 4
For PQ :
Let, P(5, 5) = (x1, y1) and Q(4, 1) = (x2, y2)
Now, PQ = x2 – x1 = 4–5 = –1
y2 – y1 1–5 –4
and – PQ = – –1 = 1
–4 4
Since, the corresponding components of AB and –CD are equal.
∴ AB = –CD.
Infinity Optional Mathematics Book - 9 253
5. If vector AB displaces the point A(2, 2) to B(4, 8) and vector CD
displaces point C(4, –2) to the point D such that AB = –CD , find the
point D.
Solution: Here,
Let, the point D(x, y)
AB displaces the point A(2, 2) to B(4, 8) and CD displaces the point C(4, –2)
to the point D(x, y) and AB = –CD
For AB :
Let, A (2, 2) = (x1, y1)
B(4, 8) = (x2, y2)
Now, AB = x2 – x1 = 4–2 = 2
y2 – y1 8–2 6
For – CD:
We have, – CD = DC
∴ Let, D(x, y) = (x1, y1)
C(4, –2) = (x2, y2)
Now, DC = x2 – x1 = 4–x
y2 – y1 –2 – y
By the question
AB = –CD
or, AB = DC
or, 2 = 4–x
6 –2 – y
Comparing the corresponding components of equal vectors, we get
2 = 4 – x and 6 = – 2 – y
or, x = 4 – 2 = 2 or, y = –2 – 6
y = – 8
Hence, the point D(x, y) = D(2, – 8)
254 Infinity Optional Mathematics Book - 9
Exercise 6.2
Section 'A'
1. (a) What do you mean by position vector of a point?
(b) Define a null vector and a unit vector.
2. (a) Define parallel vectors. Does the direction disturb the vectors to be
parallel?
(b) Define magnitude of a vector?
(c) What do you mean by direction of a vector?
3. (a) If the given vector lies in 3rd quadrant then how to find the direction of
a vector?
(b) Write the differences between like vectors and unlike vectors.
(c) Write the formula to calculate the unit vector of →a.
4. Check whether the following vectors are like or unlike.
(i) →b = 2 , →c = 4 (ii) →d = 2 , →e = –6
5 10 5 –15
(iii) →f = –4 , →g = 2
2 –1
Section 'B'
5. Find the magnitude and direction of the following vectors.
(i) →a = 7 (ii) →b = –4 (iii) AB = 70
–7 3 4
(iv) BC = –6 (v) MN = –5 (vi) PQ = –63
7 –6
6. Find the angle between the positive x-axis and each of the following
vectors.
(i) ––58 (ii) 40 (iii) (1, 1) (iv) →a = –2
2
7. Find the magnitude and direction of the vectors represented by the
directed line segment joining the following points.
(i) A(2, –2) and B(7, –5) (ii) M(4, –2) and Q(6, 1)
(iii) P(–4, 5) and B(4, –5) (iv) S(2, 4) and T(5, 7)
8. (a) If AB displaces the point A to the point B and CD displaces from C to D
then prove that |AB| = |CD|
(i) A(–5, 4), B(0, 2), C(1, –1) and D(6, –3)
(ii) A(4, 5), B(7, –3), C(–1, –3) and D(2, –11)
Infinity Optional Mathematics Book - 9 255
(b) P(–3, –2), Q(2, 4), R(x, 3) and S(2, –2) are four points. If |PQ|= |RS |
then find the value of x.
9. Show that the following vectors are unit vectors.
(i) →a = (1, 0) (ii) AB = 21, 1
2
(iii) PQ = 53, –4 (iv) →e = –51, –52
5
10. Find the unit vector along the direction of following vectors.
(i) →a = (3, –4) (ii) →b = (5, 3) (iii) →e = 6 →i + 6 →j
(iv) →c = 1 (v) →d = 2 (vi) 4 →i – 5 →j
1 3
11. If AB displaces A to B. Find AB in form of x→i + y→j . Also find its unit vector
along AB .
(i) A(5, 6) and B(–2, 0) (ii) A(–2, 1) and B(–1, –2) (iii) A(6, 2) and B(2, 5)
12. (a) If (a, 1) is an unit vector then find the value of a.
(b) If C = –133, b is an unit vector then find the value of b.
13. (a) 13
If the magnitude of BC = 3 is 7 units, find the value of 'b' and direction
b
of BC .
(b) If the magnitude of EF = a is 13 units, find the value of a and direction
7
of EF .
(c) If AB = e and its angle with x-axis is 30°, find |AB |.
4
14. (a) If A(–1, 1), B(0, 4), C(–1, 6) and D(–2, 3) are four points, prove that AB
= – CD.
(b) If AB displaces A (–5, 2) to B(3, 5) and CD displaces C(1, 0) to D(9, 3).
Show that AB = CD
(c) If A(1, 2), B(2, 5), C(1, 0) and D(0, –3) are collinear points in a plane,
show that AB = –CD . Is AD = –BC ? Write with reason.
256 Infinity Optional Mathematics Book - 9
Section 'C'
15. A vector is p and its angle with x-axis is 45° and the magnitude is 2 2 units.
q
Find the vector.
16. (a) If A (2, –3), B(6, 5), C(x, 2) and D(–3, y) are four points. Find the values
of x and y when AC = BD. Also find |CD|.
(b) The vector MN displaces the points M(2, 3) to N(3, 7) and vector RK
displaces R(1, 4) to K such that MN = RK, find the coordinates of K.
(c) If A, B, C and D are four points with coordinates (2, –2), (6, 4), (m, n) and
(3, – 5) respectively such that AB = CD, find the coordinates of C. Also
find |CD|.
17. What are the uses of vector quantity in our daily life? Investigate and make a
report and present in the classroom.
Vectors Operation
Vectors can be added, subtracted and multiplied with each other. Similarly, a vector
can be multiplied by scalar quantity.
The operation by which two or more vectors combine to give a single vector is known
as vector operation. They are
1. Multiplication of a vector by a scalar
Let, →a = x1 then the multiplication of →a →a
y1
k→a
by a scalar 'k' denoted by k→a is defined by Fig. (i)
k →a = k x1 = kx1 →a
y1 ky1
k →a is a vector parallel to →a and its
magnitude is k times the magnitude of →a .
→a and k →a have the same direction when k k→a
is positive as in fig (i). and their direction Fig. (ii)
are opposite when k is negative as in figure (ii).
Infinity Optional Mathematics Book - 9 257
Properties of scalar multiplication
→
If →a and are two vector, m and n are scalars then
b
(i) →a = 0 when m = 0 (ii) m(n→a ) = (mn) →a = n(m→a )
(iii) (m + n) →a = m→a + n→a (iv) m( →a + →b ) = m→a + m→b
2. Addition of vectors:
Addition of vectors can be expressed in terms of different principles as
described below.
(i) Triangle law of vector addition B
Let, →a and →b be two vectors. Let us take a point O and
draw the vectors OA = →a and AB = →b . Such that the
terminal point of →a is the initial point of →b . →
By joining O and B it is directed from O to B and defined
as b
OB = OA + AB O →a A
or, OB = →a + →b
∴ The displacement from O to A and again from A to B is same as
displacement from O to B. This law of addition of vectors is known as the
triangle law of vector addition.
Hence, triangle law of vector addition states that "the sum of the vectors
represented by two sides of a triangle taken in the same order is equal to the
vector represented by the third side taken in the reverse order".
Note:
(a) A vector PQ can be expressed as the sum of PR and RQ taking 'R' a
point in the same plane.
(b) The sum of vectors representing the sides of a triangle, taken in order
is zero. Let 'O' be the origin. In ∆OBC, O
OB + BC + CO = (0, 0) = O
(c) By triangle law of vector addition OB + BC = OC
or, BC = OC – OB
or, BC = position vector of the point C – position B C
vector of the point B
(ii) Parallelogram law of vector addition D C
In the adjoining figure, let, AB = →a and AD = →b be two B
→
b
vectors having same initial point A. A parallelogram ABCD A →a
258 Infinity Optional Mathematics Book - 9
is made by taking AB and AD as the adjoining sides. Then AD = BC, AD//BC,
AB = DC and AB//DC.
Now, AD = BC =→b
In ∆ABC, by triangle law of vector addition
AB + BC = AC
or, AB + AD = AC [ BC = AD being opp. sides of //gram]
or, AC = AB + AD or, AC = →a + →b
This is called the parallelogram law of vector addition.
Hence, parallelogram law of vector addition states that "if two adjacent sides
of a parallelogram through a point represents two vectors in magnitude and
direction their sum is given by the diagonal of the parallelogram through the
same point in magnitude and direction."
(iii) Polygon law of vector addition E D
C
It is the expanded form of triangle law of vector addition.
Two or more vectors can be added by the application of F
polygon law of vector addition.
In the adjoining figure, ABCDEF is a polygon. The sum A B
of vectors represented by AB , BC , CD , DE and EF
taken in order is equal to AF . Join AC, AD and AE.
In ∆ABC, by ∆ law of vector addition
AC = AB + BC ..........(i)
Similarly, in ∆ACD,
AD = AC + CD
AD = AB + BC + CD ............(ii) [ From (i)]
In ∆ADE,
AE = AD + DE
Infinity Optional Mathematics Book - 9 259
= AB + BC + CD + DE ............ (iii) [ From (ii)]
In ∆AEF,
AF = AE + EF
AF = AB + BC + CD + DE + EF [From (iii)]
∴ AF = AB + BC + CD + DE + EF
Hence, polygon law of vector addition states that "the number of vectors can be
represented in magnitude and direction by the sides of polygon taken in order, their
sum is given in magnitude and direction by the closing side of the polygon taken in
reverse order.
Addition of column vectors
Let, OA = x1 and OB = x2 be the position vectors of A and B respectively.
y1 y2
Now, constructing a parallelogram OACB as shown in the figure taking OA and OB
as adjacent sides, we get
OA + OB = OC ( By parallelogram law of vector addition)
Again, draw BN, AM and CQ perpendiculars on OX, and AP perpendicular to CQ,
we get.
DOBN ≅ DCPA [ ∠BNO = ∠CPA, ∠BON = ∠CAP and OB = CA]
∴ CP = BN and AP = ON
ON = x2, BN = y2, OM = x1, AM = y1 Y
B(x 2,y 2) C
x-component of OC = OQ
= OM + MQ
= OM + AP (x1,y1)A P
N MQ
= OM + ON X' O X
= x1 + x2 Y'
y-component of OC = CQ
= CP + PQ
= BN + AM = y2 + y1 = y1 + y2
∴ OC = OQ = x1 + x2
CQ y1 + y2
260 Infinity Optional Mathematics Book - 9
Thus, OA + AC = OC
OA + OB = OC gives x1 + x2 = x1 + x2
y1 y2 y1 + y2
Hence, the column vectors are added by adding x-components and y-components
respectively to get x and y component of the resultant vector.
Properties of vector addition:
(i) →a + → = → + →a (Commutative property)
b b
(ii) →a + (→b + →c ) = (→a + →b ) + →c (Associative property)
(iii) →a + → = → + →a = →a (Identity property)
0 0
(iv) →a + (–→a ) = → (Additive inverse property)
0
3. Subtraction of vectors
Discuss on the following questions.
Can we write →a – → as →a + (–→b ) ? What are the similarity between addition
b
and subtraction of vector.
The difference of two vectors →a and → denoted by →a B
→b
b O
– →b
– → is defined as the vector sum of →a and negative →a
D
b
of →b . So, if →a = x1 and → = x2 then →a – → is A
y1 y2
b b →a – →
x1 –x2 b
y1 –y2
defined by →a – → = →a + (– →b ) = +
b
= x1 – x2 C
y1 – y2
Let, OA = →a and OB = →
b
If OB = OD and they are opposite then OD = – OB = –→b
OACD is a parallelogram then by parallelogram law of vector addition
OC = OA + OD OC = →a – →
b
Infinity Optional Mathematics Book - 9 261
WORKED OUT EXAMPLES
1. If AB = –5 and CD = 3 , find AB + CD, AB – CD, 2AB + 3CD and
3 –3
their magnitudes.
Solution: Here,
AB = –5 and CD = 3
Now, 3 –3
(i) AB + CD = –5 + 3 = –5 + 3 = –2
3 –3 3–3 0
(ii) AB – CD = –5 – 3 = –5 – 3 = –8
3 –3 3+3 6
(iii) 2AB + 3CD =2 –5 +3 3 = –10 + 9 = –10 + 9 = –1
3 –3 6 –9 6– 9 –3
(iv) |AB + CD | = x2 + y2 = (–2)2 + (0)2 = 4 + 0 = 2 units
(v) |AB – CD | = x2 + y2 = (–8)2 + (6)2 = 64 + 36 = 100 = 10 units and
(vi) 2AB + 3CD |= x2 + y2 = (–1)2 + (–3)2 = 1 + 9 = 10 unit.
3
2. If →a = 2 and → = 2 prove that →a and → are parallel.
6 41
b b
Solution : Here, 2
3
→a = 2 and → = 2 .
6 41
b
2
Now, →a = 2 =2 1
6 3
\ →a = 1 ........... (i)
2 3
3 3 3×1
→ 2 2 2 3 1
and = 41 = 9 = 3×3 = 2 3
b
22 2
262 Infinity Optional Mathematics Book - 9
or, →b = 3 →a [ From equation (i)]
2 2
or, →b = 3 →a
4
By the definition of parallel vector
→b = k→a where k = 3
4
Hence, it is proved that →a and →b are parallel.
3. If →a = 3 , →b = –1 and →c = 0 find the scalar 'k' such that 3→a + k→b = →c
1 0 3
Solution: Here,
→a = 3 , → = –1 , →c = 0 and 3→a + k→b = →c
1 0 3
b
Since, 3→a + k→b = →c
or, 3 3 +k –1 = 0
1 0 3
or, 93 + –k = 0
0 3
or, 9 – k = 0
3 3
Comparing corresponding components, we get
9 – k = 0
\ k = 9
4. If →p = 3 and 4→p + 3→q = 2 find →q .
–2 6
Solution: Here,
→p = 3 and 4→p + 3→q = 2
–2 6
Now, 4→p + 3→q = 2
6
or, 4 3 + 3→q = 2
–2 6
or, –128 + 3→q = 2
6
Infinity Optional Mathematics Book - 9 263
or, 3→q = 2 – 12
6 –8
or, 3→q = –10
14
or, →q = 1 –10
3 14
∴ →q = –310, 14
3
5. If the position vectors of A and B are →i + 2→j and 3→i – →j respectively,
find AB and its magnitude. Also, find the unit vector along the
direction of AB .
Solution: Here,
The position vector of the point A(OA ) = →i + 2→j
The position vector of the point B(OB ) = 3→i – →j
We have,
(i) AB = OB – OA = 3→i – →j – (→i + 2→j ) = 3→i – →j – →i – 2→j = 2→i – 3→j
AB = 2→i – 3→j
∴ AB = (2, –3)
(ii) |AB | = x2 + y2 = (2)2 + (–3)2 = 5 units
(iii) Again,
Unit vector of AB = (AB) = AB = (2, – 3)
\ |AB 5
|
AB = 52, –3
5
6. The coordinates of P and Q are (5, 12) and (3, 4) respectively.
(i) Write the column vector PQ. (ii) | PQ|
(iii) Direction of PQ.
(iv) M be the midpoint of PQ find the position vector of M under the
operation of (a) vector addition (b) vector subtraction
264 Infinity Optional Mathematics Book - 9
Solution: Here, the coordinates of P and Q are (5, 12) and (3, 4) respectively.
Let, P(5, 12) = (x1, y1) Q(3, 4) = (x2, y2)
(i) By formula, PQ = x2 – x1 = 3–5 = –2
y2 – y1 4 – 12 –8
(ii) |PQ | = x2 + y2
= (–2)2 + (–8)2
= 4 + 64
= 68 = 2 17 units
(iii) For the direction of PQ
We have,
tanθ = y = –8
x –2
or, θ = tan-1 (4) = 75.96°
Since, both components of PQ are negative, so PQ lies in 3rd quadrant.
\ Required direction = 180° + 75.96° = 255.96°.
(iv) Since, M is the midpoint of PQ. P(5,12) M Q(3, 4)
Let, O be the origin.
∴ MQ = 1 PQ = 1 –2 = –1
2 2 –8 –4
O
–1 1
and – MQ = – –4 = 4
∴ QM = 1
4
(a) Under the operation of vector addition
OM = OQ + QM = 3 + 1 = 4
4 4 8
(b) Under the operation of vector subtraction
OM = OQ – QM = 3 – –1 = 3+1 = 4
4 –4 4+4 8
Infinity Optional Mathematics Book - 9 265
7. In ∆PQR, M is the midpoint of QR then show that P
PM = 1 (PQ + PR )
2
Solution: Here, in ∆PQR, M is sthe midpoint of QR.
To show: PM = 21(PQ + PR ) Q MR
In ∆PQM, by ∆ law of vector addition PQ + QM= PM
QM = PM – PQ ........... (i)
Similarly, in ∆PMR,
PM + MR = PR
or, MR = PR – PM ........... (ii)
By the question [ M is the mid point QR]
QM = MR
or, PM – PQ = PR – PM
or, PM + PM = PQ + PR
or, 2PM = PQ + PR
PM = 12(PQ + PR )
8. SILVER is a regular hexagon. If SI = V and IL = W , find SL, SV and
SE in terms of v and w. Also show that SI + IL + LV + VE + ER + RS
=0
Solution: Here,
SILVER is a regular hexagon. EV
SI = v and IL = w
To find: SL , SV and SE in terms of v and w . R L
To show: SI + IL + LV + VE + ER + RS = 0
(i) In ∆SIL by ∆ law of vector addition
SL = SI + IL SI
SL = v + w
(ii) SV = 2IL [ Being diagonal of a regular hexagon is double of its sides]
SV = 2w
266 Infinity Optional Mathematics Book - 9
(iii) In ∆SEV, by ∆ law of vector addition
SE + EV = SV
or, SE = SV – EV [ SI = EV]
= 2w – SI
SE = 2w – v
(iv) Again, In regular hexagon SILVER by polygon law of vector addition, we get
SI + IL + LV + VE + ER = SR
or, SI + IL + LV + VE + ER = – RS
or, SI + IL + LV + VE + ER + RS = 0
Exercise 6.3
Section 'A'
1. (a) State triangle law of vector addition.
(b) Define the parallelogram law of vector addition.
(c) Express the vector PQ in terms of position vectors of P and Q taking
origin as initial point.
2. (a) Define Collinear vectors. Does the direction disturb them to be parallel?
(b) What do you understand by multiplication of a vector by a scalar? Give
a geometrical representation.
(c) State the polygon law of vector addition.
3. (a) If a vector →q = 2 then find 3→q . (b) If →b = 4 then find –2→b .
–3 –3
(c) If →a = k→b and k be a scalar. What is the relation between →a and →b ?
4. (a) If AB = 4 and CD = –2 then find AB + CD .
3 7
(b) If →p = 6 and →q = 2 then find →p – →q .
–2 3
5. (a) If →p = 2→i – 3→j and →q = 6→i + 8→j then find →p + →q .
(b) If MN = 5→i + 7→j and PQ = 3→i + 4→j then find MN – PQ .
Infinity Optional Mathematics Book - 9 267
6. (a) If m→ = 2 and m→ + →n = 5 then find →n.
7 8
(b) AB – CD = 4 and AB = 2 , find CD .
10 –5
Section 'B'
7. (a) If →a = 3 and →b = –4 find →a + →b , →a – →b and their magnitudes.
4 –3
(b) If PQ = (5, 7) and RS = –2 find PQ + RS , PQ – RS and their
magnitudes. 3
(c) If MN = 4→i + 3→j and OP = –3→i – 4→j find MN + OP , MN – OP and their
magnitudes.
8. If →a = –3 , →b = –2 and →c = 2 find
4 1 –3
(i) 2→a + 3→b (ii) 3→b – 4→c (iii) →a + (→b + →c )
(iv) 2(→a + →b + →c ) (v) 2→a – 4→c + 3→b (vi) 2(→a + →b ) – 5 →c
9. (a) If →a = –5 and →b = –10 show that →a and →b are parallel.
2 4
(b) If AB = –3 and CD = – 3 show that AB and CD are parallel.
2 2
1
(c) If PQ = 4→i + 5→j and RS = –8→i – 10→j then prove that PQ and RS are
parallel.
10. (a) If →p = 3 and →q = 6 are two parallel vectors then find the value of 'k.
–2 k
(b) MN = 3→i – 2→j , PQ = 6→i + m→j and MN||PQ then find the value of
scalar 'm'.
11. (a) If the position vector of A and B are 5 and 6 respectively, find AB .
3 5
(b) If the position vector of P and Q are (7, –3) and (2, 9) respectively then
find PQ and QP .
268 Infinity Optional Mathematics Book - 9
(c) Find the position vector of N if the position vector of M is –3
7
and MN = 5 P
9
12. (a) In the given figure, alongside show that : PQ + QR + RP = 0
(b) If AB , BC , CD and DA be the vectors represent by the
sides of a parallelogram ABCD show that QR
(i) AB + BC + CD + DA = 0 (ii) AB + CD = 0
(iii) BC + DA = 0
(c) In the adjoining figure, ABCD is a quadrilateral. Express the following
vectors as a single vector. DC
E
(i) AB + BC (ii) AC – BC
(iii) BC + CD + DA (iv) BE + EC – BA
(v) AE – BE + BC
13. (a) If →a = 2 →b –3 and →c = –1 AB
3 5 8
, = . Find the scalar 'm' such that
→a + m→b = →c .
(b) If AB = 2 , CD = –1 and EF = –1 find 'k' such D C
1 2 7
B
that AB + kCD = EF . →b
A B
(c) In the given figure alongside, ABCD is a →a
parallelogram. If OA = →a , OB = →b and OC = →c find
OD in forms of →a , →b and →c . C
→c →a A
(d) In the adjoining figure, OA = 2CB = →a and
OC = →c . Find OB and BA in terms of →a and →c . O
Infinity Optional Mathematics Book - 9 269
Section 'C'
14. ( a) If the position vector of A and B are (6, 4) and (4, –2) respectively, find
AB , magnitude of AB and the direction of AB . Also find the unit vector
along the direction of AB .
(b) The position vectors of the points A and B are 5→i + 6→j and
4→i + 2→j respectively. Find, AB in column form, magnitude of AB and
the direction of AB . Also, find the unit vector along the direction of AB .
15. (a) If the position vectors of A and B are –4 and 6 respectively find
3 5
(i) AB (ii) the magnitude of AB
(iii) the position vector of the middle point M of AB using vector addition.
(b) If the position vectors of A and B are 2 and –4 . If P is the midpoint of
5 1
AB, find the position vector of P on the basis of the operations under
(i) vector addition (ii) vector subtraction
16. (a)
The position vectors of four points A, B, C and D are →a + →b +→c , 2→a +5→b ,
3→a + 2→b – 3→c and →a – 6→b – →c respectively. Prove that CD = – 2AB and
CD //AB .
(b) The position vectors of four points, A, B, C and D are 2→a + 4→c ,
5→a +3 3→b + 4→c , –2 3 →b + →c and 2→a + →c respectively. Show that CD = 2
3
AB and AB is parallel to CD .
V
(c) In the given ∆VAT, M and N are the midpoints of
sides VA and VT respectively then by using triangle M N
1
law of vector addition prove that MN = 2 AT and
MN//AT . AT
17. (a) If ABCDE is a regular pentagon than prove that:
(i) AB + BC + CD + DE + EA = 0
(ii) AB + AE + BC + DC + ED + AC = 3AC
270 Infinity Optional Mathematics Book - 9
(b) 'MONKEY' is a regular hexagon. If MO = m→, ON = n→, NK = →p and KE
= q→then,
(i) Find the vector represented by MO + ON .
(ii) Express MO + ON + NK in terms of sum of single vector.
(iii) Prove that : MN + NK = MO + OK
(iv) Prove that : m→ + n→+ →p = 2n→
(v) Prove that: m→ + →q = 0
18. (a) Prove by vector method that the points A(–2, 3), B(3, 8) and C(4, 1) are
the vertices of an isosceles triangle.
(b) Show by vector method that the points A(2, 1), B(2, 3), C(5, 3) and
D(5, 1) are the vertices of a rectangle. A
19. In the adjoining ∆ABC, D, E and F are the mid points of F E
sides BC, AC and AB respectively.
Prove that: AD + BE + CF = 0. BDC
20. (a) What are the uses of vector operation in our daily life? Investigate, make
a report and present in the classroom.
(b) Compare the addition and subtraction of vectors with the addition and
subtraction of matrix, Algebra. When the dot product of the two vectors
is a scalar quantity then what is the cross product of two vectors? Ask
with your teacher, investigate and make a report and present it in the
class-room.
Infinity Optional Mathematics Book - 9 271
UNIT
7 TRANSFORMATION
7.1 Transformation
Review
You have already learnt about the transformation in class 8 in compulsory
mathematics. Now, discuss the following questions in the class among the friends.
(i) What have you understand about the transformation?
(ii) How many types of transformation you have learnt and what are they?
(iii) Can you draw the images of the following figures after reflection through the
line l. (b) (c) P
(a) l H G
A
Q
B E F
C l
l
R
A
(iv) Are the object and its corresponding image congruent?
(v) Draw the image of DABC after rotation through +90° about B C
the centre at O. O
(vi) What are the difference found in the object and image after rotation.
(vii) When an geometrical object is rotated or reflected or translated, are the image
and object congruent or similar? Why?
Introduction
The rule or process of changing an object either in position or in size or in both
the position and size is known as transformation. In transforming an object
272 Infinity Optional Mathematics Book - 9
corresponding to a given object, there will be a definite object of the similar form
known as image of the given object under transformation.
In transformation, each point of a plane has exactly one image point and each image
point has exactly one pre-image point in the same plane. So, transformation is also
called a one to one and onto function (or mapping).
Transformation sometimes leaves certain points unchanged. These points, distance,
angles, parallelograms, triangle etc which are unchanged in transformation are
called invariants and such transformations are called identity transformations.
There are four fundamental types of transformations. They are reflection, rotation,
translation and enlargement or reduction.
In reflection, rotation and translation, there is change in the position of the object
or geometrical figure but not in size. In these three transformations, the image and
object are congruent and are also called isometric transformations.
In enlargement or reduction there is change in the position as well as in the size of
the object or geometrical figure . In this transformation, the image and object are
similar. So, this transformation is called non-isometric transformation.
Reflection Mirror Image
Reflection is a type of transformation. In this M
transformation, the shape and size of the object and A A'
image are same. The line through which the reflection
takes place is known as the axis of reflection. The P
axis of reflection acts as two way plane mirror.
Properties of the reflection of geometrical Object
objects in the axis of reflection.
When a geometrical figure is reflected in the axis of
reflection, the following properties are found.
1. The object and its image are equidistant from the axis C R C'
of reflection. In the adjoining figure, AP = PA', BQ = Q
N
QB' and CR = RC'. B B'
2. The shape of the object and image are laterally inverted. i.e. top remains at
the top, bottom remains at the bottom but left side is transformed into right
and right into left.
Infinity Optional Mathematics Book - 9 273
Object Mirror Image
B
3. The line joining the same ends of object and its A P
image are perpendicular to the axis of reflection.
B'
In the adjoining figure AA', BB' and CC' are
C
perpendicular to the axis of reflection PQ. C' A'
4. The points on the axis of reflection are invariant Q X
points i.e. their images are the same points. AA
X
B
B'
A
A'
Y TT
5. The bilaterally symmetrical figures will have similar image after
MM
refection in vertical axes. X
Method of finding the image under refection
In the adjoining figure, MN is the axis of reflection and ABC is a triangle
To obtain the image of DABC, following steps are AM
used. C' F
1. Draw AF^MN and produce it to A' making B E A'
AF=FA' C
D
2. Draw BD^MN and produce it to B' making BD N B'
=DB'
3. Draw CE^MN and produce it to C' making
CE= EC'
4. Join A'B,' B'C' and C'A' with straight lines.
DA'B'C' is obtained which is the image of DABC.
274 Infinity Optional Mathematics Book - 9
Reflection using coordinates.
We can find the image of any geometrical figure through the following axes of
reflections.
1. Reflection about X-axis (y = 0 line).
Let A(4,3), B(–1, 4) and C(1, Y
–2) are the three vertices
of a triangle ABC in the
rectangular axes of coordinates.
From A,B and C draw AM, BN B(–1, 4) X
and CR perpendiculars on X – A(4, 3)
axis. Produce AM making AM
= MA', BN making BN= NB' X' C'(1,2)
and CR making CR = RC'. Join
A', B' and C'. Then A', B' and N
C' are the images of A, B and OR M
C(1,–2) A(4, –3)
B'(–1,–4)
C of DABC under the refection
about X-axis. The co-ordinates
of A' , B' and C' are A' (4,–3) , B'
(-1,-4) and C'(1,2). Y'
We can write coordinates of
DABC and its images as
A(4,3) → A' (4, –3)
B (–1, 4) → B'(–1, –4)
C(1, –2) → C'(1, 2).
We see that when the reflection is done on X -axis there is no change in
X-coordinate but there is a change in the sign of Y-coordinate of the image
of the given point .
Hence, for the object having coordinates P(x, y), its image under the reflection
through x–axis is P(x, –y). i.e. P (x,y) → P'(x, –y).
WORKED OUT EXAMPLES
1. Find the image of A(– 7, – 6) after reflection through X–axis (y = 0 line).
Solution:
In reflection through X – axis (y = 0 line)
P(x, y) → P'(x, – y)
A(– 7, – 6) → A' (– 7, 6)
Hence, the image of A (– 7, – 6) is A' (– 7, 6).
Infinity Optional Mathematics Book - 9 275
2. A(– 1, 6), B(– 3, – 4) and Y
C(2, 5) are the three vertices
of ∆ABC. Find the coordinates A(–1, 6) C(2, 5)
C'(2,–5)
of the vertices of the image B(–3, 4)
of ∆ABC after reflection in
X – axis. Also show the given
triangle and its image on the X' O X
same graph.
Solution:
In reflection through X – axis B(–3,–4)
P(x, y) → P' (x, – y) A'(–1,–6)
A(– 1, 6) → A' (– 1, – 6)
B(– 3, – 4) → B' (– 3, 4)
C (2, 5) → C' (2, – 5) Y'
Now, ∆ABC and its image after
reflection in X – axis are shown on the graph.
2. Reflection about Y– axis (x = 0 line).
Let A(3, 2), B(6, 5) and (7, – 1) are Y
the three vertices of ∆ABC in the
rectangular axes of coordinates. B'(–6,5) M B(6,5)
From A, B and C draw AN, BM
X
and CR perpendicular to Y-axis. N A(3,2) C(7,–1)
Produce AN making AN = NA', BM X' A'(–3,2)
making BM = MB' and CR making
CR = RC'. Join A',B' and C'. Then C'(–7,–1) O
R
A', B' and C' are the image of A, B
and C of ∆ABC under the reflection
about Y–axis. The coordinates of
A', B' and C' are A'(– 3, 2), B' (– 6, Y'
5) and C' (– 7, – 1).
We can write the coordinates of ∆ABC and its image as
A (3, 2) → A' (– 3, 2)
B (6, 5) → B' (– 6, 5)
C(7, – 1) → C' (– 7, – 1)
276 Infinity Optional Mathematics Book - 9
We see that when the reflection is done on Y–axis, there is a change in sign in
x–coordinate but there is no change in y – coordinate of the image of the given
point.
Hence, for the object having coordinates P(x, y), its image under the reflection
through y – axis is P' (– x, y).
i.e. P(x, y) → P' (– x, y).
WORKED OUT EXAMPLES
1. Find the image of the point A(– 8, – 6) after reflection through y-axis
(x = 0 line)
Solution: In reflection through y – axis .
P (x, y) → P' (–x, y)
A (– 8, – 6) → A' (8, – 6)
Hence, the image of A (– 8, – 6) is A' (8, – 6)
3. Reflection about the line y = x.
In the given rectangular axes Y
of coordinates MN is a straight
line passing through origin and M
bisects ∠YOX. The equation of A(2, 5)
the line MN is y = x. D
B(–1,2) A(5,2)
E
Let A(5, 2), B (– 1, 2) and X' O X
C (– 3, – 4) are the vertices of C'(–4,–3) B(2,–1)
∆ABC. From A, B and C draw F C(–3,–4)
AD, BE and CF perpendicular
on the line MN. Produce AD N
making AD = DA', BE making
BE = EB' and CF making CF = Y'
FC'. Join A', B' and C' . Then A',
B', and C' are the image of A, B and C of ∆ABC under the reflection about
y = x line. The coordinates of A', B' and C' are A' (2, 5), B' (2, –1) and
C' (– 4, – 3).
Infinity Optional Mathematics Book - 9 277
We can write the coordinates of ∆ABC and its images as
A(5, 2) → A' (2, 5)
B (– 1, 2) → B' (2, – 1)
C (– 3, – 4) → C' (– 4, – 3).
We see that when the reflection takes place through the line y = x, the x
– coordinates and y-coordinates of the given point and its image are
interchanged.
Hence, for the object having coordinates P(x, y), its image under the reflection
through the line y = x is P' (y, x).
i.e. P (x, y) → P' (y, x)
Reflection about the line y = – x. Y
In the given rectangular axes of M
coordinates, MN is a straight line
passing through origin and bisects
∆YOX'. The equation of the line MN
is y = – x. 135° A(4,1)
X' O X
Let A (4, 1) be a point on the plane.
R
From A draw AR perpendicular to the A(–1,–4)
line MN. Produce AR making AR =
RA'. Then A' is the image of the point
A after reflection in the line y = – x. N
The coordinates of A' is (– 1, – 4)
We can write the coordinates of A and Y'
its image A' as
A(4, 1) → A (– 1, –4).
We see that when the reflection is done in the line y = – x, then x – coordinate and
y– coordinate of the given point and its image are interchanged with change in sign
of both.
Hence, for the object having coordinates P(x, y), its image under the reflection
through the line y = –x is P' (– y, – x).
i.e. P (x, y) → P' (– y, – x).
278 Infinity Optional Mathematics Book - 9
Y
Reflection about the line x = h M P(x',y')
P(x,y) T(h,y)
x = h is a line parallel to y-axis where
h is the distance between the line and
y-axis .
Let P (x, y) be any point on the plane. X' h X
From P, draw PT ⊥ MN and produce O
PT up to P' making PT = P'T . Let P'
(x', y') be the image of P(x, y). From N
figure, T is the midpoint of PP' and
the coordinates of T is (h, y).
Y'
Now, using midpoint formula,
x + x' y + y'
h= 2 and y = 2
∴ x' = 2h – x and y' = y
Hence, the coordinates of P' (x', y') = P' (2h – x, y)
∴ When the reflection takes place through the line x = h, the image of P(x, y) is
P' (2h – x, y).
i.e. P (x, y) → P' (2h – x, y).
WORKED OUT EXAMPLES
1. Find the image of the point A (– 6, – 5) after reflection about the line x
= 3.
Solution: In reflection through x = 3 (x = h, h = 3)
P (x, y) → P' (2h – x, y)
A (– 6, – 5) → A' (2 × 3 – (– 6), – 5) = A' (12, – 5).
Hence, the image of A (– 6, – 5) is A' (12, – 5).
Infinity Optional Mathematics Book - 9 279
Reflection about the line y = k. Y
y = k is a line parallel to X – axis P(x, y)
where k is the distance between the
line and X – axis . M TN X
Let P (x, y) be any point on the plane. k
From P, draw PT ⊥ MN and produce X'
PT to P' making PT = TP'. Let P' P(x', y')
O
(x', y') be the image of P(x, y). From
figure, T is the midpoint of PP' and
the coordinates of T is (x, k).
Now, using midpoint formula,
x + x' y + y'
x = 2 and k = 2
Y'
∴ x' = x and y' = 2h – y
Hence, the coordinates of P' (x', y') = P' (x, 2k – y)
∴ When the reflection takes place through the line y = k, the image of P (x, y) is
P'(x, 2k – y)
i.e. P (x, y) → P' (x, 2k – y)
WORKED OUT EXAMPLES
1. Find the image of the point B(– 6, 2) after reflection about the line
y = – 2.
Solution: In reflection about the line y = – 2 (y = k, k = – 2)
P(x, y) → P' (x, 2k – y)
B(– 6, 2) → B' (– 6, 2 × (– 2) – 2) = B' (– 6, – 6)
Hence, the image of B (– 6, 2) is B' (– 6, – 6).
2. Find the axis of reflection when
(i) the image of A (5, – 6) is A' (5, 6).
Solution: Here, the image of A (5, –6) is A' (5, 6)
Now, midpoint of AA' = 5 + 5, –6 + 6 = (5, 0)
2 2
Here, X – coordinates of P = X – coordinates of P'
So, the equation of the axis of reflection is y = b.
Since, the line passes through (5, 0) then
the equation of the axis of reflection is y = 0.
280 Infinity Optional Mathematics Book - 9
(ii) The image of B(6, 2) is B' (– 2, 2).
Solution:
Here, the image of B (6, 2) is B' (– 2, 2)
Now, midpoint of BB' = 6 – 2 , 2 + 2
2 2
= (2, 2)
Here, y – coordinates of B = y – coordinates of B'.
So, the equation of the axis of reflection is x = a.
Since, the line passes through (2, 2)
2=a
∴ a = 2.
∴ The equation of the axis of reflection is x = 2.
Exercise 7.1
Section 'A'
1. Draw the image of each of the following geometrical figure after
reflection on the line MN.
(a) A B (b) M (c) A B
MC Q P M N
N S
R N
C
(d) M (e) P (f) M
A E M A
D BD
QR
B C C
N N
N
2. Find the images of the following points after reflection about X – axis
(y = 0 line)
(i) A(2, 3) (ii) B (– 6, 5) (iii) C (– 7, – 6) (iv) D (8, – 4)
(v) P(0, 0) (vi) E (5, 0) (vii) F (– 6, 0) (viii) G (0, 4)
(ix) H (0, – 7) (x) I (– 9, 6)
Also, write down the invariant points.
Infinity Optional Mathematics Book - 9 281
3. Find the image of the following points after reflection about y – axis
(x= 0 line).
(i) A (1, 9) (ii) B (– 7, 8) (iii) C (– 2, – 6) (iv) D (3, – 5)
(v) E (5, 0) (vi) F (– 3, 0) (vii) G (0, 5) (viii) H (0, – 2)
Determine the invariant points.
4. Find the images of the following points after reflection about y = x
line.
(i) A (– 6, 4) (ii) B (– 5, – 7) (iii) C (9, – 6) (iv) D (2, 3)
(v) E (6, 0) (vi) F (0, 8) (vii) G (– 5, 0) (viii) H (0, 9).
5. Find the images of the following points after reflection about y = – x
line.
(i) A (0, 7) (ii) B (– 6, 8) (iii) C (5, – 3) (iv) D (– 1, – 2)
6. Find the image of the following points after reflection about x = 2
line.
(i) A (3, 1) (ii) B (– 6, 2) (iii) C (– 2, – 5) (iv) D (7, 6)
(v) E (– 6, 0) (vi) F (5, 0) (vii) G (0, – 7) (viii) H (0, 2).
7. Find the images of the following points after reflection about y = 3
line.
(i) A (3, 1) (ii) B (2, – 4) (iii) C (– 1, – 5) (iv) D (– 7, 8)
(v) E (7, 0) (vi) F (– 6, 0) (vii) G (0, – 5) (viii) H (0, 2)
8. Find the image of the points A (3, 2), B (– 7, 6) and C (– 5, – 8) after
reflection about the line joining the following points.
(i) (3, 3) and (– 7, – 7) (ii) (– 3, 3) and (8, – 8)
(iii) (– 3, 5) and (– 3, – 6) (iv) (7, – 2) and (– 1, – 2)
(v) (3, 0) and (– 7, 0) (vi) (0, – 8) and (0, 5)
9. Find the coordinates of the point A in each of the followings.
(i) A → A' (– 7, 6) after reflection in the line y = 0.
(ii) A → A' (3, 5) after reflection in the line x = 0
(iii) A → A' (– 7,– 6) after reflection in the line y = x
(iv) A → A' (6, – 2) after reflection in the line y + x = 0
(v) A → A' (1, 9) after reflection in the line 2y – 6 = 0
(vi) A → A' (– 3, 2) after reflection in the line 3x + 6 = 0.
10. Find the axis of reflection when
(i) A (3, 5) → A' (– 3, 5) (ii) B (– 7, 2) → B' (– 7, – 2)
(iii) C (9, – 5) → C' (– 5, 9) (iv) D (– 4, 6) → D' (– 6, 4)
(v) E (6, 4) → E' (– 2, 4) (vi) F (– 1, – 2) → F' (– 1, – 4)
Section 'B'
11. A (5, 2), B (– 3, 3) and C (1, 6) are the three vertices of ∆ABC. Find the
coordinates of the vertices of the image of ∆ABC after reflection through y = 0
line. Show ∆ABC and its image on the same graph paper.
282 Infinity Optional Mathematics Book - 9
12. A (2, 6), B (– 1, 4) and C (5, 1) are the three vertices of ∆ABC. Find the
coordinates of the vertices of the image of ∆ABC after reflection through x = –
2 line. Show ∆ABC and its image on the same graph.
13. A (– 2, – 3), B (– 1, 1), C (4, 1) and D (3, – 3) are the vertices of a parallelogram
ABCD. Find the coordinates of the vertices of the image of parallelogram
ABCD after reflection through y = x line. Also show both the abject and its
image on the same graph.
14. A (3, 2), B (– 1, 4) and C (1, 5) are the vertices of ∆ABC. It is reflected in the
line y = – x line to ∆A'B'C'. The triangle A'B'C' is then reflected in y – axis
to ∆A"B"C". Write down the coordinates of A",B" and C". Also show ∆ABC ,
∆A'B'C' and ∆A"B"C" on the same graph.
15. Draw ∆MNK having the vertices M(2, 1), N (1, 5) and K (6, 7) on the graph
paper . Then draw the image ∆M'N'K' after reflection on x-axis. Again draw
∆M"N"K" after reflection on the line y = x.
Rotation
Rotation is the process of finding the image of an object by rotating every point of
the object in the same direction through a certain angle about a fixed point.
A
A
B
A' C'
80° 100° C
O A' B'
Figure (i) O
Figure (ii)
In the above figure (i), a point A is rotated through 80º in clockwise direction about
the point 0. The image of the point A is A'. Here ∠AOA' = 80º and OA = OA'.
In figure (ii), DABC is rotated through 100º in anticlockwise direction about the
point 0. Here DA'B'C' is the image of DABC. In the figure, ∠AOA' = ∠BOB' = ∠COC'
= 100º, OA = OA', OB = OB' and OC = OC'. Here DABC and DA'B'C' have the same
shape and size.
The fixed point about which an object is rotated is called the centre of rotation. In
the above figures 0 is the centre of rotation. The angle through which every point
Infinity Optional Mathematics Book - 9 283
of the object is rotated is called the angle of rotation. The angle it rotated either in
clockwise direction or in anticlockwise direction. The rotation in clockwise direction
is negative rotation. In figure (i), the angle of rotation is –80º. The rotation in
anticlockwise direction is positive rotation. In figure (ii), the angle of rotation is +
100º.
A rotation through 90º in anticlockwise direction is called positive quarter turn
(Q+) A rotation through 90º in clockwise direction is called negative quarter turn
(Q–). A rotation through 180º in either direction is called half turn (H). The rotation
through 360º in either direction is called full turn.
Properties of rotation
(i) The image of an object can be obtained if the centre, angle and direction of
rotation are given .
(ii) A rotation transforms all the point of the geometrical figures in same direction
and same angular displacement.
(iii) The geometrical figure and its image after rotation are congruent.
(iv) The image distance from centre of rotation is equal to the object distance from
centre of rotation.
(v) The perpendicular bisector of the line segment joining a point of the object and
its corresponding image passes through the centre of rotation.
(vi) The rotation about a point through an angle a in the clockwise direction is
equivalent to the rotation about the same point through the angle (360º – a)
in anticlockwise direction and vice versa.
Method of finding the image after rotation.
When the centre of rotation, angle of rotation and the direction of rotation are given,
the image after rotation of an object can be obtained in the following ways.
∆ABC is a given triangle. O is the centre of rotation and 100º is angle of rotation and
is rotated in anticlockwise direction.
(i) Join A, B and C with O. A
C' B
C
A O
B'
284 Infinity Optional Mathematics Book - 9
(ii) Draw arcs taking O as the centre and OA, OB and OC as radii in anticlockwise
direction.
(iii) Mark a point A' on the arc through A making OA' = OA and ∠AOA' = 100º.
(iv) Similarly mark B' and C' on the arcs through B and C making OB' = OB and
OC = OC' and ∠BOB' = ∠COC' = 100º.
(v) Join A'B', B'C' and C'A' to get DA'B'C'.
Hence, DA'B'C' is the required image of DABC after rotation through 100º in
anticlockwise direction about the centre of rotation at O.
Method of finding the centre and the angle of rotation.
Let AB be a straight line and A'B' be the image of AB after rotation. To find the
centre of rotation and the angle of rotation, the following steps are followed.
(i) Join AA' and BB'. A
(ii) Find the perpendicular bisector of the two lines AA' and
BB'. B'
(ii) Here the perpendicular bisector of the two lines intersect B
at O, which is the centre of rotation. O A'
(iv) Join A and A' with O. Similarly join B and B' with O.
Then ∠AOA' or ∠BOB' is the angle of rotation.
Rotation using coordinates.
(i) Rotation through + 90º about the centre at origin (positive
quarter turn about the origin. Y
Let XOX' and YOY' be the rectangular axes of
coordinates. Let A (3, 1) be a point on the plane. A'(–1,3) A(3,1)
Join OA. Taking O as the centre and OA as O X
radius to draw an arc. Construct ∠AOA' = 90º X'
in anticlockwise direction such that OA' meets
the arc at A'. Then A' is the image of A under
rotation through + 90º. From the graph, the
coordinates of A' are (– 1, 3) . Y'
Hence, the image of A (3, 1) after rotation
through + 90º about the centre at origin is
A' (– 1, 3). i.e. A (3, 1) → A' (– 1, 3).
Similarly for P(x, y), the image after rotation through + 90º about the centre
at origin is P'(– y, x).
i.e. P (x, y) → P' (– y, x)
Infinity Optional Mathematics Book - 9 285
Note: The image obtained after rotation through + 90º about a given point is same
as the image obtained. after rotation through – 270º about the same centre
of rotation i.e. for the rotation through – 270º P (x, y) → P' (– y, x).
WORKED OUT EXAMPLES
1. Find the coordinates of the image Y
of the point A (– 7, 6) after rotation B(3,7)
C(3,5)
through 90º in anticlockwise
A(1,3)
direction about the centre at C(–5,3) OX
origin. X' A'(–3, 1) B(7,–3)
Solution: Y'
In rotation through 90º in
anticlockwise direction about the
centre at origin.
P (x, y) → P' (– y, x)
A(–7,6) → A'(–6,–7)
Hence, the image of A (– 7, 6) is A' (– 6, – 7)
2. A (1, 3), B (7, – 3) and C (3, 5) are the vertices of ∆ABC. Find the
coordinates of the vertices of the image of ∆ABC after rotation
through 270º in clockwise direction about the centre at origin.
Solution:
In rotation through 270º in clockwise direction about the centre at origin.
(i. e. – 270º)
P (x, y) → P' (– y, x) A (1, 3) → A' (– 3, 1)
B (7, – 3) → B' (3, 7) C (3, 5) → C' (– 5, 3).
286 Infinity Optional Mathematics Book - 9
(ii) Rotation through 90º about the centre at origin (Negative
quarter turn about the centre at origin)
Let XOX' and YOY' be the rectangular axes of Y
coordinates. Let A (2, 4) be a point on the plane.
Join OA. Taking O as the centre and OA as A(2, 4)
radius, draw an arc. Construct ∠AOA' = 90º in
clockwise direction such that OA' meets the arc X' OX
at A'. Then A' is the image of A under rotation A'(4, –2)
through – 90º. From the graph, the coordinates
of A' is (4, – 2).
Hence, the image of A (2, 4) after rotation Y'
through – 90º about the centre at origin is
A' (4, – 2).
i.e. A (2, 4) → A' (4, – 2).
Similarly, for P (x, y), the image after rotation through – 90º about the centre
at origin is P' (y, – x)
i.e. P (x, y) → P' (y, – x)
Note:
The image obtained after rotation through – 90º about a given point is
same as the image obtained after rotation through + 270º about the same
centre of rotation
i.e., for the rotation through + 270º,
P(x, y) → P'(y, – x)
WORKED OUT EXAMPLES
1. Find the coordinates of the image of the point A (– 5, – 7) after rotation
through quarter turn in clockwise direction about the centre at
origin.
Solution:
In rotation through quarter turn in clockwise direction about the centre at
origin (i.e. – 90º)
P (x, y) → P' (y, – x)
A (– 5, – 7) → A' (– 7, 5)
Hence, the image of A (– 5, – 7) is A' (– 7, 5).
Infinity Optional Mathematics Book - 9 287
Rotation through 180º about the centre at origin (Halt turn
about the centre at origin)
Let XOX' and YOY' be the rectangular axes of coordinates.
Let B (4, 2) be a point on the plane. Join Y
OB. Taking O as the centre and OA as
radius, draw an arc. Construct ∠BOB' =
180º, such that OB' meets the arc at B'. X' B'(4,2)
Then B' is the image of B under rotation OX
through 180º. From the graph, the
coordinates of B' is (– 4, –2). Hence the A'(4,–2)
B'(–4,–2)
image of B (4, 2) after rotation through
180º about the centre at origin is B' (– 4,
– 2).
i.e. B (4, 2) → B' (– 4, – 2). Similarly, for
P (x, y), the image after rotation through 180° about the centre at origin is
P' (–x, –y)
i.e. P (x, y) → P' (–x, –y)
Note:
The image obtained after rotation through + 180º about a given point is
same as the image obtained after rotation through – 180º about the same
centre of rotation. i.e. P(x, y) → P' (–x, –y)
WORKED OUT EXAMPLES
1. Find the coordinates of the image of A (6, 4) after rotation through
half turn about the centre at origin.
Solution: in rotation through half turn (180º) about the centre at origin,
P (x, y) → P' (– x, – y)
A (6, 4) → A' (– 6, – 4)
Hence, the image of A (6, 4) is A' (– 6, – 4).
2. A (– 3, 2), B (2, 3), C (0, 8) and D (– 4, 6) are the vertices of a quadrilateral
ABCD. Find the coordinates of the vertices of the image of quadrilateral
ABCD after rotation through half turn about the centre at origin.
Solution:
In rotation through half turn (180º) about the centre at origin,
288 Infinity Optional Mathematics Book - 9
P (x, y) → P' (– x, – y) B (2, 3) → B' (– 2, – 3)
A (– 3, – 2) → A' (3, 2) D (– 4, 6) → D' (4, – 6)
C (0, 8) → C' (0, – 8)
X
C(0, 8)
D(–4, 6)
X' B(2,3) X
A'(–3,–2) A(3,2)
O
B'(–2,–3)
D'(4,–6)
C'(0,–8)
Y'
Note:
1. Rotation through + 90° about the centre at (a, b)
P(x, y) → P' (–y + a + b, x – a + b)
2. Rotation through – 90° about the centre at (a,b)
P(x, y) → P' (y + a – b, –x + a + b)
3. Rotation through ± 180° about the centre at (a,b)
P(x, y) → P' (–x + 2a, –y + 2b)
4. Rotation through ± 360° about the centre at (a,b)
P(x, y) → P' (x, y)
Infinity Optional Mathematics Book - 9 289
Exercise 7.2
Section 'A'
1. Draw the image of the following objects with the help of given centre
of rotation, angle of rotation and the direction of rotation.
(a) A (b) A (c) A
O(+90°) B O(–90°)
BC
O(180°)
(d) P (e) A (f) A
QR BD B
O(+120°) C
O(–100°) O
C
2. Find the coordinates of the image of the following points after rotation through
90º in anticlockwise direction about the centre at origin.
(i) A (3, 4) (ii) B (– 6, 4) (iii) C (– 6, – 8) (iv) D (1, – 5)
(v) E (0, 8) (vi) F (– 6, 0) (vii) G (0, – 7) (viii) H (5, 0)
3. Find the coordinates of the image of the following points after rotation through
90º in clockwise direction about the centre at origin.
(i) M (– 6, 4) (ii) N (7, 5) (iii) P (2, – 6) (iv) K (– 1, – 5)
(v) S (0, 6) (vi) T (7, 0) (vii) U (0, – 8) (viii) V (– 9, 0).
4. Find the image of the following points under the rotation through 180º about
the origin.
(i) A (– 6, – 7) (ii) B (8, 2) (iii) C (9, - 5) (iv) D(–1, 3)
(v) E (– 7, 0) (vi) F (0, – 5) (vii) G (8, 0) (viii) H (0, 3)
5. Find the image of the following points under the rotation through 270º about
the origin.
(i) M (0, 5) (ii) N (– 7, 6) (iii) R (5, – 3) (iv) S (1, 7)
6. Find the image of the following points under rotation through – 270º about the
origin.
(i) A (– 6, 3) (ii) B (7, – 5) (iii) C (– 1, – 2) (iv) D (7, 8)
Section 'B'
7. A (2, 5), B (–1, 3) and C (4, 2) are the vertices of ∆ABC. Find the coordinates of
the vertices of the image of ∆ABC after positive quarter turn about the origin.
290 Infinity Optional Mathematics Book - 9
Also draw∆ABC and its image on the same graph.
8. A (0, – 2), B (2, 1), C (– 1, 2) and D (– 3, – 1) are the vertices of a quadrilateral
ABCD. Find the coordinates of the vertices of the image of the quadrilateral
after half turn about the centre at origin. Also plot both the object and the
image on the same graph.
9. A (2, 3), B (5, 1) and C (4, – 3) are the three vertices of ∆ABC. A', B' and C' are
the vertices of the image of ∆ABC after rotation through + 90º. A", B" and C"
are the vertices of the image of ∆A'B'C' after rotation through 180º. Find the
coordinates of vertices of ∆A'B'C' and ∆A"B"C".
10. A (– 2, 3), B (1, 6) and C (6, 1) are the vertices of ∆ABC. ∆A'B'C' is the image
of ∆ABC after rotation through half turn about the origin. Again ∆A'B'C'
is rotated through negative quarter turn about the centre at origin to get
∆A"B"C". Find the coordinates of the vertices of ∆A'B'C' and ∆A"B"C". Plot
∆ABC, ∆A'B'C' and ∆A"B"C" on the same graph.
Translation
Translation is an isometric transformation is which all the points of the object lying
on the plane are displaced from one position to another position through a fixed
distance in a definite direction.
Consider the following transformations
(a) A' (b) Q Q' (c) A' (d) D C
A A
ab B' A C
P P' B d
c D'
C'
C' A' B'
C
In figure (a), the point A is displaced to the point A' in the direction of →a at a
distance equal to the magnitude of →a . i.e. AA' = |→a |. The point A' is the image of
the point A after translation by →a .
In figure (b), the two points P and Q of the line segment PQ are displaced in the
direction of →b with the distance equal to the magnitude of →b .
Infinity Optional Mathematics Book - 9 291
In figure (c), ∆ABC is displaced by to ∆A'B'C' in the direction of →c with the distance
equal to the magnitude of →c . In the figure, AA' = BB' = CC' = |→c |.
Similarly, in figure (d), Parallelogram ABCD is displaced by →d to parallelogram
A'B'C'D' in the direction of →d with the distance equal to the magnitude of →d. In the
figure, AA' = BB' = CC' = DD' = |→d|.
The vector which is used to translate a given object is called the translation vector.
In the above examples, →a , →b , →c and →d are the translation vectors.
Properties of translation
(i) In translation, an object and its image are congruent. i.e. the object and image
have the same shape and size.
(ii) Lines joining the corresponding points of the object and its image are equal
and parallel.
Method of finding the image under translation
In translation, the image of an object is obtained with the help of translation vector.
To translate ∆ABC by translation vector →a , following steps are used.
A A'
B B'
C C'
(i) Through the point A, draw a line segment in the direction of →a and mark A'
on it making AA' = |→a |.
(ii) Through the point B, draw a line segment in the direction of →a and mark B'
on it making BB' = |→a |.
(iii) Through the point C', draw a line segment in the direction of →a and mark C'
on it making CC' = |→a |.
(iv) Join A'B', B'C' and C'A' to form ∆A'B'C'. ∆A'B'C' is the image of ∆ABC after
translation by translation vector →a .
292 Infinity Optional Mathematics Book - 9
Translation using co-ordinates
Let XOX' and YOY' be the rectangular axes of coordinates. Now plot A(1, 3), B (5,
2) and C (– 1, – 4) of ∆ABC on the graph. Y
Let T = 3 be the translation vector. The A'(4,7)
4 B'(8,6)
given translation vector is also shown on T= 3
4
the graph. A(1,3)
B(5,2)
X' O C'(2,0) X
C(–1,–4)
Now, through the points A, B and C
draw line segments in the direction of
→
translation vector equal in length of the
T
vector T. From the graph ∆A'B'C' is the
image of ∆ABC after translation by the
vector T.
Y'
From the graph, ∆A'B'C' is the image of
∆ABC after translation by the vector T. The vertices of ∆A'B'C' are A' (4, 7), B' (8,
6) and C' (2, 0).
i.e. A (1, 3) → A' (4, 7) = A' (1 + 3, 3 + 4)
B (5, 2) → B' (8, 6) = B' (5 + 3, 2 + 4)
C (– 1, – 4) → C' (2, 0) = C' (– 1 + 3, – 4 + 4)
Similarly, when T = a is a translation vector then the image of P (x, y) is
P' (x + a, y + b) b
i.e. P (x, y) → P' (x + a, y + b).
WORKED OUT EXAMPLES
1. Find the coordinates of the image of A(5, – 7) under the translation.
→T = 3 = a .
Solution 1 b
→T 3 a
: Here, the translation vector = 1 . = b .
In translation
P (x, y) → P' (x + a, y + b)
A (5, – 7) → A' (5 + 3, – 7 + 1) = A' (8, – 6)
Hence, the image of A (5, – 7) is A' (8, – 6).
Infinity Optional Mathematics Book - 9 293
2. A (4, 3), B(– 3, 6) and C(2, - 1) are the three vertices of a ∆ABC. Determine
the coordinates of the vertices of the image of ∆ABC after translation
by translation vector –3 . Also plot ∆ABC and its image on the same
2
graph.
Solution: Here, Y
the translation vector → = –3 = a . B'(–6,8)
2 b B(–3,6)
T A'(1, 5)
In translation,
P (x, y) → P' (x + a, y + b)
A (4, 3) → A' (4 – 3, 3 + 2) = A' (1, 5) T = –3 C'(–1,1) A(4,3)
B (– 3, 6) → B' (–3 –3, 6+2) = B' (–6, 8) 2 C(2,–1)
X' O X
C (2, –1) → C' (2 – 3, –1 + 2) = C' (–1, 1)
Hence, the coordinates of the image
of the vertices of ∆ABC are A' (1, 5),
B' (– 6, 8) and C' (– 1, 1).
Y'
3. Find the translation vector which translate a point A (5, 6) to the
point A' (– 1, 2).
Solution: Here, the image of A (5, 6) after translation is A' (– 1, 2)
Let, the translation vector be a .
b
Now, in translation,
P (x, y) → P' (x + a, y + b)
A (5, 6) → A' (5 + a, 6 + b)
By question, (5 + a, 6 + b) = (– 1, 2)
Now, equating the corresponding coordinates, we get
5 + a = – 1 and 6 + b = 2
∴ a = – 6 ∴ b=–4
Hence, translation vector → = a = – 6 .
b – 4
T
294 Infinity Optional Mathematics Book - 9
Exercise 7.3
Section 'A'
1. Draw the image of the following geometrical figures under the
translation by the given translation vector.
(a) A (b) A (c) A
B BC B
(d) P (e) M (f) C
A
QS NK
R R BD
C
2. Find the coordinates of the images of the following point under the
translation → = 3 .
2
T
(a) A (5, 2) (b) B (– 7, 3) (c) C (– 6, – 4)
(d) D (8, – 1) (e) E (0, 6) (f) F (8, 0)
(g) G (0, – 3) (h) H (– 6, 0)
3. Find the coordinates of the images of A (6, 3) B (– 3, – 5), C (9, – 6) and
D (7, – 8) with the following translation vectors.
(a) T1 = 0 (b) T2 = –3 (c) T3 = 5 (d) T4 = –5
2 4 –3 –6
4. Find the translation vector which maps a points A (– 1, 2) to A' (– 4, – 6).
Using the same translation vector, find the images of the points B (– 2, 4) and
C (7, 5).
Infinity Optional Mathematics Book - 9 295
5. Find the translation vector which translates a point M (6, – 3) to the point
M' (8, 4). Also find the image of A (9, 2) and B (– 1, 2) under this translation.
6. A(2, 6), B (– 1, 4) and C (6, 2) are the vertices of ∆ABC. Find the coordinates
of the vertices of image of DABC by the translation vector 3 . Also, plot ∆ABC
2
and its image on the same graph.
7. A (– 1, 2), B (5, 3) C (4, – 3) and D (– 3, – 2) are the vertices of quadrilateral
ABCD. Find the coordinates of the vertices of the image of quadrilateral ABCD
by the translation vector –1 . Also plot both the object and image on the same
–3
graph.
8. A translation T maps a point B (3, 7) to B' (1, – 2). Find the vertices of ∆PQR
with the vertices P(– 1, –2), Q(2, 5) and R (3,2) under the same translation T.
Section 'B'
9. A (2, 4), B (5, 2) and C (8, 7) are the vertices of ∆ABC. Find the coordinates of
→→
the image after translating in the magnitude and direction of AC. (ii) 2BA.
10. If the points P(3, 3), Q (2, 5), R (– 3, – 5) and S (7, – 1) form a quadrilateral
PQRS, find its translated image P'Q'R'S' in the magnitude and direction of 1
2
→
PR. Also show quadrilateral PQRS and its image P'Q'R'S' on the same graph.
296 Infinity Optional Mathematics Book - 9
Enlargement
Enlargement is a non–isometric transformation in which the size of an object or
geometrical figure is changed without changing its shape. When the size of an object
increases, then it is called enlargement and when the size of the object decreases,
it is called reduction.
Consider the following transformation
(a) A' (b) R' Q' (c) A F
M
O A P R A' F'
C' Q E' E
B B B' N
B'
C' D'
C
C D
P'
Figure (i) Figure (ii) Figure (iii)
In figure (i), O is a fixed point and ABC is a triangle. Here OA, OB and OC are
joined and produced them up to A', B' and C' such that OA' = 2OA, OB' = 2OB and
OC' = 2OC. ∆A'B'C' is the image of ∆ABC whose shape is same as the shape of ∆ABC
and is two times bigger in size. This is the enlargement. Here O is called the centre
of enlargement.
In figure (ii), ∆P'Q'R' is the image of ∆PQR. They have the same shape and size. But
are on the opposite side of centre of enlargement M.
In figure (iii), Hexagon A'B'C'D'E'F' is the image of hexagon ABCDEF, where the
size of image is half of the size of object. Here, N is the centre. This is the reduction.
Centre of enlargement and scale factor
The enlargement (or reduction) is made with the help of a fixed point. The fixed
point is called the centre of enlargement. In the above figures O, M and N are the
centre of enlargement.
The ratio of the corresponding sizes of the image and the object is called the scale
factor. It is denoted by k.
In figure (i),
Scale factor (k) = A'B' = B'C' = C'A' = 2.
AB BC CA
Infinity Optional Mathematics Book - 9 297
In figure (ii)
Scale factor (k) = P'Q' = Q'R' = R'P' = 2.
PQ QR RP
Hence, scale factor (k) = size of image
size of object
Scale factor can also be defined as the ratio of image distance and object distance
from the centre of enlargement (or reduction).
From figure (i), Scale factor (k) = OA' = OOBB'= OOCC'.
OA
For enlargement (or reduction), centre of enlargement (or reduction) and scale
factor must be given. An enlargement with centre at point O and scale factor k is
represented by E [Q, k].
Properties of enlargement.
1. The object and its image are similar.
2. If k > 1, then the image is larger in size than the given object.
3. If k = 1, then the size of image and object are same.
4. If k < 1 the image is smaller in size than the given object.
5. If k is positive the image and object lie on the same side of the centre of
enlargement.
6. If k is negative, the object and image lie on the opposite of centre of enlargement.
The image will be inverted.
7. The corresponding sides of object and image are parallel.
8. The object point, its image point and the centre of enlargement lie on a straight
line.
Method of finding the image under an enlargement.
1. When the scale factor is positive. A'
Here, centre of enlargement is O and scale factor k = 2. A
The image of ∆ABC is obtained by the following
procedures: O B B'
C
(i) Join OA and produce it to A' making OA' =
2OA. C'
(ii) Join OB and produce it to B' making OB' =
2OB.
(iii) Join OC and produce it to C' making OC' = 2OC
(iv) Join A' B' and C'.
Hence, ∆A'B'C' is the image of ∆ABC under the enlargement E [0, 2].
298 Infinity Optional Mathematics Book - 9
2. When the scale factor is negative
Here O is the centre of enlargement and - 1 is a scale factor. When the scale
factor is negative, the image and object are formed on the opposite sides of the
centre. The image of ∆ABC is obtained by the following procedures. C'
A
(i) Join AO and produce it to A' making OA' = AO
(ii) Join BO and produce it to B' making OB' = BO. B o B'
(iii) Join OC and produce it to C' making OC' = CO. C A'
(iv) Join A', B' and C'
∴ ∆A'B'C' is the image of ∆ABC under the enlargement
E [0, – 1].
Eenlargement Using coordinates Y
1. When the centre of enlargement C(12,8)
is at origin and scale factor is k, A'(2,6)
E[O, k]. A(1,3) C(6,4)
In the given figure, XOX' and
YOY' are the rectangular axes of X' X
co-ordinates. OB(4,–1)
Let A (2, 6), B (4,–1) and C (6, 4) B'(8,–2)
be the vertices of a ∆ABC. Let the
centre of enlargement be origin
(0, 0) and scale factor (k) be 2.
Join OA and produce it to A' Y'
making OA' = 2OA. Also, join OB
and produce it to B' making OB' = 2OB.
Similarly join OC and produce it to C' making OC' = 2OC.
Then A', B' and C' are the images of A, B and C respectively. The co-ordinates
of the images are A'(2, 6), B' (8, – 2) and C' (12, 8). Hence, when the centre of
enlargement is at origin and scale factor is 2, then
A (1, 3) → A' (2, 6) = A' (2 × 1, 2 × 3)
B = (4, – 1) → B' (8, – 2) = B' (2 × 4, 2 × – 1)
C (6, 4) → C' (12, 8) = C' (2 × 6, 2 × 4)
From the above examples, we can say that when the centre of enlargement is
taken as origin and scale factor as k, then the coordinates of the image of P (x,
y) is P' (kx, ky).
i.e. P (x, y) → P' (kx, ky).
Infinity Optional Mathematics Book - 9 299
WORKED OUT EXAMPLES
1. Find the image of A (3, 5) under the enlargement with the centre at
origin and scale factor – 3.
Solution: Here, the centre of enlargement is at (0, 0).
Scale factor (k) = – 3
We know that, for centre at origin and scale factor k.
p (x, y) → P' (kx, ky)
A (3, 5) → A' (– 3 → 3, – 3 → 5) = A' (– 9, – 15).
2. A (2, 2), B (5, 1) and C (3, 5) are the vertices of ∆ABC. Find the
coordinates of the image of ∆ABC after enlargement with centre at
origin and scale factor 2. Present ∆ABC and its image on the same
graph.
Solution: Y C'(6,10)
Here, centre of enlargement is at C'(3,5)
origin and scale factor (k) = 2. A(4,4)
We know that, for centre at origin A(2,2)
and scale factor k. A(5,1) B'(10,2)
P (x, y) → P' (kx, ky) X' OX
A (2, 2) → A' (2 × 2, 2 × 2) = A' (4, 4)
B (5, 1) → B' (2 × 5, 2 × 1) = B' (10,
2)
C (3, 5) → C' (2 × 3, 2 × 5) = C' (6,
10)
Y'
2. When the centre of enlargement is at (a, b) and scale factor k,
E [(a, b), k].
In the given figure, XOX' and YOY' are the rectangular axes of co-ordinates.
Let A (a, b) be the centre of enlargement and k be the scale factor.
Let P (x, y) be the given point.
Join AP and produce it to P' making AP' = kAP.
Let the coordinates of P' be (x', y')
300 Infinity Optional Mathematics Book - 9
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