246 PRIME Opt. Maths Book - IX 2. If A(1, 2) and B(7, 10) are any two points, find components of AB & vector AB . Solution : Initial point is A(1, 2) Final point is B(7, 10) Then x - component = x2 – x1 = 7 – 1 = 6 y - component = y2 – y1 = 10 – 2 = 8 \ AB = x comp y comp - - d n = 6 8 d n 3. If A(–3, 2), B(1, 4), P(5, 1) and Q(1, –1) are the four points, prove that AB = –PQ . Also prove that they are parallel to each others. Solution : The given points are A(–3, 2), B(1, 4), P(5, 1) and Q(1, –1) Then, AB = y x x y – – 2 1 2 1 d n = 1 3 ( ) 4 2 – – – d n = 4 2 d n PQ = y x x y – – 2 1 2 1 d n = 1 5 1 1 – – – d n = 4 2 – – d n Here, AB = 4 2 d n = – 4 2 – – d n = –PQ \ AB = – PQ 4. If P(a, 3), Q(7, 1), R(5, 3), S(2, 5) and PQ = RS , find the value of ‘a’. Solution: Taking the points P(a, 3) & Q(7, 1) PQ = y x x y – – 2 1 2 1 d n = 7 a 1 3 – – c m = 7 a 2 – – c m Taking the points R(5, 3) & S(2, 5) RS = y x x y – – 2 1 2 1 d n = 2 5 5 3 – – d n = 3 2 – c m Also, Taking PQ = – RS or, 7 a 2 – – c m = – 3 2 – c m or, 7 a 2 – – c m = 3 2 – c m Again, AB = – PQ is same as a = mb for a ||b \ AB ||PQ proved.
PRIME Opt. Maths Book - IX 247 Exercise 6.1 1. i) Differentiate between vector and scalar in two points. ii) Which of the following quantities are the scalar quantities? Length, work, mass, density, force, acceleration, power, time iii) Which of the quantities given above are the vector quantities? iv) Which type of line segments given below ? For what purpose are they used? A B P Q v) Write down the components of vector AB from the given diagram. Also write down the column vector AB . B 4 3 A 2. Study the given graph and write down the components and vector for the followings. i) AB ii) CD iii) PQ iv) RS v) MN D C B P S R M Q N A 3. Show the following vectors in directed line segment. i) PQ 3 5 = d n ii) AB 2 4 – = d n iii) CD 3 4 – – = d n iv) RS 3 –5 = d n v) a 6 8 – = d n Equating the corresponding x-component. 7 – a = 3 or, 7 – 3 = a or, 4 = a \ a = 4
248 PRIME Opt. Maths Book - IX 4. i) If A(3, 2) and B(5, 6) are the any two points, find the components and vector of AB . ii) If A(1, 4) is a point, find the components and vector of OA . iii) If P(3, 2) is a point find the position vector of P. iv) If P(–1, 4) and Q(3, 1) are any two points, find PQ . v) If M(3, –2) and N(–1, – 5) are any two points, find MN . 5. PRIME more creative questions: i) A(4, –5) displaced to B(3, –6) in the vector AB and G(2, 5) displaced to H(1, 4) in the vector . Prove that AB = GH . ii) D(–3, 2) displaced to E(1, –3) in vector DE and K(1, –1) displaced to L(–3, 4) in the vector KL . Prove that KL = – DE iii) If A(3, 2), B(4, 5), C(1, 7) and D(2, 10) are the four points. Prove that AB = CD . Also prove that AB | |CD . iv) If P(3, 7), Q(1, 4), R(5, –1) and S(3, –4) are the four points, prove that PQ = RS . Also prove that they are parallel to each other. v) If A(2, 4), B(7, 8), P(3, –1) and Q(–2, –5) are the four points, prove that AB = –PQ . Also prove that they are parallel to each other. 6. i) If A(3, x), B(1, 4), C(5, –1) and D(3, –4) and AB = CD , find the value of ‘x’. ii) If P(–2, –5), Q(3, –1), R(7, 8), S(m, 4) and PQ = –RS , find the value of ‘m’. iii) If A(2, 4), B(6, 3), C(–3, 5) and D(a, b) are any four points where AB = CD , find the co-ordinate of D. iv) If P(–3, 2), Q(–2, 4), R(0, –2), S and PQ = RS , find the co-ordiante of S. v) If CD = –RS in the points C(P, 2), D(1, –3), R(1, –1) and S(–3, q), find the value of p and q. Answer 1. Show to your teacher. 2. Show to your teacher. 3. Show to your teacher. 4. i) 2, 4, 2 4 d n ii) 1, 4, 1 4 d n iii) 3 2 d n iv) 4 –3 d n v) 4 3 – – d n 6. i) x = 7 ii) m = 2 iii) (1, 4) iv) (1, 7) v) (–3, 4)
PRIME Opt. Maths Book - IX 249 6.2 Vector operations 6.2.1 Magnitude of a vector Let us consider the two points A(1, 2) and B(4, 6). AB = . . x comp y comp - - e o = AR RB d n = x x y y – – 2 1 2 1 d n = 4 1 6 2 – – d n = 3 4 d n Where, Length of AB = d(AB) = AR RB 2 2 + = 3 4 2 2 + = 5 units \ Magnitude of AB = 5 units. It is written as modulus of vector AB = | AB | = 5 units. The modulus of a vector which denotes the length of the vector is called magnitude of the vector. i.e. If AB = . . x comp y comp - - e o | AB | = (x - comp.)2 + (y - comp.)2 It is called the absolute value of vector too. 6.2.2 Direction of a vector The angle made by the directed line segment for a vector with positive direction of x - axis (taking the ratio of y - component and x-component as the slope of a line segment) is called direction of the vector. i.e. for direction q of a vector we have slope = Tanq = . comp comp x y - - R Y X A(1, 2) B(4, 6) O M N
250 PRIME Opt. Maths Book - IX Let us consider: Here, AB = comp comp x y - - d n = AR BR d n = PQ BQ – RQ d n = OQ OP BQ AP – – d n = x x y y – – 2 1 2 1 d n = 2 1 2 3 3 – – d n = 1 3 d n Then, Slope of AB = . comp comp x y - - or, Tanq = 1 3 or, Tanq = Tan60° \ q = 60° \ Direction of AB is 60° The angle made by a vector with x-axis in positive direction which is calculated by taking the tangent ratio is called direction of a vector. i.e. Tanq = . comp comp x y - - where ‘q’ is the direction Important points for direction of vector: 1. If x-component and y- component both are positive, direction= q. 2. If x-component is –ve and y- component is positive (+ve), direction = 180° – q. • If x-component is -ve and y-component -ve, direction = 180° + q which is same as 1. • If x-component is +ve and y-component -ve, direction = 360° – q which is same as z. Note : As domain for Tan is –90° < q < 90° so we find most of the angles written in both negative and positive sign. O (+, +) direction = q (+, –) direction = 360° – q (–, –) direction = 180° + q (–, +) direction = 180° – q Y’ Y X’ X A B (2, 2 3 ) (1, 3 ) R P Q X Y O q q
PRIME Opt. Maths Book - IX 251 6.2.3 Unit vector The vector having magnitude one is called unit vector. i.e. AB = 1 Where, If AB = 1 0 d n Here, AB = x y 2 2 + = 1 0 2 2 + = 1 \ AB is the unit vector. Taking another vector PQ , If PQ = 3 5 4 5 J L K K K K K KK N P O O O O O OO Here, PQ = x y 2 2 + = = 25 9 1 + 6 = 1 \ PQ is a unit vector. Formula calculating of unit vector. If a is a vector. Unit vector along a (a) = a a 1 ^ h = its u mod lus vector If a = 6 8 d n Then, |a | = x y 2 2 + = 6 8 2 2 + = 10 units Then, Unit vector along a = a a 1 ^ h \ a = 10 1 6 8 d n a = 3 5 4 5 J L K K K K K KK N P O O O O O OO
252 PRIME Opt. Maths Book - IX 6.2.4 Addition of vectors Let us consider AB & BC are any two vectors having same direction, they can be combined as AB + BC which will be the new vector AC . A B A C C Let us consider, a = 1 2 d n and b = 3 4 c m Then, a + b = 1 2 d n + 3 4 c m = 4 6 c m The sum of any two vectors is the resultant vector obtained by adding their corresponding components. If, a = x y 1 1 d n & b = x y 2 2 d n Resultant vector of their sum = a + b = x y 1 1 d n + x y 2 2 d n = x x y y 1 2 1 2 + + d n Note : Difference of any two vectors is also same as the addition. i and j vector in the form of column vector The unit vectors are the vectors having magnitude one where such unit vectors taken along x-axis and y-axis are denoted by i and j vectors. i.e. i = unit vector along x-axis = 1 0 d n j = unit vector along y-axis = 0 1 d n In the given diagram, O Y B(0, 1) A(1, 0) X
PRIME Opt. Maths Book - IX 253 i = OA = 1 0 d n j = OB = 0 1 d n By using such unit vectors any kind of column vector AB , OA , OP , PQ , d can be expressed in i and j vectors where, a x y = d n = x 0 y 0 d d n n + = x y 1 0 0 1 d d n n + = xi + yj 6.2.5 Properties of vector addition i) Commutative property: a b + = b a + ii) Associative property: a b + + ( ) c = ( ) a b + + c iii) Addition with negative vector: a a + (– ) = (–a a ) + = 0 iv) Addition identity: a o + = o a + = a 6.2.6 Multiplication of a vector by scalar. Let us consider, a = 4 6 c m = 2 2 2 3 × × d n = 2 2 3 d n Here, 2 is the common constant value of both components, which is the scalar value. Multiplication of a vector by a scalar results the new vector obtained by multiplying each component with the given scalar. It multiples the magnitudes without changing the direction of the vector if scalar is positive. Example: If a = x y d n is a vector. k is a scalar. Then, ka = k x y d n = kx ky d n
254 PRIME Opt. Maths Book - IX Example : If a = 2 1 c m & b = 0 1 c m, find the value of 2 a + b . Solution: Here, a = 2 1 c m , b = 0 1 c m Then, 2a + b = 2 2 1 c m + 0 1 c m = 4 2 c m + 0 1 c m = 4 3 c m 1. If a =( , 3 1), find its magnitude. Solution, a = ( , 3 1) Magnitude of a is, |a | = x y 2 2 + = 3 1 2 2 ^ h + ^ h = 4 = 2 units 2. Find the direction of a = 3 1 e o Solution : a = 3 1 e o For the direction of a q, Tanq = component component x y - - Tanq = 3 1 or, Tanq = Tan30° \ q = 30° Worked out Examples
PRIME Opt. Maths Book - IX 255 3. Find the magnitude and direction of AB where A(1, –4) and B(5, 0) are any two points. Solution : Taking the points A(1, 4) and B(5, 0) AB = x x y y – – 2 1 2 1 d n = 5 1 0 4 4 4 – + d d n n = Then, Magnitude of AB is, AB = x y 2 2 + = 4 4 2 2 + = 32 = 4 2 units Again, For the direction of AB ‘q’, Tanq = component component x y - - = 4 4 = 1 or, Tanq = Tan45° \ q = 45° 4. If a = 6 8 d n, find the unit vector of a . Solution : a = 6 8 d n |a | = x y 2 2 + = 6 8 2 2 + = 100 = 10 unit Then, unit vector of a is, a = a a 1 ^ h U = 10 1 6 8 d n = 6 10 8 10 J L K K K K K KK N P O O O O O OO = 3 5 4 5 J L K K K K K KK N P O O O O O OO
256 PRIME Opt. Maths Book - IX 5. If direction of a = 2 3 i p – j is 330°, find the value of ‘p’. Also find the magnitude of a . Solution : a = 2 3 i p – j = –p 2 3 e o For direction q = 330° Tanq = x component y component - - or, Tan 330° = p 2 3 – or, – 3 1 = p 2 3 – \ p = 2 Again, magnitude of a is, |a | = (x - comp.)2 + (y - comp.)2 = (2 3 )2 + (–p)2 = 12 + 4 = 4 units. 6. If a 4 2 = d n, b 1 3 = d n, find the magnitude of a b + 2 . Solution : a 4 2 = d n b 1 3 = d n Then, a 2 + b = 4 2 2 1 3 d d n n + = 4 2 2 6 + + e o = 6 8 d n \ Magnitude of a 2 + b is, | a 2 + b | = x y 2 2 + = 6 8 2 2 + = 100 = 10 units.
PRIME Opt. Maths Book - IX 257 Exercise 6.2 1. i) What do you men by magnitude of a vector? ii) Define the term direction of a vector. iii) What do you mean by position vector? iv) What do you men by unit vector? v) Is a 1 0 = d n a unit vector? 2. Find the magnitude of the following vectors. i) a 3 4 = d n ii) b =( , 8 17) iii) AB 6 8 = d n iv) AB for A(1, 1) and B(–3, –2) v) PQ for P(3, 7) and Q(–1, 7) 3. Find the direction of the following vectors. i) a =( , 2 3 2) ii) b 3 – 27 = d n iii) AB 4 0 = d n iv) PQ for P(–2, –1) and Q(–6, –5) iii) CD for C(–2, 5) and D(–2, 8) 4. Find the unit vector of the following vectors. i) a 2 1 = d n ii) b 6 8 – = d n iii) AB = ^ h 2 2, 2 iv) PQ for P(3, –2) & Q(–1, 1) v) RS for R(–1, –3) & S(7, 3) 5. Write down the followings in i and j vectors. i) a 2 3 = d n ii) b 3 4 – – = d n iii) 3a where a 2 –3 = d n iv) PQ where p(3, – 2) and Q(1, –5) v) 2AB where A(–3, 1) and B(1, 4) 6. If a 3 2 = d n and b 1 2 = d n, find the followings. i) a + b ii) a – b iii) 2a + b iv) 3a – 2b v) 3 2 a + b
258 PRIME Opt. Maths Book - IX 7. PRIME more creative questions: a. i) If a = 1 2 c m and b = 2 3 d n, find the magnitude of a + b . ii) If a = 2 3 d n and b = 4 0 c m, find the magnitude of 2a + b . iii) If a = 2 4 c m and b = 3 4 c m, find the magnitude of a + 2b . iv) If A(–3, –1) and B(2, 4), find the magnitude and direction of AB . v) If magnitude of a x 4 = d n is 5 units, find the value of ‘x’. b. i) If a i = 3 3 – j , find the magnitude and direction of a . ii) If magnitude of a m= i j – 8 is 10 units, find the value of ‘m’. iii) If direction of a = pi – 27 j is 60°, find the value of ‘p’. iv) If a 0 – 3 = d n, k b –2 3 = d n and a b + = 6 units, find the positive value of ‘k’. Also find the direction of a b + . v) If a i = 5 – j , b m = – – i j 3 and magnitude of a b – 2 is 5 2 units, find the negative value of ‘m’. Also find the direction of a b – 2 . Answer 1. Show to your teacher. 2. i) 5 units ii) 5 units iii) 10 units iv) 5 units v) 4 units 3. i) 30° ii) 300° iii) 0° iv) 225° v) 90° 4. i) 5 2 5 1 J L K K K K K K K KK N P O O O O O O O OO ii) 5 3 5 4 – J L K K K K K K KK N P O O O O O O OO iii) , 5 2 5 1 c m iv) 5 4 5 3 – J L K K K K K K KK N P O O O O O O OO v) 5 4 5 3 J L K K K K K K KK N P O O O O O O OO 5. i) 2 3 i j + ii) – – 3 4 i j iii) 6 9 i j – iv) – – 2 3 i j v) 8 6 i j + 6. i. 4 4 ii. 2 0 d n iii. 7 6 d n iv. 7 2 d n v. 9 10 d n 7. a. i. 34 units ii. 10 units iii. 4 13 iv. 5 2 units, 45° v. +3 b. i. 2 3 units 330° ii. 6 iii. –3 iv. 3, 300° v. –5, 135°
PRIME Opt. Maths Book - IX 259 6.3 Concepts of vector geometry: 6.3.1 Triangle law of vector addition If the point A is displaced to B by AB and the point B is displaced by C by BC , then the sum of AB and BC (which is denoted by AB + BC ) is given by the vector AC where the point A is displaced to C. The vector AC is called the resultant vector of the vectors AB and BC . i.e. AB + BC = AC . Also, AB + BC = AC . or, AB + BC – AC = 0 or, AB + BC + CA = 0 Note : As the application of triangle law resultant vector can be calculated in different polygons like quadrilateral pentagon, hexagon etc. 6.3.2 Quadrilateral law of vector addition In a quadrilateral ABCD, AC is a diagonal. By using triangle law of vector addition, AB + BC = AC ^In TABCh ............................ (i) AC + CD = AD ^In TACDh Then, AC + CD = AD or, AB + BC + CD = AD (from equation ‘i’) [It is quadrilateral law of vector addition] or, AB + BC + CD – AD = 0 or, AB + BC + CD + DA = 0 6.3.3 Parallelogram law of vector addition: C B D A In a parallelogram ABCD, AC is a diagonal where, AB + BC = AC [a using triangle law] or, AB + AD = AC [a AD = BC \ AB + AD = AC is the parallelogram law of vector addition. i.e. C A B C D A B
260 PRIME Opt. Maths Book - IX C c b a O A B The sum of any two vectors having same initial point is equal to the diagonal vector represented by the parallelogram completed in the given vectors is called parallelogram law of vector addition. OA and OB are any two vectors having same initial point ‘0’ where, OA = a and OB = b OC = c is a diagonal vector completed in the given vectors OA and OB . \ a + b = c . 6.3.4 Addition of column vectors: Let OA = y x 1 1 a k and OB = y x 2 2 a k the position vector of points A(x1 , y1 ) and B(x2 , y2 ) respectively construct a parallelogram OACB as shown in diagram to presents the given vectors. Draw the perpendiculars, AM⊥OX, BN⊥OX, CP⊥OX and BQ⊥CP. Then, DAOM , DCQB. \ OM = BQ = NP = x1 CQ = AM = y1 Where, OP = ON + NP = x2 + x1 PC = PQ + QC = NB + QC = y2 + y1 . \ Co-ordinate of c is (OP, CP) = (x1 + x2 , y1 + y2 ) According to parallelogram law of vector addition, OA + OB = OC i.e. y x 1 1 a k + y x 2 2 a k = y y x x 1 2 1 2 + + a k. B(x2 , y2 ) A(x1 , y1 ) Q M N P X Y O C
PRIME Opt. Maths Book - IX 261 6.3.5 Theorems related to position vector Theorem 1: Position vector of a point which divides a line segment at mid-point. Let, a point M cuts a line segment AB at mid-point where, the position vectors are, OA = a OB = b OM = ? Now, taking AM = MB or, AM = MB [\Being in same direction] or, AO + OM = MO + OB [\Using triangle law] or, –OA + OM – MO = b or, –a + OM + OM = b or, 2OM = a + b or, OM = a b 2 + \ OM = OA OB 2 + Theorem 2: Position vector of a point which cuts the line segment in the ratio m:n internally. Solution : Let, a point ‘P’ cuts a line segment AB in the ratio m:n internally, where position vectors of A and B are OA = a OB = b OP = ? Here, PB AP n m= or, n AP = m PB or, n AP = mPB or, n( ) AO + OP = m( ) PQ + PB [\ Using triangle law ] or, n( ) –a O+ P = m(–OP + b) or, –n a n + OP = mOP + mb or, nOP + mOP = na + mb or, (m + n)OP = na + mb \ OP = m n mb na + + i.e. OP = m n m n OB OA + + O B P m n a b ? A ? B M A O b a
262 PRIME Opt. Maths Book - IX 1. Express the sides of regular hexagon ABCDEF, in terms of a and b where AB = a and BC = b . Solution: In the regular hexagon ABCDEF, AB = a and BC = b Join CF & AD which are intersected At ‘0’. Then, CD = BO = BA + BC = –a + b DE = OF = BA = – a EF = DO = CB = – b FA = OB = OC + OA = AB + CB = a – b . Worked out Examples B C O E D F A b a Theorem 3 : Position vector of a point which cuts a line segment in the ratio m:n externally. Solution : Let, a point P cuts the line segment AB in the ratio m:n externally, where the position vectors are, OA = a OB = b OP = ? Here, from the concept of section point, BP AP n m= or, n AP = mBP or, n( ) AO + OP = m( ) BO + OP or, n( ) –a O+ P = m(–b O+ P) or, –n –na n + OP = m – ) mb + mOP or, mb – na = OP( ) m n – \ OP = m n – mb – na i.e. OP = m n – m n OB – OA O P B m n a b ? A
PRIME Opt. Maths Book - IX 263 2. In a parallelogram ABCD, OA = a , OB = a and OC = c for any point ‘0’. Find OD in terms of a , b & c . Solution: In a parallelogram ABCD, OA = a OB = b OC = c OD = ? Then, using triangle law, OD = OA + AD = OA + BC = OA + BO + OC = a – b + c \ OD = a – b + c 3. In the parallelogram ABCD. AB = a . Find AC + DB in terms of a . Solution: ABCD is parallelogram, AB = a Now, Using triangle law, AC + DB = ^AB + BCh + ^DA + ABh = AB + BC + CB + AB = 2AB + BC – BC = 2AB = 2a \ AC + DB = 2a 4. Find the position vector of a point p under a & b from the given diagram where 2AP = 3PB. Solution : OA = a OB = b 2AP = 3PB Now, using triangle law in the given condition, 2 AP = 3PB or, 2^ h AO + OP = 3^ h –OP + b or, ^ h –a O+ P = 3 –^ h OP + b or, 2OP + 3OP = 2 3 a b + or, 5OP = 2 3 a b + \ OP = 5 1 (2 3 a b + ) C a B D A O B P a b A ? B C A D O a b c
264 PRIME Opt. Maths Book - IX Exercise 6.3 1. i) Explain triangle law of vector addition. ii) What do you mean by i and j vectors? iii) Write down the formula of position vector of mid-point of a line segment AB. iv) Write down the formula of position vector of section point of a pine segment PQ in the ratio. v) State parallelogram law of vector addition. 2. Find the following: i) If a = 3 2` j and b = 2 1 – ` j, find a b + and write down in i and j vector. ii) If OA = 4i + 3j and OB = 5i – 2j , find AB . iii) If position vectors of P and Q are 2i – j and 5i – 3j , find PQ . iv) If 2a + 3b = (0, –7) and b = (2, –3), find a . v) If a b + = 1 5` j, a = 4 0` j, find b in i and j form. 3. Prove that the followings: i) AB + BC + CA = 0 in DABC. ii) PQ + QR + RS + SP = 0 in quadrilateral PQRS. iii) AB + BC + CD + DE = AE in a pentagon ABCDE. iv) DE + EF + FG + GH + HI + ID = 0 in the given diagram. E F H G I D v) AC + BD = 2BC in a parallelogram ABCD. 4. i) Express OP in terms of a and b where P is the mid- point of AB in the adjoining diagram. ? B P A O b a
PRIME Opt. Maths Book - IX 265 ii) Prove that 2AR = AP + AQ in the given diagram where R is the mid-point of PQ. Q R P A iii) Find OC in terms of p and q where AC = 2CB in the adjoining diagram. B C A O p q iv) Find OP in terms of m and n where 3AP = 2PB . B P A O n m v) Prove that AB + AC + AD + EA + FA = 4AB in a regular hexagon ABCDEF. D E A F B C
266 PRIME Opt. Maths Book - IX 5. PRIME more creative questions: a. i) If A and B are the mid points of sides QR and RS of a parallelogram PQRS respectively, prove that 2^PA + PBh = 3PR . ii) In a parallelogram ABCD, AB = a & AD = b . Find the vectors CB , DB & AP in terms of a & b where P is the intersecting point of diagonals. iii) In the given diagram ABCD is a parallelogram and P is the intersecting point of diagonals. Prove that OA + OB + OC + OD = 4OP for any point ‘O’. C P B D A O iv) Prove that AD + BE + CF = 0 from the given diagram where D, E, & f are the mid-point of sides of DABC. C B D A E F v) Find the followings from the given diagram in the single vector. AB + BC DO + OC DC + CA DO + AD A D B O C vi) Prove that the position vector of centroid of a triangle is g 3 1 = ^ h a b + + c . b. i) If 4 3 a b – and 2 5 a b + are the position vectors of p and Q, find PQ . ii) If a = 1 –2 ` j, b = 3 7 – ` j & m = 2, prove that m^a b + h = ma + mb .
PRIME Opt. Maths Book - IX 267 Answer 1. Show to your teacher 2. i. 3i j + ii. i j – 5 iii. 3 2 i j – iv. (–6, 2) v. 5 3 i j – 4. i. OP = ( ) a b 2 1 + iii. OC = ( ) p q 3 1 + 2 iv. OP = ( ) m n 5 1 3 2 + 5.a. ii. b a, – b , ( ) a b 2 1 + v. AC, , DC DA, AC b. i. – a2 3 + b iii. 2 6 i j – v. a b a b + , a b a b – D C B A , a b b a – D C B A a b + 2 a + b a A C B O iii) If AB = –5 7 i j – and BC = 7i j + , find AC . iv) Where and how vector operations are used in our daily life? Prepare report and present in your class. v) Present a b + , a b – , b a – and 2a b + in arrow diagrams. 6. Project work Collects the formula and operation system of vectors in a chart paper and present into the classroom.
268 PRIME Opt. Maths Book - IX Vector Unit Test Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. Define the term position vector of a point. 2. a. If a = 3 –3 c m, find the direction of a . b. Find the unit vector of 3i – 4j . c. If magnitude of a vector a = m –6 ` j is 10 units, find the value of ‘m’. 3. a. If a = 1 2 –a k and b = 2 3a k, find the magnitude and direction of a + 2b . b. Prove that AB + BC + CD +DA = 0 in a quadrilateral ABCD. 4. If M is the mid - point of a line segment AB where OA = a and OB = b , find the position vector of M in terms of a and b.
PRIME Opt. Maths Book - IX 269 Unit 7 Transformation Specification Grid Table K(1) U(2) A(4) HA(5) TQ TM Periods No. of Questions 1 – 1 1 3 10 18 Weight 1 – 4 5 K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total Question, TM = Total Marks Objectives : At the end of the lesson • Students are able to know the concept of transformation. • Students are able to identify the types of isometric and non- isometric transformations. • Students are able to know the formula to find image of a point. • Students are able to find the image of an object for different transformations and can plot them in graph. Materials Required: • Model of transformation. • Geo-board. • Flash card. • Graph paper. • Chart paper. • Formula chart of transformations.
270 PRIME Opt. Maths Book - IX 7.1 Transformation Let us see the shape of fishes of same size as well as the printed clothes as shown in the diagram. Here one fish is transferred to the another fish continuously and size of fishes can be seen which is the concept of transformation in the first example. One flower is transferred to the another continuously and shape of flowers are printed in the clothes as shown in diagram in second example which give the concept of transformation. 7.1.1 Types of transformation : Isometric transformations : The transformation in which objects and its image are always congruent is called isometric transformation. Example : Reflection, Rotation and translation. Non-Isometric transformations : The transformation in which object and image are not congruent but only similar is called non-isometric transformation. Example: Enlargement 7.1.2 Reflection on mirror line : Image of an object is shifted from one place to another place by the mirror line as the reflection axis where image and object are always congruent in reflection. The process of changing the position, shape and size of objects under any geometrical conditions is called transformation.
PRIME Opt. Maths Book - IX 271 Here, M is the mirror line. DABC is an object. Draw, BP⊥M, BP = PB’ AQ ⊥M, AQ = QA’ CR⊥M, CR = RC’ Then, DA’B’C’ so formed is called image of DABC. 7.1.3 Reflection using co-ordinate: i) Reflection on x-axis. Theoretical proof Let us consider a point p(x , y) is reflected to p’(x’ , y’) by the reflection axis as x-axis (y=0). Draw PM⊥OX, \ PM = MP’ where M is (x , 0) which is the midpoint of PP’. Then, Using mid-point formula, x = x x 2 1 2 + , y = y y 2 1 2 + or, x = x x 2 + 1 , 0 = y y 2 + 1 or, 2x = x + x’ , y + y’ = 0 or, x’ = x , y’ = -y \ p(x , y) " p’(x’ , y’) = p’(x , -y) ii) Reflection on y-axis: Theoretical proof Let us consider a point p(x, y) is reflected to p’(x’, y’) under the reflection axis y - axis (x = 0) Draw PM =OY, Then, PM = MP’ where M (0, y) is the mid - point of PP’. Then, Using mid - point formula, x = x x 2 1 2 + , y = y y 2 1 2 + or, x = x x' 2 + , y = y y' 2 + or, x + x’ = 0 y + y’ = 2y ` x’ = – x y’ = y ` p(x, y) $ p’(x’, y’) = p’(–x, y) X’ X O Y Y’ P(x, y) P’(x’, y’) (x, 0) M B’ B C Q P M R A C’ A’ X’ X O Y Y’ P(x, y) P’(x’, y’) M(0, y)
272 PRIME Opt. Maths Book - IX iii) Reflection on y = x Theoretical proof : Let us consider a point p(x, y) is reflected to p’(x’, y’) by the a reflection axis as y = x Draw PM⊥AB \ PM = MP’ where M (x, y) is the mid point of pp’ Then, Using mid - point formula, x = x x 2 1 2 + , y = y y 2 1 2 + Here, x = y or, x x 2 1 2 + = y y 2 1 2 + Here, x = y or ' x x y y' 2 2 + = + [ x x 2 1 2 + = y y 2 1 2 + ] or, 2x + 2x’ = 2y + 2y’ or, x – y = y’ – x’ ............................ (i) Again, Line pp’ and y = x are perpendicular to each other, so, m1 × m2 = –1 or, x x y y – – 2 1 2 1 × . . coeff of y coeff of x c m – = –1 or, ' ' x x y y – – × 1 = –1 or, y’ – y = –x’ + x or, x + y = x’ + y’ Adding equation (i) and (ii) 2x = 2y’ ` y’ = x Subtracting equation (i) and (ii), or, 2y = 2x’ or, y = x’ ` x’ = y ` p(x, y) $ p’(x’, y’) = p’(y, x) iv) Reflection on y = –x Theoretical proof : Let us consider a point p(x. y) is reflected to p’(x’, y’) under a reflection axis y= –x Draw PM⊥AB DPM = DMP’ ` M is the mid - point of pp’. X’ X O Y Y’ P(x, y) A P’(x’, y’) x = y M B
PRIME Opt. Maths Book - IX 273 Then, Using mid - point formula, x = x x 2 1 2 + , y = y y 2 1 2 + or, x = x x' 2 + , y = y y' 2 + Taking y = – x or, ' y y x x' 2 2 + = + or, y + x = – x’ – y’ ........................... (i) Again, The lines pp’ and y = – x are perpendicular lines, So, m1 × m2 = – 1 or, x x y y – – 2 1 2 1 × . . coeff of y coeff of x c m – = –1 or, ' ' x x y y – – × (–1) = – 1 or, x’ – x = y’ – y or, x – y = x’ – y’ ............... (ii) Now, adding equation (i) and (ii), 2x = –2y’ \ y’ = – x Subtracting equation (i) and (ii), 2y = – 2x’ \ x’ = – y \ p(x, y) $ p’(x’, y’) = p’(–y, –x) v) Reflection on y = k (parallel to x - axis) Theoretical proof : Let us consider a point p(x, y) is reflected to p’(x’, y’) under reflection axis y=k. Draw PM =AB \ PM=MP’ where, M(x, k) is the mid-point of PP’. Then, Using mid-point formula, x = x x 2 1 2 + , y = y y 2 1 2 + or, x = x x' 2 + , y = y y' 2 + or, x + x’ = 2x y + y’ = 2k \ x’ = x y’ = 2k - y \ p(x, y) " p (x’, y’) = p’ (x, 2k - y) X’ X O Y Y’ P(x, y) A A P’(x’, y’) y = –x M B B X’ X O Y Y’ P(x, y) P’(x’, y’) M A B y = k
274 PRIME Opt. Maths Book - IX vi) Reflection on x = h (parallel to y-axis) Theoretical proof: Let us consider a point p(x, y) is translated to p’(x’, y’) under a reflection axis x = h. Draw PM =AB \ PM = MP’ ie. M( x, y) is the mid-point of pp’. Then, Using mid-point formula, x = x x 2 1 2 + , y = y y 2 1 2 + or, x = x x' 2 + , y = y y' 2 + or, x + x’ = 2h y + y’ = 2y \ x’ = 2h – x y’ = y \ p(x, y) $ p’(x’, y’) = p’ (2h – x, y) S.No Reflection axis Object Image 1. About x - axis (y = 0) (x. y) (x, – y) 2. About y-axis (x = 0) (x. y) (–x, y) 3. About y = x (x = y) or x – y = 0 (x. y) (y, x) 4. About y = – x (x = – y) or x + y = 0 (x. y) (– y, – x) 5. About x = h (parallel to y - axis) (x. y) (2h – x, y) 6. About y = k (parallel to x - axis) (x. y) (x, 2k – y) 7.1.4 Properties of reflection : i) The object and image are congruent. ii) The object and image are equidistance from the axis of reflection. iii) The line joining the object point A and image point A’ is perpendicular to the mirror line. iv) For the objects point A, B and C, AA’' BB’' CC’. X’ X O Y Y’ P(x, y) P’(x’, y’) M B A x = h
PRIME Opt. Maths Book - IX 275 1. Find the image of the plane figure given below under the given mirror line M. M B A C Solution: Here, DABC is the given object mirror line M is the reflection axis. Draw, BP⊥M, BP = PB’ AQ⊥M, AQ = QA’ CR⊥M, CR = RC’ B M A C’ B’ C R A’ Q P Then, Join A’, B’ & C’ so DA’B’C’ is the image of DABC after reflection under a mirror line M. 2. Find the image of a point A(–2, 3) under reflection about x-axis. Solution: Under reflection about x-axis. P(x, y) → P’(x, –y) A(–2, 3) → A’(–2, –3) 3. Find the image of DABC having vertices A(2, 3), B(4, 6) & C(7, 1) under reflection about x = 0. Also plot them in graph. Solution : Under reflection about x = 0 (y-axis) P(x, y) → P’(–x, y) A(2, 3) → A’(–2, 3) B(4, 6) → B’(–4, 6) C(7, 1) → C’(–7, 1) Worked out Examples
276 PRIME Opt. Maths Book - IX X’ X O A’ C’ B’ C B A Y Y’ Here, DABC is the given object. Y = axis (x = 0) is the mirror line. DA’B’C’ is the image. 4. A reflection axis transform A(3, 2) to A’(1, 2). Find the axis. Also find the image of B(5, 5) and C(6, –3) with it and plot the DABC and DA’B’C’ in graph. Solution : The point A(3, 2) → A’(1, 2) with an axis of reflection x = h. \ P(x, y) → P’(2h – x, y) A(3, 2) → A’(2h – 3, 2) By equating A’, we get 2h – 3 = 1 or 2h = 4 \ h = 2 \ The reflection axis is x = 2. Again, B(5, 5) → B’(–1, 5) C(6, –3) → C’(–2, –3) X’ X O A’ C’ B’ C B A Y Y’ Here, DABC is an object x = 2is axis. DA’B’C’ is the
PRIME Opt. Maths Book - IX 277 5. If A(2a – 1, – 10) → A’(3b + 1, – 3) under a reflection axis y = – x. Find the value of a and b. Solution : Under reflection about y = –x. P(x, y) P’(–y, –x) \ A(2a – 1, –10) A’(10, 1 – 2a) By the question A(2a –1, –10) A’(3b + 1, – 3) By equating the corresponding elements of A’. 3b + 1 = 10 and 1 – 2a = – 3 or, 3b = 9 and 4 = 2a \ – b = 3 and a = 2 \ a = 2 b= 3 6. Transform a triangle having vertices A(1, 2), B(–2, 4) and C(0, – 4) under reflection about x = 2 followed by reflection about y = –3. Also plot the object and image in graph. Solution: Under reflection about x = 2, P(x, y) →p(2h – x, y) = p(4 – x, y) \ A(1, 2) → A’(3, 2) B(–2, 4) → B’ (6, 4) C(0, – 4) → C’ (4, –4) Again, under reflection about y = –3 P(x, y) → P’(x, 2k – y) =P(x, –6 – y) \ A1 (3, 2) → A”(3, – 8) B1 (6,4) → B” (6, – 10) C1 (4, – 4) → C” (4, –2) Here, DABC is an object DA’B’C’ is first image DA”B”C” is final image. X’ X O A’ C’ C’’ B’’ A’’ B’ C B A Y Y’
278 PRIME Opt. Maths Book - IX Exercise 7.1 1. i) What is transformation? Write down the name of transformations. ii) What is reflection in transformation. iii) Write down the formula of a point P(a, b) under reflection about the axis x = h. iv) Write down the formula of a point A(p, q) and reflection about the axis y = k. v) What will be the image of a point P(x, y) under reflection about x-axis followed by reflection about y - axis. 2. Find the image of the following plane figures under the given reflection axis ‘M’. i) M B A C ii) B A M iii) M S P R Q iv) M Q R P v) M R Q P T S vi) M 3. Find the image of point A(3, –2) under the following reflection axis. i) about x-axis ii) about x = 2 iii) about y = x iv) about y – 3 = 0 v) about y = – x 4. Find the image of object p(–3, 5) under the following reflection axis. i) under x + y = 0 ii) under x + 2 = 0 iii) under y = 0 iv) under x = 0 v) under y = – 2
PRIME Opt. Maths Book - IX 279 5. Find the co-ordinate of image for the followings. i) Find the image of the points A(2, 2) and B(6, 7) under reflection about x-axis. Also plot them in graph. Also join AB and A’B’. ii) Find the image of DPQR having vertices P(–2, 1), Q(–5, 4) & R(–6, –2) under reflection about y-axis. Also plot them in graph. iii) Find the image of quadrilateral having vertices A(1, 2), B(3, 5), C(6, 6) and D(7, 1) under reflection about y = –2. Also plot them in graph. iv) Find the image of quadrilateral having vertices P(3, 1), Q(4, 5), R(7, 4) and S(8, –2) under reflection about X = –1. Also plot them in graph. v) Find the coordinate of image of triangle having vertices A(2, 1), B(4, 5) & C(5, –3) under y = x to DA’B’C’. Also plot DABC and DA’B’C’ in graph. 6. Find the co-ordinate of image for the followings. i) If a point A(3, 7) is transferred to A’(–7, –3) with a reflection axis, find the axis. ii) A(5, –2) is transferred to A’(–1, –2) under a reflection axis, find the axis. iii) P’(–3, –4) is the image of P(–3, –2) under a reflection axis. Find the axis. iv) A’(–1, 5) is the image of A(1, 5) under a reflection axis, find the axis. v) If a point A(2x + 1, y – 3) is reflected to A’(1, 5) under reflection about y = x. Find the value of ‘x’ and ‘y’. 7. PRIME more creative questions: i) A reflection axis gives image A’(–3, 2) of an object of A(–1, 2). Find the reflection axis. Also find the image of points B(2, 5) and C(–2, –4) with it and plot the object DABC and its image in graph. ii) A’(2a + 1, 5 – b) is the image of A(1 – 3b, – 7) under reflection about x + y = 0. Find the value of ‘a’ and ‘b’. iii) Find the image of DABC having vertices A(1, 4), B(–3, –1) and C(0, –5) under reflection about x + 2 = 0 to DA’B’C’. Again transform DA’B’C’ to DA’’B’’C’’ under reflection about x = – y. Also plot them in graph. iv) Transform a triangle having vertices A(–2, 1), B(2, 5) and C(4, 0) under reflection about x-axis followed by reflection about x – y = 0. Also plot the object and image in graph. v) Transform a triangle having vertices P(2, –1), Q(4, 4) and R(6, 1) under reflections about x = 0 following by y = 0. Also plot the object and image in graph.
280 PRIME Opt. Maths Book - IX Answer 1. Show to your teacher. 2. Show to your teacher. 3. i. A’ (3, 2) ii. A’ (1, –2) iii. A’(–2, 3) iv. A’ (3, 8) v. A’ (2, –3) 4. i. P’(–5, 3) ii. P’(–1, 5) iii. P’(–3, –5) iv. P’ (3, 5) v. P’ (–3, –9) 5. i. A’ (2, –2) B’ (6, –7) ; graph ii. P’(2, 1), Q’ (5, 4), R’ (6, –2) ; graph. iii. A’ (1, –6), B’(3, –9), C’(6, –10), D’(7, –5) ; graph. iv. P’ (–5, 1) Q’ (–6, 5), R’ (–9, 4), S’ (–10, –2) ; graph. v. A’ (1, 2), B’ (5, 4), C’(–3, 5) ; graph. 6. i. Reflection about y = –x. ii. Reflection about x = 2. iii. Reflection about y = –3. iv. Reflection about y – axis. v. x = 2, y = 4 7. i. Reflection about x = –2 ; B’ (–6, 5) and C’ (–2, –4) ; graph. ii. a = –4, b = –2. iii. A’ (–5, 4), B’ (–1, –1), C’(–4, –5) ; A” (–4, 5), B” (1, 1), C” (5, 4) ; graph. iv. A’ (–2, –1), B’(2, –5), C’(4, 0) ; A” (–1, –2), B”(–5, 2), C” (0, 4) ; graph. v. P’ (–2, –1), Q’ (–4, 4), R’ (–6, 1) ; P” (–2, 1), Q” (–4, –4), R”(–6, –1) ; graph.
PRIME Opt. Maths Book - IX 281 7.2 Rotation: Let us consider a point P is transferred to P’ with an angle 90° in clockwise direction about a fixed point called centre of rotation, where OP = OP’ and \POP’ = 90° The transformation of an object from one position to another place depands on centre, direction and angle is called rotation. Congruent image of an object is formed in rotation. w Rotation takes place about a fixed point, called the centre of rotation. – Clockwise (negative) direction. – Anti clockwise (positive) direction. w Angle of rotation – Quarter turn (90°) – Half turn (180°) – Full turn (360°) w Full turn gives invariant image as an object. w Let us consider a DABC is rotated to DA’B’C’ under [O, – 90°). O B’ C’ A’ A B C Here, DAOA’ = \BOB’ = COC’ = 90° Anti clockwise direction is used to rotate the DABC to DA’B’C’. Centre of rotation O is taken for the rotation. Then, Image of DABC is drawn under the above conditions. Where DA’B’C’ is the image. P P’ O 90°
282 PRIME Opt. Maths Book - IX 7.2.1 Rotation using co-ordinates: Taking an object point p(x, y) which transferred to p’(x’, y’) under rotation according to the following conditions. Then the image p’(x’, y’) can be expressed in formula as follows. i) Quarter turn about the centre as origin: ( + 90°) Theoretical proof : Here, \POP’ = 90° (+ve and –ve) i.e. m1 × m2 = – 1 or, × ' ' x y x y 0 0 0 0 – – – – = – 1 or, yy’= – xx’ or, y’ = – ' y xx ...........................(i) Again, we have, OP2 = OP’2 = r2 or, (x – o)2 + (y –0)2 = (x’ – 0)2 + (y’ – 0)2 or, x2 + y2 = x2 + ' y x x2 2 2 or, y2 (x2 + y2 ) = x’2 (x2 + y2 ) or, y2 = x’2 ` x’ = ±y Again, From equation (i), y’ = – y x y × or – y x y × – = – x or + x \ P(x, y) $ P’(x’, y’) = P’ (–y, x) or (y, –x) i.e. For anticlockwise (+ 90°) P(x, y) $ P’(–y, x) For clockwise (–90°) P(x, y) $ P’(y, –x) ii) Half turn (+ 180°) about centre origin. Theoritical proof : Here, ]POP’ = 180° (+ve and –ve) Where O is the mid - point of line PP’. Using mid - point formula x = x x 2 1 2 + , y = y y 2 1 2 + or, x = x x' 2 + , y = y y' 2 + O P(x, y) –180° +180° X’ X Y Y’ O P(x, y) P’(x’, y’) X’ X Y Y’ –ve 90° +ve 90° P’(x’, y’)
PRIME Opt. Maths Book - IX 283 or, x + x’ = 0 y + y’ = 0 \ x’ = – x, y’ = – y \ P(x, y) $ P’(x’, y’) = P’(–x, –y). iii) Quarter turn about centre (a, b). Let us consider A(a, b) is the centre of rotation and P(x, y) be the object point. The image P(x’, y’) can be calculated by using the following steps : Step - I Firstly a and b which are the components of the centre of rotation should be to the x - components and y - component of the object point P(x, y). i.e. P(x, y) ( , a b) centre (x – a, y – b) Step - II The new point (x – a, y –b) have to be rotated as for the centre origin. (x, y) +90° (–y, x) ` (x – a, y –b) $ (–y + b, x –a) (x, y) – ° 90 (y, – x) ` (x – a, y – b) $ (y – b, –x + a) Step - III The components of the centre (a, b) should be added to the corresponding elements of newly occurred points which will be the formula of rotation for the point p(x, y) about the centre of rotation (a, b) as (–y + b, x – a) +90° (–y + b + a, x – a + b) (y – b, –x + a) – ° 90 (y – b + a, –x + a + b) \ P(x, y) +90° P’(x’, y’) = P’(–y + a + b, x – a + b) = p’{–(y – b) + a, (x – a) + b} P(x, y) +90° P’(x’, y’) = P’(y + a – b, –x + a + b) = p’{(y – b) + a, – (x – a) + b} Examples : Image of P (2, –3) under rotation through +90° with centre (1, 2). We have P(x, y) centre a( , b) +90° P’(–y + a + b, x – a + b) P’(–y + 1 + 2, x – 1 + 2) P’(– y + 3, x + 1) \ P(x, y) P’(3 + 3, 2 + 1) P’(6, 3) P(x, y) P’(x’, y’) –90° +90° P’(x’, y’) A(a, b)
284 PRIME Opt. Maths Book - IX X’ X O Y Y’ P’(6, 3) P(2, –3) A(1, 2) iv) Half turn about centre (a, b) Here , \ PAP’ = 180° (+ve or – ve) i.e. A(a, b)e is the mid - point of PP’. Then, Using mid - point P’(x’, y’) formula, x = x x 2 1 2 + , y = y y 2 1 2 + or, x = x x' 2 + , y = y y' 2 + or, x + x’ = 2a, y + y’ = 2b \ x’ = –x + 2a y’ = –y + 2b \ P(x, y) $ P’(x’, y’) = P’(–x + 2a, –y + 2b) = P’{–(x – a) + a, – (y – b) + b} S.No. Centre Direction Angle Object Image 1. (0, 0) +ve 90° (x, y) (–y, x) 2. (0, 0) +ve 90° (x, y) (y, –x) 3. (0, 0) ––––– 180° (x, y) (–x, –y) 4. (a, b) +ve 90° (x, y) (–y + a + b, x – a + b) 5. (a, b) –ve 90° (x, y) (y + a – b, –x + a + b) 6. (a, b) ––––– 180° (x, y) (–x + 2a, – y + 2b) [(2a – x), (2b – y)] 7.2.2 Properties of rotation: i) Three informations are given to rotate the objects which are direction, angle and centre. ii) Object and image are congruent in it. iii) Centre of rotation is the intersecting point of perpendicular bisectors of line joining the object and corresponding images. iv) Anticlockwise direction is called positive direction and clockwise is called negative direction. v) If centre of rotation is inside the object, the object and image will be invariant. P(x, y) P’(x’, y’) A(a, b) –180° +180°
PRIME Opt. Maths Book - IX 285 1. Rotate the given triangle under positive quarter turn with center ‘O’. O B C A Solution: O B C A A’ B’ C’ Here, DABC is an object. O is center of rotion. \AOA’ – \BOB’ = \COC’ = 90° in positive direction (Anticlockwise). DA’B’C’ is the image. 2. Find the image of a point A(–3, 2) under rotation about –90° with centre origin. Solution : Under rotation through –90° about (0, 0) P(x, y) → P’(y, –x) \ A(–3, 2) → A’(2, 3) 3. Find the image of DPQR under rotation about positive quarter turn. Also plot the object and image in graph. Where P(1, 2) Q(3, –2) & R(5, 4) are the vertices. Solution: Under rotation through +90° about (0, 0) P(x, y) → P’(–y, x) \ P(1, 2) → P’(–2, 1) Q(3, –2) → Q’(2, 3) R(5, 4) → R’(–4, 5) Worked out Examples
286 PRIME Opt. Maths Book - IX X’ X R Q O R’ P’ P Q’ Y Y’ Here, DPQR is the object. under rotation about +90°. DP’Q’R’ is the image. 4. Find the co-ordinate of image of a point P(3, –2) under rotation about half in anti clockwise direction with centre origin and then about + 90° with centre (–1, 2). Solution: The given point is P(3, –2) under rotation through 180°, about (0, 0) P(x, y) →p’(–x, –y) \ P(3, –2) →p’(–3, 2) Again, under +90° with centre (–1, 2) P(x, y) → p’(–y + b + a, x – a+ b) → p’ (–y + 1, x + 3) \ P(–3, 2) → p” (–2 + 1, x + 3) → p” (–1, 5). 5. Transform A(2, 4), B(5, 1) & C(–3, 2) under rotation about positive quarter turn with centre origin followed by rotation about 180° with centre (3, –1). Also plot the object and image in graph. Solution: The given vertices of DABC are A(2, 4), B(5, 1) and C(–3, 2) Under rotation about +90° with centre (0, 0). P(x, y) → p’(–y, x) \ A(2, 4) → A’ (–4, 2) B (5, 1) → B’ (–1, 5) C (–3, 2) → C’ (–2, –3). Again, Under rotation about 180° with centre (3, –1). P(x, y) → p’ (–x + 2a, – y + 2b) → p’ (–x + 6, – y – 2) \ A’(–4, 2) → A’’ (4 + 6, –2 – 2) = A’’ (10, – 4) B’(–1, 5) → B’’ (1 + 6, – 5 – 2) = B’’ (7, – 7) C’(–2, – 3) → C’’ (2 + 6, 3, 2) = C’’(8, 1)
PRIME Opt. Maths Book - IX 287 X’ X A O C’ A’ A’’ B’’ C’’ B’ B C Y Y’ Here, DABC is the object. DA’B’C’ is first image. DA’’B’’C’’ is final image. Exercise 7.2 1. i) What is rotation in transformation? ii) Write down the formula for rotation about 180° with centre (a, b). iii) Write down the image of p(x, y) under rotation about –90° followed by rotation about +90° with centre origin. iv) Find the image of A(a, b) under reflection about y = –x followed by rotation about –270° with centre origin. v) Find the image of a point m(p, q) under rotation about 180° with centre origin followed by reflection about y-axis. 2. Draw the image of the following diagrams under the following conditions given in question. i) A C B O (+90°) ii) O (–90°) P R Q
288 PRIME Opt. Maths Book - IX iii) O (–90°) P S Q R iv) O (–90°) E A D B C v) O (+270°) E F G H D A C B vi) O (–270°) A S T U W V Q R 3. Find the image of the following points under rotation about negative quarter turn with centre origin. i) (–5, –3) ii) (5, 2) iii) (–2, –7) iv) (–6, 3) v) (–7, –1) vi) (5, –3) 4. Find the image of the following points under rotation about positive quarter turn. i) (–7, 2) ii) (–5, –3) iii) (8, –2) iv) (3, 7) v) (–6, 2) vi) (–2, –8) 5. Find the image of the following points under rotation about half turn. i) (–2, –6) ii) (5, –3) iii) (3, 6) iv) (–5, 4) v) (–4, –5) vi) (2, –5) 6. i) Find the co-ordinate of image of DABC having vertices A(–2, 3), B(–3, –1) and C(–5, 4) under rotation about –270°. Also plot the object and image in graph. ii) Find the co-ordinate of the image of DPQR having vertices P(2, 1), Q(3, 5) & Q(5, 0) under rotation about +270°. Also plot the DPQR and DP’Q’R’ in graph. iii) Find the co-ordinate of the image of quadrilateral having vertices A(–3, 2), B(–1, 5), C(2, 4) & D(3, –1) under rotation about positive quarter turn. Also plot them in graph. iv) Find the image of quadrilateral having vertices A(3, 4), B(4, 6), C(6, 2) & D(4, 1) under rotation about 180° with center origin. Also plot the object and image in graph. v) Transform a triangle having vertices A(1, –2), B(4, –5), C(6, 0) under rotation about +90° with centre origin to DA’B’C’ and then rotation about 180° with
PRIME Opt. Maths Book - IX 289 center origin to DA’’B’’C’’. Also plot the object and image in graph. 7. Find the image under the following. i) P(3, –2) under +90° with center (1, 2). ii) A(3, 4) under –90° with center (–1, 2). iii) M(–3, 3) under 180° with center (2, –1). iv) Triangle having A(1, 1), B(3, 5) and C(5, 0) under +ve quarter turn with center (–2 –1). Also plot them in graph. 8. i) Which rotation gives the image of point A(2, –1) to A’(1, 2)? Also find the image of the points B(3, 2) and C(5, –4). Also plot DABC and DA’B’C’ in graph. ii) Find centre of rotation, angle of rotation and direction of rotation which gives A(2, –3) to A’(–3, –2) and B(4, 1) to B’(1, –4) by plotting the points in graph. iii) Find the image of P(a, b) under rotation about +90° with center origin and again under 180° with center origin. iv) Find the centre, direction and angle of rotation where DABC having A(3, 2), B(5, 1) and C(4, –2) is rotated to A’(–2, 3) B’(–1, 5) and C’(2, 4) by plotting in graph. v) Find the center direction and angle of rotation which transformed DABC having A(3, –4), B(5, –6) & C(7, –1) to A’(1, 4), B’(3, 6), C’(–2, 8) by plotting in graph. 9. PRIME more creative question: i) Transform a quadrilateral having vertices A(2, 1), B(1, –2), C(–3, –2) and D(–5, 1) under rotation about 180° with centre origin followed by rotation about + 90° centre (3, –4). Also plot the object and image in graph. ii) Find the image of DPQR having vertices P(–4, 6), Q(–1, –2) and R(3, –5) under negative quarter turn with centre origin followed by rotation about 180° with centre origin Also plot the object and image in graph. iii) Transform a triangle having vertices A(2, 3), B(1,5) and C(–2, 4) under reflection about y = x. Following by rotation about rotation about +90° with centre in origin. Also plot the image and object in graph. iv) Triangle having vertices P(3, 4), Q(–2, 6) and R(1, – 5) is reflected under x = –2 to DP’Q’R’ and rotated again to DP’’Q’’R’’ under +270° with centre origin. Also plot them in graph. v) If a point P(x, y) is reflected under x-axis and then under y-axis continuously is there any difference it with rotation about 180° with centre origin? Discuss the application of rotation in group and a report.
290 PRIME Opt. Maths Book - IX Answer 1. Show to your teacher. 2. Show to your teacher. 3. i. (–5, –3) ii. (2, –5) iii. (–7, 2) iv. (3, 6) v. (–1, 7) vi. (–3, –5) 4. i. (–2, –7) ii. (3, – 5) iii. (2, 8) iv. (–7, 3) v. (–2, –6) vi. (8, –2) 5. i. (2, 6) ii. (–5, 3) iii. (–3, –6) iv. (5, –4) v. (4, 5) vi. (–2, 5) 6. i. A’(–3, –2), B’(1, –3), C’(–4, –5) ; graph. ii. P’(1, –2), Q’(5, –3), R’(0, –5) ; graph. iii. A’(–2, –3), B’(–5, –1), C’(–4, 2), D’(1, 3) ; graph. iv. A’(–3, –4), B’(–4, –6), C’(–6, –2), D’(–4, –1) ; graph. v. A’(2, 1), B’(5, 4), C’(0, 6) ; A”(–2, –1), B”(–5, –4), C”(0, –6) ; graph. 7. i. (5, 4) ii. (1, –2) iii. (7, –5) iv. A’(–4, 2) , B’(–8, 4) , C’(–3, 6) ; graph. v. P’(0, 5) , Q’(–4, 6) , R’(2, 8) ; graph. 8. i. Rotation about +90° with centre origin, B’(–2, 3) , C’(4, 5) ; graph. ii. Centre (0, 0), angle 90°, negative direction. iii. P(a, b) →P’(–b, a) → P”(b, –a) iv. Centre (0, 0) angle 90°, positive direction v. Show to your teacher. 9. i. A’ (–2, –1), B’(–1, 2), C’(3, 2) D’(5, – 1); A”(0, 5), B”(–3, 6), C”(–3, 10), F’(0, 12); graph. ii. P’(6, 4), Q’(–2, 1) R’(–5, –3) ; P”(–6, –4), Q”(2, –1), R”(5, 3); graph. iii. A’(3, 2), B’(5, 1), C’(4, –2) ; A”(–2, 3), B”(–1, 5), C”(2, 4); Graph. iv. P’(–7, 4), Q’(–2, 6), R’(3, –5) ; P”(–4, – 7), Q”(–6, –2) & R”(5, 3) ; v. P(x, y) → P’ (x, – y) → P” (–x, – y). It is same as rotation about 180° with centre
PRIME Opt. Maths Book - IX 291 7.3 Translation Here, vector AB is the magnitude & direction of translation. DPQR is, an object. PP’ = QQ’ = RR’ = AB PP’ || QQ’ || RR’ ' AB DP’Q’R’ is the image of DPQR The transformation of an object from one place to another place by the magnitude and direction of a given vector is called translation. 7.3.1 Translation using co-ordinate Let us consider P(x, y) is an object point and OA is the translation vector a b < F. According to properties of translation, P is shifted to P’ where direction is according to OA and PP’'OA. Also PP’ = OA i.e. P’ is defined as ‘a’ units more in horizontal direction than of p and ‘b’ units more in vertical direction than p. \ x’ = x + a and y’ = y + b \ P(x, y) $ P’(x’, y’) = P’(x + a, y + b) Also, If a point P is translated to P’ under a displacement and direction of AB having co - ordinates A(x1 , y1 ) and B(x2 , y2 ) then the translation component can be defined as, a = x2 – x1 b = y2 – y1 \ P(x, y) AB P’(x + a, y + b) where a = x2 – x1 and b = y2 – y1 O P(x, y) A(a, b) M P’(x’, y’) b a X’ X Y Y’ P R P’ R’ Q A B Q’
292 PRIME Opt. Maths Book - IX i) Translation using co-ordinate: X’ X B(3, -1) C(6, -1) O C’(9, 1) A’(7, 5) A(4, 3) B’(6, 1) Y Y’ Here, A(4, 3) is translated to A’(7, 5) B(3, –1) is translated to B’(6, 1) C(6, –1) is translated to C’(9, 1) i.e. All the points are translated with constant number 3 for x-component and 2 for y - component. i.e. A (4, 3) → A’(4 + 3, 3 + 2) = A’(7, 5) i.e. Translation vector is T = 3 2 < F i.e. under translation T = a b < F P(x, y) → P’(x + a, y + b) iii) Translation using vector: Let us consider a translation vector is AB where A(1, 2) and B(3, 5) are any two points. Then, according to the concept of column vector AB AB = x x y y – – 2 1 2 1 < F = 3 1 5 2 – – < F = 2 3 < F \ Translation vector T = AB = 2 3 < F Properties of translation: i) Object and image are congruent in it. ii) For the object points A & B and A’ & B’ for vector T //AA’//BB’. iii) Length of AA’ and BB’ is always equal to magnitude of T . iv) Image is always in the same direction of vector T .
PRIME Opt. Maths Book - IX 293 1. Find the image of a point A(3, –2) under a translation vector of T = 1 3 < F. Solution: Under translation vector T = 1 3 < F P(x, y) → P’(x + a, y + b) → P’(x + 1, y + 3) A(3, –2) → A’(3 + 1, –2 + 3) → A’(4, 1) \ A(3, –2) T 1 3 = < F A’(4, 1) 2. If A (3, –1) and B(1, 2) are any two points, find the image of point A and B under AB . Solution : Here, Translation vector AB from the given points A(3, –1) & B(1, 2) is, AB = x x y y – – 2 1 2 1 < F = 1 3 2 1 – + < F = 2 3 – < F = T Then, under translation vector T = 2 3 – < F p (x, y) → P’(x + a, y + b) = P(x – 2, y + 3) A(3, –1) → A’(3 – 2, –1 + 3) = A’(1, 2) \ A(3, –1) T 2 3 – = < F A’(1, 2) B(1, 2) → B’(1 – 2, 2 + 3) = B’(–1, 5) 3. Find the image of DABC having vertices A(3, 1), B(–1, 5) & C(1, –3) under a translation vector T = 3 2 < F. Also draw the object and image in graph. Solution: Under a translation vector T = 3 2 < F P(x, y) → P’(x + a, y + b) → P’(x + 3, y + 2) Then, A(3, 1) → A’(3 + 3, 1 + 2) = A’(6, 3) B(–1, 5) → B’(–1 + 3, 5 + 2) = B’(2, 7) C(1, –3) → C’(1 + 3, –3 + 2) = C’(4, –1) Here, DABC is the object. DA’B’C’ is the image of DABC Worked out Examples X’ X B C O C’ A’ A B’ Y Y’
294 PRIME Opt. Maths Book - IX 4. Draw the image of given triangle under the given vector AB . P R Q A B Solution: Here, Draw PP’, QQ’, RR’ || AB . Then, Taking PP’ = QQ’ - RR’ = AB DPQR is the object. DP’Q’R’ is the image. 5. Translate a triangle having vertices A(1, –1), B(–2, 2) and C(3, 3) under T1 = 3 2 < F and again translate under T = 1 –3 < F. Also plot the object and image in graph. Solution : We have, Under translation about T1 = 3 2 < F P(x, y) → P’(x + a, y + b) = P’(x + 3, y + 2) \ A(1, –1) → A’(1 + 3, –1 + 2) = A’(4, 1) B(–2, 2) → B’(–2 + 3, 2 + 2) = B’(1, 4) C(3, 3) → C’(3 + 3, 3 + 2) = C’(6, 5) Again, under translation about T2 = 1 –3 < F P(x, y) → P’(x + 1, y – 3) \ A’(4, 1) → A’’(5, –2) B’(1, 4) → B’’(2, 1) C’(6, 5) → C’’(7, 2) Here, DABC is the object. DA’B’C’ is the first image. DA’’B’’C’’ is the final image. P R R’ Q Q’ A B P’ X’ X B C O C’ A’ A’’ C’’ B’’ A B’ Y Y’
PRIME Opt. Maths Book - IX 295 Exercise 7.3 1. i) What do you mean by translation in transformation? ii) What is the formula of p(a, b) under translation about T = m n ? iii) Find the image of a point p(x, y) under translation about T1 = 1 2 followed by translation about T2 = 2 3 < F iv) Find translation vector T1 Which gives A(2, 3) → A’(5, 2). v) If A’(3, 5) is the image after translation T = 2 3 < F, find the point A. 2. Find the image of the following figures under the given vector. i) P Q R A B ii) C M N A B iii) S R B A Q T iv) B P C D A Q v) N F M H A G B C E D 3. Find the image of the following points under the translation vector T = 2 1 < F. i) A(3, 2) ii) P(–2, 5) iii) M(–3, –1) iv) N(4, –6) v) O(–3, –4) 4. Find the image of the points P(3, –2) under the translation vector AB for the following points A and B. i) A(3, 2) and B(1, 0) ii) A(1, –2) & B(3, 1) iii) A(2, 1) and B(4, 5) iv) A(–1, 2) & B(2, 1)