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Bckground knowledge Contents: A B C D E F G H I J K L Surds nd rdicls Scientific nottion (stndrd form) Numer systems nd set nottion Algeric simplifiction Liner equtions nd inequlities Modulus or solute vlue Product expnsion Fctoristion Formul rerrngement Adding nd sutrcting lgeric frctions Congruence nd similrity Coordinte geometry

BACKGROUND KNOWLEDGE This chpter contins mteril tht is normlly covered prior to this course. It is ssumed knowledge. Not ll preliminries re covered within the chpter. However, other necessry work is revised within the chpters. A SURDS AND RADICALS Rel numers like p, p 3, p 5, p 6, etc., re clled surds or rdicls. Surds re present in solutions to some qudrtic equtions. p 4 is rdicl ut is not surd s it simplifies to. Surds re irrtionl rel numers. Definition: p is the non-negtive numer such tht p p =. Properties: ² ² ² ² p p is never negtive, so > 0. p is meningful only for > 0. p p p = for > 0 nd > 0. r p = p for > 0 nd >0. EXERCISE A Exmple 1 Write s single surd: p p 3 p 18 p 6 p p 3 = p 3 = p 6 = p p 18 6 q 18 6 = p 3 or p 18 p 6 = p 6 p 3 p 6 = p 3 1 Write s single surd or rtionl numer: p p p 3 5 ( 3) c p p d 3 p p e 3 p 7 p 7 f p 1 p g p 1 p 6 h p 18 p 3 Exmple Simplify: 3 p 3+5 p 3 p 5 p Compre with x 5x = 3x 3 p 3+5 p 3 p 5 p =(3+5) p 3 = ( 5) p =8 p 3 = 3 p

BACKGROUND KNOWLEDGE 3 Simplify the following mentlly: p +3 p p 3 p c 5 p 5 3 p 5 d 5 p 5+3 p 5 e 3 p 5 5 p 5 f 7 p 3+ p 3 g 9 p 6 1 p p p p 6 h + + Exmple 3 Write p 18 in the form p where nd re integers nd is s lrge s possile. p 18 = p 9 f9 is the lrgest perfect squre fctor of 18g = p 9 p =3 p 3 Write the following in the form p where nd re integers nd is s lrge s possile: p p p p 8 1 c 0 d 3 p p p p e 7 f 45 g 48 h 54 p p p p i 50 j 80 k 96 l 108 Exmple 4 Simplify: p 75 5 p 7 p 75 5 p 7 = p 5 3 5 p 9 3 = 5 p 3 5 3 p 3 =10 p 3 15 p 3 = 5 p 3 4 Simplify: 4 p 3 p 1 3 p + p 50 c 3 p 6+ p 4 d p 7 + p p p p p p 1 e 75 1 f + 8 3 Exmple 5 Write 9 p 3 without rdicl in the denomintor. 9 p3 = 9 p 3 p 3 p 3 = 9p 3 3 =3 p 3

4 BACKGROUND KNOWLEDGE 5 Write without rdicl in the denomintor: 1 p 6 p3 c 7 p d 10 p 5 e 10 p f 18 p 6 g 1 p 3 h 5 p7 i 14 p 7 j p 3 p B SCIENTIFIC NOTATION (STANDARD FORM) Scientific nottion (or stndrd form) involves writing ny given numer s numer etween 1 nd 10, multiplied y power of 10, i.e., 10 k where 1 6 <10 nd k Z. EXERCISE B Exmple 6 Write in stndrd form: 37 600 0:000 86 37 600 = 3:76 10 000 fshift deciml point 4 plces to the =3:76 10 4 left nd 10 000g 0:000 86 = 8:6 10 4 fshift deciml point 4 plces to the =8:6 10 4 right nd 10 000g 1 Express the following in scientific nottion: 59 59 000 c :59 d 0:59 e 0:000 59 f 40:7 g 4070 h 0:0407 i 407 000 j 407 000 000 k 0:000 040 7 Express the following in scientific nottion: The distnce from the Erth to the Sun is 149 500 000 000 m. c d e Bcteri re single cell orgnisms, some of which hve dimeter of 0:0003 mm. A speck of dust is smller thn 0:001 mm. The core temperture of the Sun is 15 million degrees Celsius. A single red lood cell lives for out four months nd during this time it will circulte round the ody 300 000 times. Exmple 7 Write s n ordinry numer: 3: 10 5:76 10 5 3: 10 =3:0 100 = 30 5:76 10 5 = 000005:76 10 5 =0:000 057 6

3 Write s n ordinry deciml numer: BACKGROUND KNOWLEDGE 5 4 10 3 5 10 c :1 10 3 d 7:8 10 4 e 3:8 10 5 f 8:6 10 1 g 4:33 10 7 h 6 10 7 4 Write s n ordinry deciml numer: 4 10 3 5 10 c :1 10 3 d 7:8 10 4 e 3:8 10 5 f 8:6 10 1 g 4:33 10 7 h 6 10 7 5 Write s n ordinry deciml numer: The wvelength of light is 9 10 7 m. The estimted world popultion for the yer 000 ws 6:130 10 9. c The dimeter of our glxy, the Milky Wy, is 1 10 5 light yers. d The smllest viruses re 1 10 5 mm in size. 6 Find, correct to deciml plces: (3:4 10 5 ) (4:8 10 4 ) (6:4 10 ) c 3:16 10 10 6 10 7 d (9:8 10 4 ) (7: 10 6 ) e 1 3:8 10 5 f (1: 10 3 ) 3 7 If missile trvels t 5400 km h 1 how fr will it trvel in: 1 dy 1 week c yers? Give your nswers in scientific nottion correct to deciml plces, nd ssume tht 1 yer ¼ 365:5 dys. 8 Light trvels t speed of 3 10 8 metres per second. How fr will light trvel in: 1 minute 1 dy c 1 yer? Give your nswers with deciml prt correct to deciml plces, nd ssume tht 1 yer ¼ 365:5 dys. C NUMBER SYSTEMS AND SET NOTATION NUMBER SYSTEMS We use: ² R to represent the set of ll rel numers. These include ll of the numers on the numer line. negtives ² N to represent the set of ll nturl numers. N = f0, 1,, 3, 4, 5,...g ² Z to represent the set of ll integers. Z = f0, 1,, 3, 4,...g ² Z + is the set of ll positive integers. Z + = f1,, 3, 4,...g ² Q to represent the set of ll rtionl numers which re ny numers of the form p where p nd q re integers, q q 6= 0. 0 positives

6 BACKGROUND KNOWLEDGE SET NOTATION fx j 3 <x<g reds the set of ll vlues tht x cn e such tht x lies etween 3 nd. the set of ll such tht Unless stted otherwise, we ssume tht x is rel. 3 <x< cn lso e written s x ] 3, [. EXERCISE C 1 Write verl sttements for the mening of: fx j x>5, x R g fx j x 6 3, x Z g c fy j 0 <y<6g d fx j 6 x 6 4, x Z g e ft j 1 <t<5g f fn j n< or n > 6g Exmple 8 Write in set nottion: included 0 4 3 4 not included fx j 1 6 x 6 4, x N g or fx j 1 6 x 6 4, x Z g fx j 3 6 x<4, x Z g Write in set nottion: c 0 1 5 3 d e f 0 5 0 5 3 Sketch the following numer sets: fx j 4 6 x<10, x N g fx j 4 <x6 5, x Z g c fx j 5 <x6 4, x R g d fx j x> 4, x Z g e fx j x 6 8, x R g D ALGEBRAIC SIMPLIFICATION To nswer the following questions, you will need to rememer: ² the distriutive lw ( + c) = + c ² power nottion =, 3 =, 4 =, nd so on. EXERCISE D 1 Simplify if possile: 3x +7x 10 3x +7x x c x +3x +5y d 8 6x x e 7 +5 f 3x +7x 3

Remove the rckets nd then simplify: 3(x + 5) + 4(5 + 4x) 6 (3x 5) BACKGROUND KNOWLEDGE 7 c 5( 3) 6( ) d 3x(x 7x +3) (1 x 5x ) 3 Simplify: x(3x) E 3 3 9 4 c p 16x 4 d ( ) 3 3 4 When deling with inequlities: ² multiplying or dividing oth sides y negtive reverses the inequlity sign. ² do not multiply or divide oth sides y the unknown or term involving the unknown. EXERCISE E 1 Solve for x: x +5=5 3x 7 > 11 c 5x +16=0 x d 3 7=10 e 6x +11< 4x 9 f 3x =8 5 1 g 1 x > 19 h x +1= 3 x i 3 3x 4 = 1 (x 1) Solve simultneously for x nd y: ½ x +y =9 x y =3 F d LINEAR EQUATIONS AND INEQUALITIES ½ 5x 4y =7 3x +y =9 e ½ x +5y =8 x y = ½ x +y =5 x +4y =1 c f ½ 7x +y = 4 3x +4y =14 8 x >< + y 3 =5 >: x 3 + y 4 =1 The modulus or solute vlue of rel numer is its size, ignoring its sign. For exmple: the modulus (or solute vlue) of 7 is 7, nd the modulus (or solute vlue) of 7 is lso 7. GEOMETRIC DEFINITION The modulus of rel numer is its distnce from zero on the numer line. Becuse the modulus is distnce, it cnnot e negtive. We denote the modulus of x s MODULUS OR ABSOLUTE VALUE 7 7-7 jxj. jxj is the distnce of x from 0 on the rel numer line.

8 BACKGROUND KNOWLEDGE x If x>0 If x<0 0 x x x 0 ALGEBRAIC DEFINITION jx j cn e considered s the distnce of x from : jxj = ½ x if x > 0 x if x<0 or jxj = p x MODULUS EQUATIONS It is cler tht jxj = hs two solutions, x = nd x =, since jj = nd j j =. EXERCISE F 1 Find the vlue of: If jxj = where > 0, then x =. 5 ( 11) j5j j 11j c j5 ( 11)j d ( ) + 11( ) e j 6j j 8j f j 6 ( 8)j If =, =3, nd c = 4 find the vlue of: jj jj c jjjj d jj e j j f jj jj g j + j h jj + jj i jj j k c jcj l jj 3 Solve for x: jxj =3 jxj = 5 c jxj =0 d jx 1j =3 e j3 xj =4 f jx +5j = 1 g j3x j =1 h j3 xj =3 i j 5xj =1 G PRODUCT EXPANSION y =(x 1)(x +3) cn e expnded into the generl form y = x + x + c. Likewise, y =(x 3) +7 cn lso e expnded into this form. Following is list of expnsion rules you cn use: ² ( + )(c + d) =c + d + c + d ² ( + )( ) = fdifference of two squresg ² ( + ) = + + fperfect squresg

BACKGROUND KNOWLEDGE 9 EXERCISE G Exmple 9 Expnd nd simplify: (x + 1)(x +3) (3x )(x +3) (x + 1)(x +3) (3x )(x +3) =x +6x + x +3 =3x +9x x 6 =x +7x +3 =3x +7x 6 1 Expnd nd simplify using ( + )(c + d) =c + d + c + d: (x + 3)(x +1) (3x + 4)(x +) c (5x )(x +1) d (x + )(3x 5) e (7 x)( + 3x) f (1 3x)(5 + x) g (3x + 4)(5x 3) h (1 3x)( 5x) i (7 x)(3 x) j (5 x)(3 x) k (x + 1)(x +) l (x 1)(x +3) Exmple 10 Expnd using the rule ( + )( ) = : (5x )(5x +) (7 + x)(7 x) (5x )(5x +) (7 + x)(7 x) =(5x) =7 (x) =5x 4 = 49 4x Expnd using the rule ( + )( ) = : (x + 6)(x 6) (x + 8)(x 8) c (x 1)(x +1) d (3x )(3x +) e (4x + 5)(4x 5) f (5x 3)(5x +3) g (3 x)(3 + x) h (7 x)(7 + x) i (7 + x)(7 x) j (x + p )(x p ) k (x + p 5)(x p 5) l (x p 3)(x + p 3) Exmple 11 Expnd using the perfect squre expnsion rule: (x +) (3x 1) (x +) (3x 1) = x +(x)() + =(3x) + (3x)( 1) + ( 1) = x +4x +4 =9x 6x +1 3 Expnd nd simplify using the perfect squre expnsion rule: (x +5) (x +7) c (x ) d (x 6) e (3 + x) f (5 + x) g (11 x) h (10 x) i (x +7) j (3x +) k (5 x) l (7 3x)

10 BACKGROUND KNOWLEDGE 4 Expnd the following into the generl form y = x + x + c: y =(x + )(x +3) y =3(x 1) +4 c y = (x + 1)(x 7) d y = (x +) 11 e y =4(x 1)(x 5) f y = 1 (x +4) 6 g y = 5(x 1)(x 6) h y = 1 (x +) 6 i y = 5 (x 4) Exmple 1 Expnd nd simplify: 1 (x +3) (3 + x) ( + x)(3 x) The use of rckets is essentil! 1 (x +3) =1 [x +6x +9] =1 x 1x 18 = x 1x 17 (3 + x) ( + x)(3 x) =6+x [6 x +3x x ] =6+x 6+x 3x + x = x + x 5 Expnd nd simplify: 1+(x +3) +3(x )(x +3) c 3 (3 x) d 5 (x + 5)(x 4) e 1 + (4 x) f x 3x (x + )(x ) g (x +) (x + 1)(x 4) h (x +3) +3(x +1) i x +3x (x 4) j (3x ) (x +1) H FACTORISATION Algeric fctoristion is the reverse process of expnsion. For exmple, (x + 1)(x 3) is expnded to x 5x 3, wheres x 5x 3 is fctorised to (x + 1)(x 3). Notice tht x 5x 3=(x + 1)(x 3) hs een fctorised into two liner fctors. Flow chrt for fctorising: Expression Difference of two squres =( + )( ) Tke out ny common fctors Recognise type Perfect squre + + =( + ) Sum nd Product type x + x + c, 6= 0 ² find c ² find the fctors of c which dd to ² if these fctors re p nd q, replce x ypx + qx ² complete the fctoristion Sum nd Product type x + x + c x + x + c =(x + p)(x + q) where p + q = nd pq = c

BACKGROUND KNOWLEDGE 11 Exmple 13 Fully fctorise: 3x 1x 4x 1 c x 1x +36 3x 1x f3x is common fctorg =3x(x 4) 4x 1 =(x) 1 =(x + 1)(x 1) fdifference of two squresg Rememer tht ll fctoristions cn e checked y expnsion! c x 1x +36 = x +(x)( 6) + ( 6) fperfect squre formg =(x 6) EXERCISE H 1 Fully fctorise: 3x +9x x +7x c 4x 10x d 6x 15x e 9x 5 f 16x 1 g x 8 h 3x 9 i 4x 0 j x 8x +16 k x 10x +5 l x 8x +8 m 16x +40x +5 n 9x +1x +4 o x x + 11 Exmple 14 Fully fctorise: 3x +1x +9 x +3x +10 3x +1x +9 f3 is common fctorg =3(x +4x +3) fsum =4, product =3g =3(x + 1)(x +3) x +3x +10 = [x 3x 10] fremoving 1 s common fctor to mke the coefficient of x e 1g = (x 5)(x +) fsum = 3, product = 10g Fully fctorise: x +9x +8 x +7x +1 c x 7x 18 d x +4x 1 e x 9x +18 f x + x 6 g x + x + h 3x 4x +99 i x 4x j x +6x 0 k x 10x 48 l x +14x 1 m 3x +6x 3 n x x 1 o 5x +10x +40

1 BACKGROUND KNOWLEDGE FACTORISATION BY SPLITTING THE x-term Using the distriutive lw to expnd we see tht: (x + 3)(4x +5) =8x +10x +1x +15 =8x +x +15 We will now reverse the process to fctorise the qudrtic expression 8x +x +15: 8x +x +15 Step 1: Split the middle term =8x +10x +1x +15 Step : Group in pirs =(8x +10x) + (1x + 15) Step 3: Fctorise ech pir seprtely =x(4x + 5) + 3(4x +5) Step 4: Fctorise fully =(4x + 5)(x + 3) The trick in fctorising these types of qudrtic expressions is in Step 1 where the middle term needs to e split into two so tht the rest of the fctoristion proceeds smoothly. Rules for splitting the x-term: The following procedure is recommended for fctorising x + x + c : ² Find c: ² Find the fctors of c which dd to : ² If these fctors re p nd q, replce x y px + qx: ² Complete the fctoristion. Exmple 15 Fully fctorise: x x 10 6x 5x +14 x x 10 hs c = 10 = 0: The fctors of 0 which dd to 1 re 5 nd +4: ) x x 10 = x 5x +4x 10 = x(x 5) + (x 5) =(x 5)(x +) 6x 5x +14 hs c =6 14 = 84: The fctors of 84 which dd to 5 re 1 nd 4: ) 6x 5x +14 =6x 1x 4x +14 =3x(x 7) (x 7) =(x 7)(3x )

BACKGROUND KNOWLEDGE 13 3 Fully fctorise: x +5x 1 3x 5x c 7x 9x + d 6x x e 4x 4x 3 f 10x x 3 g x 11x 6 h 3x 5x 8 i 8x +x 3 j 10x 9x 9 k 3x +3x 8 l 6x +7x + m 4x x +6 n 1x 16x 3 o 6x 9x +4 p 1x 10 9x q 8x 6x 7 r 1x +13x +3 s 1x +0x +3 t 15x x +8 u 14x 11x 15 Exmple 16 Fully fctorise: 3(x +)+(x 1)(x +) (x +) 3(x +)+(x 1)(x +) (x +) =(x + )[3 + (x 1) (x + )] fs (x +) is common fctorg =(x + )[3 + x x ] =(x + )(x 1) 4 Fully fctorise: 3(x +4)+(x + 4)(x 1) 8( x) 3(x + 1)( x) c 6(x +) +9(x +) d 4(x +5)+8(x +5) e (x + )(x + 3) (x + 3)( x) f (x + 3) + (x + 3) x(x + 3) g 5(x ) 3( x)(x +7) h 3(1 x)+(x + 1)(x 1) Exmple 17 Fully fctorise using the difference of two squres : (x +) 9 (1 x) (x +1) (x +) 9 (1 x) (x +1) =(x +) 3 = [(1 x) (x + 1)][(1 x)+(x + 1)] =[(x + ) + 3][(x +) 3] = [1 x x 1][1 x +x +1] =(x + 5)(x 1) = 3x(x +) 5 Fully fctorise: (x +3) 16 4 (1 x) c (x +4) (x ) d 16 4(x +) e (x +3) (x 1) f (x + h) x g 3x 3(x +) h 5x 0( x) i 1x 7(3 + x)

14 BACKGROUND KNOWLEDGE INVESTIGATION 1 ANOTHER FACTORISATION TECHNIQUE Wht to do: (x + p)(x + q) h pq i 1 By expnding, show tht = x +[p + q]x + : If x (x + p)(x + q) + x + c =, show tht p + q = nd pq = c. 3 Using on 8x +x +15, we hve 8x (8x + p)(8x + q) +x +15= 8 ) p =1, q =10, or vice vers ) 8x +x +15= (8x + 1)(8x + 10) 8 4(x + 3)(4x +5) = 8 =(x + 3)(4x +5) where ½ p + q = pq =8 15 = 10 Use the method shown to fctorise: i 3x +14x +8 ii 1x +17x +6 iii 15x +14x 8 Check your nswers to y expnsion. I FORMULA REARRANGEMENT In the formul D = xt + p we sy tht D is the suject. This is ecuse D is expressed in terms of the other vriles x, t nd p. We cn rerrnge the formul to mke one of the other vriles the suject. We do this using the usul rules for solving equtions. Whtever we do to one side of the eqution we must lso do to the other side. EXERCISE I Exmple 18 Mke x the suject of D = xt + p. If D = xt + p ) xt + p = D ) xt + p p = D p fsutrct p from oth sidesg ) xt = D p xt ) t = D p t ) x = D p t fdivide oth sides y tg

BACKGROUND KNOWLEDGE 15 1 Mke x the suject of: + x = x = c x + = d d c + x = t e 5x +y =0 f x +3y =1 g 7x +3y = d h x + y = c i y = mx + c Exmple 19 Mke z the suject of c = m z. c = m z c z = m z z ) cz = m ) cz c = m c fmultiply oth sides y zg fdivide oth sides y cg ) z = m c Mke z the suject of: z = c z = d c 3 d = z d z = z 3 Mke: the suject of F = m r the suject of C =¼r c d the suject of V = ldh d K the suject of A = K : Exmple 0 Mke t the suject of s = 1 gt where t>0. 1 gt = s frewrite with t on LHSg ) 1 gt = s fmultiply oth sides y g ) gt =s ) gt g = s g ) t = s g r s ) t = g fdivide oth sides y gg fs t>0g

16 BACKGROUND KNOWLEDGE 4 Mke: r the suject of A = ¼r if r>0 x the suject of N = x5 c r the suject of V = 4 3 ¼r3 d x the suject of D = n x 3 : 5 Mke: p the suject of d = l the suject of T = 1 n 5p l c the suject of c = p r l d l the suject of T =¼ g e the suject of P =( + ) f h the suject of A = ¼r +¼rh g r the suject of I = E R + r 6 Mke the suject of the formul k = d. Find the vlue for when k = 11, d =4, =. h q the suject of A = B p q : 7 The formul for determining the volume of sphere of rdius r is V = 4 3 ¼r3. Mke r the suject of the formul. Find the rdius of sphere which hs volume of 40 cm 3. 8 The distnce (in cm) trvelled y n oject ccelerting from sttionry position is given y the formul S = 1 t where is the ccelertion in cm s nd t is the time in seconds. Mke t the suject of the formul. Consider t>0 only. Find the time tken for n oject ccelerting t 8 cm s to trvel 10 m. 9 The reltionship etween the oject nd imge distnces (in cm) for concve mirror 1 cn e written s f = 1 u + 1 where f is the focl length, u is the oject distnce v nd v is the imge distnce. Mke v the suject of the formul. Given focl length of 8 cm, find the imge distnce for the following oject distnces: i 50 cm ii 30 cm. 10 According to Einstein s theory of reltivity, the mss of prticle is given y the m 0 formul m = r ³, where m v 0 is the mss of the prticle t rest, 1 c v is the speed of the prticle, nd c is the speed of light. c Mke v the suject of the formul given v>0. Find the speed necessry to increse the mss of prticle to three times its rest mss, i.e., m =3m 0. Give the vlue for v s frction of c. A cyclotron incresed the mss of n electron to 30m 0 : With wht velocity must the electron hve een trvelling, given c =3 10 8 ms 1?

J BACKGROUND KNOWLEDGE 17 ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS To dd or sutrct lgeric frctions, we comine them into single frction with the lest common denomintor (LCD). x 1 For exmple, 3 EXERCISE J Exmple 1 x +3 hs LCD of 6, so we write ech frction with denomintor 6. Write s single frction: + 3 x x 1 3 x +3 + 3 x ³ x = + x 3 x x 1 x +3 3 = µ x 1 3 µ x +3 3 3 = x +3 x = = = (x 1) 3(x +3) 6 x 3x 9 6 x 11 6 1 Write s single frction: 3+ x 5 d 3 x 4 e 1+ 3 x +x 3 + x 4 5 c 3+ x x +5 f x 1 4 6 Exmple 3x +1 Write x s single frction. = = 3x +1 x µ 3x +1 x (3x +1) (x ) x 3x +1 x +4 = x = x +5 x µ x x fthe LCD is (x )g

18 BACKGROUND KNOWLEDGE Write s single frction: 1+ 3 x + d + 3 x 4 x 1 x +1 +3 e 3 x x +1 c 3 x 1 f 1+ 4 1 x 3 Write s single frction: 3x x +5 + x 5 x K c 5x x 4 + 3x x +4 CONGRUENCE d 1 x 1 x 3 x +1 x 3 x +4 x +1 CONGRUENCE AND SIMILARITY Two tringles re congruent if they re identicl in every respect prt from position. The tringles hve the sme shpe nd size. There re four cceptle tests for the congruence of two tringles. Two tringles re congruent if one of the following is true: ² corresponding sides re equl in length (SSS) ² two sides nd the included ngle re equl (SAS) ² two ngles nd pir of corresponding sides re equl (AAcorS) ² for right ngled tringles, the hypotenuse nd one other pir of sides re equl (RHS). Exmple 3 A Explin why ABC nd DBC re congruent: B C A B D C D s ABC nd DBC re congruent (SAS) s: ² AC = DC ² ACB = DCB, nd ² [BC] is common to oth.

BACKGROUND KNOWLEDGE 19 If congruence cn e proven then ll corresponding lengths, ngles nd res must e equl. EXERCISE K.1 1 Tringle ABC is isosceles with AC = BC. [BC] nd [AC] re produced to E nd D respectively so tht CE = CD. E C D Prove tht AE = BD. A B Point P is equidistnt from oth [AB] nd [AC]. Use congruence to show tht P lies on the isector of B AC. A P B C 3 Two concentric circles re drwn. At P on the inner circle, tngent is drwn which meets the other circle t A nd B. Use tringle congruence to prove tht P is the midpoint of [AB]. A O P B SIMILARITY Two tringles re similr if one is n enlrgement of the other. Similr tringles re equingulr, nd hve corresponding sides in the sme rtio. Exmple 4 Estlish tht pir of tringles is similr nd find x if BD =0cm: 1 cm B A x cm C E ( x + ) cm D 1 cm B A E x cm C 0 cm ( x + ) cm D ² - x + x smll - 0 1 lrge The tringles re equingulr nd hence similr. ) x + = x fsides in the sme rtiog 0 1 ) 1(x +)=0x ) 1x +4=0x ) 4 = 8x ) x =3

0 BACKGROUND KNOWLEDGE EXERCISE K. 1 In ech of the following, estlish tht pir of tringles is similr, nd hence find x: A c U P cm B C 5cm X x cm 3cm 6cm x cm Q S 7cm R x cm cm D E 6cm T V 5cm Y 6cm Z d D e X f A 4cm E B 3cm 5cm C x cm Y cm U 5cm x cm Z V S P 8cm x cm 10 cm 50 Q 50 9cm R T A fther nd son re stnding side-y-side. The fther is 1:8 m tll nd csts shdow 3: m long, while his son s shdow is :4 m long. How tll is the son? L COORDINATE GEOMETRY THE NUMBER PLANE The position or loction of ny point in the numer plne cn e specified in terms of n ordered pir of numers (x, y), where x is the horizontl step from fixed point O, nd y is the verticl step from O. The point O is clled the origin. Once O hs een specified, we drw two perpendiculr xes through it. The x-xis is horizontl nd the y-xis is verticl. The numer plne is lso known s either: ² the -dimensionl plne, or ² the Crtesin plne, nmed fter René Descrtes. (, ) is clled n ordered pir, where nd re the coordintes of the point. is clled the x-coordinte. is clled the y-coordinte. y-xis DEMO P(, ) x-xis

BACKGROUND KNOWLEDGE 1 THE DISTANCE FORMULA If A(x 1, y 1 ) nd B(x, y ) re two points in plne, then the distnce etween these points is given y AB = p (x x 1 ) +(y y 1 ). Exmple 5 Find the distnce etween A(, 1) nd B(3, 4). A(, 1) B(3, 4) " " " " x 1 y 1 x y AB = p (3 ) +(4 1) = p 5 +3 = p 5 + 9 = p 34 units THE MIDPOINT FORMULA If M is hlfwy etween points A nd B then M is the midpoint of [AB]. A M B If A(x 1, y 1 ) nd B(x, y ) re two points then µ x1 + x the midpoint M of [AB] hs coordintes, y 1 + y : Exmple 6 Find the coordintes of the midpoint of [AB] for A( 1, 3) nd B(4, 7). The x-coordinte of the midpoint = 1+4 The y-coordinte of the midpoint = 3+7 ) the midpoint of [AB] is (1 1, 5). =5 = 3 =11 THE GRADIENT OR SLOPE OF A LINE When looking t line segments drwn on set of xes, it is cler tht different line segments re inclined to the horizontl t different ngles. Some pper to e steeper thn others. The grdient or slope of line is mesure of its steepness. If A is (x 1, y 1 ) nd B is (x, y ) then the grdient of [AB] is y y 1 x x 1 :

BACKGROUND KNOWLEDGE Exmple 7 Find the grdient of the line through (3, ) nd (6, 4). (3, ) (6, 4) " " " " x 1 y 1 x y grdient = y y 1 = 4 x x 1 6 3 = Note: ² horizontl lines hve grdient of 0 (zero) ² verticl lines hve n undefined grdient ² forwrd sloping lines hve positive grdients ² prllel lines hve equl grdients ² ckwrd sloping lines hve negtive grdients ² the grdients of perpendiculr lines re negtive reciprocls, i.e., if the grdients re m 1 nd m then m = 1 m 1 or m 1 m = 1. This is true except when the lines re prllel to the xes. EQUATIONS OF LINES The eqution of line sttes the connection etween the x nd y vlues for every point on the line, nd only for points on the line Equtions of lines hve vrious forms: ² All verticl lines hve equtions of the form x = where is constnt. ² All horizontl lines hve equtions of the form y = c where c is constnt. ² If stright line hs grdient m nd psses through (, ) then it hs eqution y = m or y = m(x ) fpoint-grdient formg x which cn e rerrnged into y = mx + c fgrdient-intercept formg or Ax + By = C fgenerl formg Exmple 8 Find, in grdient-intercept form, the eqution of the line through ( 1, 3) with slope of 5. To find the eqution of line we need to know its grdient nd point on it. The eqution of the line is y 3 x 1 =5 i.e., y 3 x +1 =5 ) y 3=5(x +1) ) y =5x +8

BACKGROUND KNOWLEDGE 3 Exmple 9 Find, in generl form, the eqution of the line through (1, 5) nd (5, ). The slope = = 3 4 5 5 1 So, the eqution is y x 5 = 3 4 y + ) x 5 = 3 4 ) 4y +8=3x 15 ) 3x 4y =3 AXES INTERCEPTS Axes intercepts re the x- nd y-vlues where grph cuts the coordinte xes. The x-intercept is found y letting y =0. The y-intercept is found y letting x =0. y y-intercept x-intercept x Exmple 30 For the line with eqution x 3y =1, find the xes intercepts. When x =0, 3y =1 ) y = 4 So, the y-intercept is 4 nd the x-intercept is 6. When y =0, x =1 ) x =6 DOES A POINT LIE ON A LINE? A point lies on line if its coordintes stisfy the eqution of the line. Exmple 31 Does (3, ) lie on the line with eqution 5x y =0? Sustituting (3, ) into 5x y =0 gives 5(3) ( ) = 0 i.e., 19 = 0 which is flse ) (3, ) does not lie on the line.

4 BACKGROUND KNOWLEDGE WHERE GRAPHS MEET Exmple 3 Use grphicl methods to find where the lines x + y =6 nd x y =6 meet. For x + y =6: when x =0, y =6 when y =0, x =6 For x y =6: when x =0, y =6, ) y = 6 when y =0, x =6, ) x =3 The grphs meet t (4, ). Check: 4+=6 X nd 4 =6 X x 0 6 y 6 0 x 0 3 y 6 0 y!-@=6 (4, ) x!+@=6 There re three possile situtions which my occur. These re: Cse 1: Cse : Cse 3: The lines meet in The lines re prllel nd The lines re coincident. single point of never meet. There is There re infinitely mny intersection. no point of intersection. points of intersection. INVESTIGATION FINDING WHERE LINES MEET USING TECHNOLOGY Grphing pckges nd grphics clcultors cn e used to plot grphs nd hence find their point(s) of intersection. This cn e useful if the solutions re not integer vlues. One downside of using grphics clcultors is tht most require the equtions to e entered in the form y = f(x). So, if the eqution of stright line is given in generl form, it must e rerrnged into grdient-intercept form. Suppose we wish to use technology to find the point of intersection of 4x +3y =10 nd x y = 3: GRAPHING PACKAGE If you re using the grphing pckge, click on the icon to open the pckge nd enter the two equtions.

BACKGROUND KNOWLEDGE 5 If you re using grphics clcultor, follow the following steps: Step 1: We rerrnge ech eqution into the form y = mx + c: 4x +3y =10 ) 3y = 4x +10 ) y = 4 3 x + 10 3 x y = 3 ) y = x 3 ) y = x + 3 Step : Enter the functions Y 1 = 4X=3+10=3 nd Y = X=+3=. Step 3: Drw the grphs of the functions on the TI sme set of xes. You my hve to chnge the viewing window. C Step 4: Use the uilt in functions to clculte the point of intersection. Thus, the point of intersection is (1, ). Wht to do: 1 Use technology to find the point of intersection of: y = x +4 5x 3y =0 x +y =8 y =7 x d x + y =7 3x y =1 e y =3x 1 3x y =6 c x y =5 x +3y =4 f y = x 3 + x +3y =6 Comment on the use of technology to find the point(s) of intersection in 1end 1f. EXERCISE L 1 Use the distnce formul to find the distnce etween the following pirs of points: A(1, 3) nd B(4, 5) O(0, 0) nd C(3, 5) c P(5, ) nd Q(1, 4) d S(0, 3) nd T( 1, 0). Find the midpoint of [AB] for: A(3, 6) nd B(1, 0) A(5, ) nd B( 1, 4) c A(7, 0) nd B(0, 3) d A(5, ) nd B( 1, 3). 3 By finding y-step nd n x-step, determine the slope of ech of the following lines: c d e f

6 BACKGROUND KNOWLEDGE 4 Find the grdient (slope) of the line pssing through: (, 3) nd (4, 7) (3, ) nd (5, 8) c ( 1, ) nd ( 1, 5) d (4, 3) nd ( 1, 3) e (0, 0) nd ( 1, 4) f (3, 1) nd ( 1, ). 5 Clssify the following pirs of lines s prllel, perpendiculr or neither. Give resons for your nswers. c d e f 6 Stte the slope of the line which is perpendiculr to the line with grdient (slope): 3 4 11 3 c 4 d 1 3 e 5 f 0 7 Find, in grdient-intercept form, the eqution of the line through: (4, 1) with slope (1, ) with slope c (5, 0) with slope 3 d ( 1, 7) with slope 3 e (1, 5) with slope 4 f (, 7) with slope 1: 8 Find, in generl form, the eqution of the line through: (, 1) with slope 3 (1, 4) with slope 3 c (4, 0) with slope 1 3 d (0, 6) with slope 4 e ( 1, 3) with slope 3 f (4, ) with slope 4 9 : 9 Find the equtions of the lines through: (0, 1) nd (3, ) (1, 4) nd (0, 1) c (, 1) nd ( 1, 4) d (0, ) nd (5, ) e (3, ) nd ( 1, 0) f ( 1, 1) nd (, 3) 10 Find the equtions of the lines through: (3, ) nd (5, ) (6, 7) nd (6, 11) c ( 3, 1) nd ( 3, 3) 11 Copy nd complete: Eqution of line Grdient x-intercept y-intercept x 3y =6 4x +5y =0 c y = x +5 d x =8 e y =5 f x + y =11 g 4x + y =8 h x 3y =1 If line hs eqution y mx c then the grdient of the line is m.

1 Does (3, 4) lie on the line with eqution 3x y =1? Does (, 5) lie on the line with eqution 5x +3y = 5? c Does (6, 1 ) lie on the line 3x 8y =? 13 Use grphicl methods to find where the following lines meet: BACKGROUND KNOWLEDGE 7 x +y =8 y = 3x 3 c 3x + y = 3 y =x 6 3x y = 1 x 3y = 4 d x 3y =8 e x +3y =10 f 5x +3y =10 3x +y = 1 x +6y = 11 10x +6y = 0 Exmple 33 A stright rod is to pss through the points A(5, 3) nd B(1, 8). c Find where this rod meets the rod given y: i x =3 ii y =4 If we wish to refer to the points on the rod (AB) etween A nd B, how cn we indicte this? Does C(3, 0) lie on the rod? B(1' 8) y 4 x 3 A(5' 3) First we must find the eqution of the line representing the rod. Its grdient is m = 3 8 5 1 = 5 4 y 3 So, its eqution is x 5 = 5 4 ) 4(y 3) = 5(x 5) ) 4y 1 = 5x +5 ) 5x +4y =37 i When x =3, 5(3) + 4y =37 ) 15 + 4y =37 ) 4y = ) y =5 1 ) meet t (3, 5 1 ). We restrict the possile x-vlues to 1 6 x 6 5. c ii When y =4, 5x + 4(4) = 37 ) 5x +16=37 ) 5x =1 ) x =4: ) they meet t (4:, 4). If C(3, 0) lies on the line, its coordintes must stisfy the line s eqution. Now LHS = 5(3) + 4( 0) = 115 80 =35 6= 37 ) C does not lie on the rod.

8 BACKGROUND KNOWLEDGE 14 Find the eqution of the: horizontl line through (3, 4) verticl line with x-intercept 5 c verticl line through ( 1, 3) d horizontl line with y-intercept e x-xis f y-xis. 15 Find the eqution of the line which is: through A( 1, 4) nd with grdient 3 4 through P(, 5) nd Q(7, 0) c prllel to the line with eqution y =3x nd psses through (0, 0) d prllel to the line with eqution x +3y =8 nd psses through ( 1, 7) e perpendiculr to the line with eqution y = x + 5 nd psses through (3, 1) f perpendiculr to the line with eqution 3x y =11 nd psses through (, 5). 16 A is the town hll on Scott Street nd D is Post Office on Kech Avenue. Digonl Rod intersects Scott Street t B nd Kech Avenue t C. c Find the eqution of Kech Avenue. Find the eqution of Pecock Street. Find the eqution of Digonl Rod. (Be creful!) A(3,17) d Plunkit Street lies on the mp reference line x =8. Where does Plunkit Street intersect Kech Avenue? Pecock St Scott St Digonl Rd B(7,0) Kech Av D(13,1) E C(5,11) Exmple 34 Find the eqution of the tngent to the circle with centre (, 3) t the point ( 1, 5). C(, 3) The grdient of [CP] is 3 5 ( 1) = 3 = 3 ) the grdient of the tngent t P is 3. Since the tngent psses through ( 1, 5), P( 1, 5) its eqution is y 5 x 1 = 3 ) (y 5) = 3(x +1) ) y 10 = 3x +3 ) 3x y = 13 The tngent is perpendiculr to the rdius t the point of contct. 17 Find the eqution of the tngent to the circle with centre: (0, ) t ( 1, 5) (3, 1) t( 1, 1) c (, ) t(5, ).

BACKGROUND KNOWLEDGE 9 Exmple 35 Mining towns re situted t B(1, 6) nd A(5, ). Where should the rilwy siding S e locted so tht ore trucks from either A or B would trvel equl distnces to rilwy line with eqution x =11? B(1, 6) A(5, ) S x 11 rilwy line 18 A(5, 5) nd B(7, 10) re houses nd y =8 is gs pipeline. Where should the one outlet from the pipeline e plced so tht it is the sme distnce from oth houses? 19 (CD) is wter pipeline. A nd B re two towns. A pumping sttion is to e locted on the pipeline to pump wter to A nd B. Ech town is to py for their own service pipes nd they insist on equlity of costs. Where should C e locted to ensure equlity of costs? Wht is the totl length of service pipe required? c Suppose S hs the coordintes (11, ). Now BS = AS ) p (11 1) +( 6) = p (11 5) +( ) ) 10 +( 6) =6 +( ) fsquring oth sidesg ) 100 + 1 + 36 = 36 + 4 +4 ) 1 +4 =4 100 ) 8 = 96 ) =1 So, the rilwy siding should e locted t (11, 1). Exmple 36 A(, 3) B A If the towns gree to py equl mounts, would it e cheper to instll the service pipeline from D to B to A? C D B(5, 4) y Scle: 1 unit 1 km y P N Q(, 4) x x Scle: ech grid unit is 1 km. A tunnel through the mountins connects town Q(, 4) to the port t P. P is on grid reference x =6nd the distnce etween the town nd the port is 5 km. Assuming the digrm is resonly ccurte, wht is the horizontl grid reference of the port?

30 BACKGROUND KNOWLEDGE Suppose P is t (6, ). Now PQ =5 ) p (6 ) +( 4) =5 ) p 16 + ( 4) =5 ) 16 + ( 4) =5 ) ( 4) =9 ) 4= 3 ) =4 3=7or 1 But from the digrm, P is further north thn Q ) >4 So, P is t (6, 7) nd the horizontl grid reference is y =7. 0 y 8 Clifton Json s girlfriend lives in house on Clifton Highwy Highwy which hs eqution y = 8. The distnce s the crow flies from Json s house to his girlfriend s house is 11:73 km. If Json lives t (4, 1), wht re the coordintes of his girlfriend s house? J(4, 1) Scle: 1 unit 1 km. 1 A circle hs centre (, ) nd rdius r units. P(x, y) moves on the circle. Show tht (x ) +(y ) = r. Find the eqution of the circle with: i centre (4, 3) nd rdius 5 units ii centre ( 1, 5) nd rdius units iii centre (0, 0) nd rdius 10 units iv ends of dimeter ( 1, 5) nd (3, 1). Find the centre nd rdius of the circle: (x 1) +(y 3) =4 x +(y +) =16 c x + y =7 3 Consider the circle with eqution (x ) +(y +3) =0: Stte the circle s centre nd rdius. Show tht (4, 1) lies on the circle. c Find the eqution of the tngent to the circle t the point (4, 1). 4 Recll tht the perpendiculr isector of chord of circle psses through the centre of the circle. Find the centre of circle pssing through points P(5, 7), Q(7, 1) nd R( 1, 5) y finding the perpendiculr isectors of [PQ] nd [QR] nd solving them simultneously.

BACKGROUND KNOWLEDGE (Answers) 31 EXERCISE A p p p p 1 15 3 c 4 d 1 e 4 f 6 g h 6 5 p p c p 5 d 8 p 5 e p 5 f 9 p 3 g 3 p 6 h 3 p 3 p p 3 c p 5 d 4 p e 3 p 3 f 3 p 5 g 4 p 3 h 3 p 6 i 5 p j 4 p 5 k 4 p 6 l 6 p 3 4 p 3 8 p c 5 p 6 d 10 p 3 e 3 p 3 f p p 5 p 7 p 3 c d p 5 e 5 p f 3 p 6 g 4 p 3 h 5 p 7 7 i p 7 j EXERCISE B 1 :59 10 :59 10 5 c :59 10 0 d :59 10 1 e :59 10 4 f 4:07 10 1 g 4:07 10 3 h 4:07 10 i 4:07 10 5 j 4:07 10 8 k 4:07 10 5 1:495 10 11 m 3 10 4 mm c 1 10 3 mm d 1:5 10 7 o C e 3 10 5 3 4000 500 c 100 d 78 000 e 380 000 f 86 g 43 300 000 h 60 000 000 4 0:004 0:05 c 0:001 d 0:000 78 e 0:000 038 f 0:86 g 0:000 000 433 h 0:000 000 6 5 0:000 000 9 m 6 130 000 000 c 100 000 light yers d 0:000 01 mm 6 1:64 10 10 4:1 10 3 c 5:7 10 18 d 1:36 10 e :63 10 6 f 1:73 10 9 7 1:30 10 5 km 9:07 10 5 km c 9:47 10 7 km 8 1:8 10 10 m :59 10 13 m c 9:47 10 15 m EXERCISE C 1 The set of ll rel x such tht x is greter thn 5. The set of ll integers x such tht x is less thn or equl to 3. c The set of ll y such tht y lies etween 0 nd 6. d The set of ll integers x such tht x is greter thn or equl to, ut less thn or equl to 4. x is, 3 or 4. e The set of ll t such tht t lies etween 1 nd 5. f The set of ll n such tht n is less thn or greter thn or equl to 6 fx j x>g fx j 1 <x6 5g c fx j x 6 or x > 3g d fx j 16 x 6 3, x Z g e fx j 0 6 x 6 5, x Z g f fx j x<0g 3 c d p 6 3 4 5 6 7 8 9 10 x x x x e x EXERCISE D 1 10x 10 9x c 5x +5y d 8 8x e 1 f cnnot e simplified x +35 16 6x c 4 3 d 3x 3 16x +11x 1 3 18x 3 c 4x d 4 10 3 EXERCISE E 1 x =10 x>6 c x = 4 d x =51 e x< 10 5 f x =14 g x 6 9 h x =18 i x = 3 x =5, y = x = 3, y = 8 3 c x =, y =5 d x = 45 11, y = 18 11 e no solution f x =66, y = 84 EXERCISE F 1 16 6 c 16 d 18 e f 3 c 6 d 6 e 5 f 1 g 1 h 5 i 4 j 4 k l 3 x = 3 no solution c x =0 d x =4 or e x = 1 or 7 f no solution g x =1 or 1 3 h x =0 or 3 i x = or 14 5 EXERCISE G 1 x +5x +3 3x +10x +8 c 10x + x d 3x +x 10 e 6x +17x+14 f 6x 13x+5 g 15x +11x 1 h 15x 11x+ i x 17x+1 j 4x 16x +15 k x 3x l 4x x +6 x 36 x 64 c 4x 1 d 9x 4 e 16x 5 f 5x 9 g 9 x h 49 x i 49 4x j x k x 5 l 4x 3 3 x +10x +5 x +14x +49 c x 4x +4 d x 1x +36 e x +6x +9 f x +10x +5 g x x +11 h x 0x+100 i 4x +8x+49 j 9x +1x +4 k 4x 0x +5 l 9x 4x +49 4 y =x +10x +1 y =3x 6x +7 c y = x +6x +7 d y = x 4x 15 e y =4x 4x +0 f y = 1 x 4x 14 g y = 5x +35x 30 h y = 1 x +x 4 i y = 5 x +0x 40 5 x +1x +19 3x +3x 16 c x +6x 6 d x x +5 e x 16x +33 f 3x +4 g 7x +8 h 7x +18x +1 i x +19x 3 j 7x 16x + EXERCISE H 1 3x(x+3) x(x+7) c x(x 5) d 3x(x 5) e (3x 5)(3x+5) f (4x+1)(4x 1) g (x )(x+) h 3(x+ p 3)(x p 3) i 4(x+ p 5)(x p 5) j (x 4) k (x 5) l (x ) m (4x +5) n (3x +) o (x 11) (x + 8)(x +1) (x + 4)(x +3) c (x 9)(x +) d (x + 7)(x 3) e (x 6)(x 3) f (x + 3)(x ) g (x )(x +1) h 3(x 11)(x 3) i (x +1) j (x+5)(x ) k (x 8)(x+3) l (x 6)(x 1) m 3(x 1) n (x +1) o 5(x 4)(x +) 3 (x 3)(x+4) (3x+1)(x ) c (7x )(x 1) d (3x )(x+1) e (x 3)(x+1) f (5x 3)(x+1) g (x+1)(x 6) h (3x+7)(x 4) i (4x+3)(x 1) j (5x+3)(x 3) k (3x 1)(x+8) l (3x+)(x+1) m (x + 3)(x 1) n (6x + 1)(x 3) o 3(x + 7)(x ) p (3x )(3x 5) q (4x 9)(x +3) r (4x + 3)(3x +1)

3 BACKGROUND KNOWLEDGE (Answers) s (6x + 1)(x +3) t (5x 4)(3x ) u (7x + 5)(x 3) 4 (x+4)(x+1) ( x)(5 3x) c 3(x+)(x+7) d 4(x + 5)(x + 11) e x(x +3) f 5(x +3) g (x )(3x + 6) h (x 1)(x 1) 5 (x + 7)(x 1) (x + 1)(3 x) c 1(x +1) d 4x(x +4) e (3x + )(x +4) f h(x + h) g 1(x+1) h 5(3x 4)(x 4) i 3(x+9)(5x+9) EXERCISE I 1 x = x = 0 y e x = f x = 5 h x = c y z = c i z = d c x = y c m 1 3y c z = d 3 x = d g d x = d 3y 7 x = t c 3 = F m r = C ¼ c d = V lh d K = A r r A 4 r = x = 5p q 3V n N c r = 3 d x = 3 ¼ 4¼ D 5 = d n l =5T c = p + c d l = gt 4¼ e = P f h = A ¼r g r = E ¼r I R h q = p B A r 6 = d 3V 1:9 7 r = 3 :1 cm k 4¼ r S 8 t = 15:81 sec 9 v = uf i 9:5 cm ii 10:9 cm u f µ 10 v = sc 1 m 0 = c p m m m 0 m v = p 8 3 c EXERCISE J 1 15 + x x +3 5 x e 8x 4x +17 f 15 1 e 3 c x +5 x + x +3 x +1 f 11 x x 4 x +3 1 x 7x 6x 5 (x 5)(x ) 8x +6x +8 x 16 c :998 10 8 ms 1 EXERCISE K.1 1 Hint: Consider s AEC, BDC d c c x +4 1 3x x 1 d d 1 (x )(x 3) 3x +3x +13 (x 3)(x +1) 14 x 4 5x + x +1 Hint: Let M e on [AB] so tht [PM]? [AB]. Let N e on [AC] so tht [PN]? [AC]. Join [PM], [PN] nd consider the two tringles formed. EXERCISE K. 1 x =:4 x =:8 c x =3 3 d x =6 11 3 e x =7 f x =7: 1:35 m tll EXERCISE L 1 p 13 units p 34 units c p 0 units d p 10 units (, 3) (, 1) c (3 1, 1 1 ) d (, 1 ) 3 c 1 d 0 e 4 f undefined 3 5 4 3 c undefined d 0 e 4 f 5 prllel, slopes 5 prllel, slopes 1 7 c perpendiculr, slopes 1, d neither, slopes 4, 1 3 e neither, slopes 7, 1 5 f perpendiculr, slopes, 1 6 4 3 3 11 c 1 1 4 d 3 e 5 f undefined 7 y =x 7 y = x +4 c y =3x 15 d y = 3x +4 e y = 4x +9 f y = x +5 8 3x y =4 3x +y =11 c x 3y =4 d 4x + y =6 e 3x y =0 f 4x +9y = 9 x 3y = 3 5x y =1 c x y =3 d 4x 5y =10 e x y = 1 f x +3y = 5 10 y = x =6 c x = 3 11 Eqution of line Slope x-int. y-int. x 3y =6 3 3 4x +5y =0 4 5 4 5 c y = x +5 5 5 d x =8 undef. 8 no y-int. e y =5 0 no x-int. 5 f x + y =11 1 11 11 g 4x + y =8 4 8 h x 3y =1 1 3 1 4 1 yes no c yes 13 (4, ) (, 3) c ( 3, 6) d (4, 0) e prllel lines do not meet f coincident lines 14 y = 4 x =5 c x = 1 d y = e y =0 f x =0 15 3x 4y = 19 x y =7 c y =3x d x +3y =19 e x y =5 f x +3y =13 16 x 8y = 83 8x + y =41 c 9x y =3 for 5 6 x 6 7 d (8, 11 3 8 ) 17 x 3y = 16 x y = 3 c x =5 18 (4 3 4, 8) 19 ( 1, 7) 3 8:03 km c yes (6:16 km) 0 (13:41, 8) or( 5:41, 8) 1 Hint: Use the distnce formul to find the distnce from the centre of the circle to point P. i (x 4) +(y 3) =5 ii (x +1) +(y 5) =4 iii x + y = 100 iv (x 1) +(y 3) =8 centre (1, 3), rdius units centre (0, ), rdius 4 units c centre (0, 0), rdius p 7 3 centre (, 3), rdius p units 0 units Hint: Sustitute (4, 1) into eqution of circle. c x +y =6 4 (3, 3) 1 4