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P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 CHAPTER 1 Reviewing Linear Equations Objectives To solve linear equations in one unknown. To transpose and solve formulae. To construct linear equations . To solve simultaneous linear equations by substitution and elimination methods. To use linear equations to solve problems . 1.1 Linear equations The solutions to many problems may be found by translating the problems into mathematical equations which may then be solved using algebraic techniques. The equation is solved by finding the value or values of the unknowns that would make the statement true. Consider the equation 2 x+ 5= 7. Ifx= 1, the true statement 2(1) + 5= 7 is obtained. The solution to the equation therefore is x= 1. In this case there is no other value of xthat would give a true statement. One way of solving equations is to simply substitute numbers in until the correct answer is found. This method of ‘guessing’ the answer is a very inefficient way of solving the problem. There are a number of standard techniques that can be used for solving linear equations algebraically. It is often helpful to look at how the equation has been constructed so that the steps necessary to ‘undo’ the equation can be identified. It is most important that the steps taken to solve the equation are done in the correct order. Essentially it is necessary to do ‘the opposite’ in ‘reverse order’. 1
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 2 Essential Mathematical Methods Unit s1&2 Example 1 Solve the equation 3 x+ 4= 16 for x. Solution 3x+ 4= 16 3x= 16 − 4 Subtract 4 from both sides. = 12 x= 12 3 Divide both sides by 3. = 4 Once a solution has been found it may be checked by substituting the value back into both sides of the original equation to ensure that the left-hand side (LHS) equals the right-hand side (RHS). LHS = 3(4) + 4= 16 RHS = 16 ∴solution is correct. Example 2 Equations with the unknown on both sides Solve 4 x+ 3= 3x–5. Solution 4x+ 3= 3x− 5 4x− 3x+ 3− 3= 3x− 3x− 5− 3 Group all the terms containing x=− 8 the unknown on one side of the equation and the remaining terms on the other. Check: when x=− 8, LHS =− 29 RHS =− 29 ∴solution is correct. Example 3 Equations containing brackets Solve 3(2 x+ 5) = 27.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 3 Solution 3(2 x+ 5) = 27 6x+ 15 = 27 First remove the brackets and then 6x= 27 − 15 use the previous rules. 6x= 12 x= 12 6 x= 2 Check: when x= 2, LHS = 3(2 × 2+ 5) = 27 RHS = 27 ∴solution is correct. Example 4 Equations containing fractions Solve x 5− 2= x 3. Solution x 5− 2= x 3 x 5× 15 − 2× 15 = x 3× 15 Multiply both sides of the equation by the lowest common multiple of 3 and 5. 3x− 30 = 5x 3x− 5x= 30 −2x= 30 x= 30 −2 x=− 15 Check: RHS =− 15 5 − 2=− 3− 2=− 5 LHS =− 15 3 =− 5 ∴the solution is correct. Example 5 Solve x− 3 2 − 2x− 4 3 = 5. Solution Remember that the line separating the numerator and the denominator (the vinculum) acts as a bracket.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 4 Essential Mathematical Methods Unit s1&2 Multiply by 6, the lowest common denominator. x− 3 2 × 6− 2x− 4 3 × 6= 5× 6 3(x− 3) − 2(2 x− 4) = 5× 6 3x− 9− 4x+ 8= 30 3x− 4x= 30 + 9− 8 −x= 31 x= 31 −1 =− 31 Check: LHS = −31 − 3 2 − 2×− 31 − 4 3 = −34 2 − −66 3 =− 17 + 22 = 5 RHS = 5 ∴solution is correct. An equation for the variable xin which the coefficients of x,including the constants, are pronumerals is known as a literal equation . Example 6 Literal equation Solve ax + b= cx + dfor x. Solution ax + b= cx + d ax − cx = d− b (a− c)x= d− b x= d− b a− c Using a graphics calculator Solving a linear equation can be accomplished using a graphics calculator, either by a suitable graphical representation or by using the Solver screen from the MATH menu. Students should be able to solve equations both by hand and by using forms of this technology.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 5 Solving an equation Using the y = and graph windows Fo rthe equation 3x− 1 4 = 5 Enter Y1 = 3x− 1 4 and Y2 = 5in the Y = window. Choose 6: ZStandard from the ZOOM menu. Press TRACE and take the cursor to a point near the intersection as shown. Choose 5: intersect from the CALC menu and press ENTER three times to obtain the coordinates of the point of intersection. The intersection point has coordinates (7, 5) and the solution of the equation 3x− 1 4 = 5is x= 7. Using solver Choose 0: Solver from the MATH menu. The equation must be entered as 0 = 3x− 1 4 − 5. Press ENTER or use the ‘down’ arrow key to move to the second screen. Ta ke the cursor to the position to the right of the equals sign and enter a guess, for example X = 5. This is necessary to ‘seed’ the numerical process the calculator undertakes. Press ALPHA ENTER to obtain the result shown. Exercise 1A 1 Solve each of the following equations for x: a x+ 3= 6 b x− 3= 6 c 3− x= 2 d x+ 6=− 2 e 2− x=− 3 f 2x= 4 g 3x= 5 h −2x= 7 i −3x=− 7 j 3x 4 = 5 k −3x 5 = 2 l −5x 7 =− 2
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 6 Essential Mathematical Methods Unit s1&2 2 Solve each of the following literal equations for x: a x− b= a b x+ b= a c ax = b d x a= b e ax b = c 3 Solve the following linear equations: Example 1 a y–4 = 3 b t+ 2= 7 c y+ 5= 2 d x–9 = 5 e 2a= 7 f 3a= 14 g y 8= 6 h t 3= 1 2 i 2x+ 5= 9 j 5y− 3= 12 k 3x− 7= 14 l 14 − 3y= 8 4 Example 2 a 6x− 4= 3x b x− 5= 4x+ 10 c 3x− 2= 8− 2x 5 Example 3 a 2(y+ 6) = 10 b 2y+ 6= 3(y− 4) c 2(x+ 4) = 7x+ 2 d 5(y− 3) = 2(2 y+ 4) Example 4 e x− 6= 2(x− 3) f y+ 2 3 = 4 g x 2+ x 3= 10 h x+ 4= 3 2x i 7x+ 3 2 = 9x− 8 4 Example 5 j 2(1 − 2x) 3 − 2x=− 2 5+ 4(2 − 3x) 3 k 4y− 5 2 − 2y− 1 6 = y 6 Solve the following linear literal equations for x: Example 6 a ax + b= 0 b cx + d= e c a(x+ b)= c d ax + b= cx e x a+ x b= 1 f a x+ b x= 1 g ax − b= cx − d h ax + c b = d 7 Solve each of the following for x: a 0.2x+ 6= 2.4 b 0.6(2 .8− x)= 48 .6 c 2x+ 12 7 = 6.5 d 0.5x− 4= 10 e 1 4(x− 10) = 6 f 6.4x+ 2= 3.2− 4x 1.2 Constructing linear equations As stated earlier, many problems can be solved by translating them into mathematical language and using an appropriate mathematical technique to find the solution. By representing the unknown quantity in a problem with a symbol and constructing an equation from the information, the value of the unknown can be found by solving the equation. Before constructing the equation, each symbol and what it stands for (including the units) should be stated. It is essential to remember that all the elements of the equation must be in units of the same system. Example 7 A chef uses the following rule for cooking a turkey: ‘Allow 30 minutes for each kilogram weight of turkey and then add an extra 15 minutes.’ If the chef forgot to weigh a turkey before cooking it, but knew that it had taken 3 hours to cook, calculate how much it weighed.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 7 Solution Let the weight of the turkey = xkilograms Then the time taken = (30 x+ 15) minutes ∴ 30 x+ 15 = 180 (3 hours is 180 minutes) x= 5.5 The turkey weighed 5.5 kilograms. Check: LHS = 30(5 .5) + 15 = 180 RHS = 180 Example 8 Find the area of a rectangle whose perimeter is 1.08 m, if it is 8 cm longer than it is wide. Solution Let length = lcm Then the width = (l–8)cm Perimeter = 2× length + 2× width = 2l+ 2(l− 8) = 4l− 16 cm Perimeter = 108 cm ∴ 4l− 16 = 108 4l= 124 l= 31 cm Therefore the length = 31 cm and the width = 23 cm. Check: LHS = 4(31) − 16 = 108 RHS = 108 Area = 31 × 23 = 713 cm 2 Example 9 Adam normally takes 5 hours to travel between Higett and Logett. One day he increases his speed by 4 km/h and finds the journey from Higett to Logett takes half an hour less than the normal time. Find his normal speed. Solution Let xkm/h be his normal speed. The distance from Higett to Logett isx× 5= 5xkilometres. (distance = speed × time)
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 8 Essential Mathematical Methods Unit s1&2 Adam’s new speed is ( x+ 4) km/h. Hence ( x+ 4) × 9 2= 5x 9(x+ 4) = 10 x 9x+ 36 = 10 x 36 = x His normal speed is 36 km/h. Exercise 1B 1 Fo reach of the following write an equation using the pronumeral xthen solve the equation for x: a A number plus two is equal to six. b A number multiplied by three is equal to ten. c Six is added to a number multiplied by three and the result is twenty-two. d Five is subtracted from a number multiplied by three and the result is fifteen. e Three is added to a number. If the result of this is multiplied by six, fifty-six is obtained. f Five is added to a number and the result divided by four gives twenty-three. 2 $48 is divided among three students, A,Band C.If Breceives three times as much as A and Creceives twice as much as A,h ow much does each receive? 3 The sum of two numbers is 42, and one number is twice the other. Find the two numbers. 4 A chef uses the following rule for cooking on a spit: ‘Allow 20 minutes for each kilogram Example 7 weight and then add an extra 20 minutes.’ If the chef forgot to weigh the food before cooking it but knew that it had taken 3 hours to cook, calculate how much it weighed. 5 Find the area of a rectangle whose perimeter is 4.8 m, if it is 0.5 m longer than it is wide. Example 8 6 Find three consecutive whole numbers with a sum of 150. 7 Find four consecutive odd numbers with a sum of 80. 8 Tw otanks contain equal amounts of water. They are connected by a pipe and 3000 litres of water is pumped from one tank to the other. One tank then contains 6 times as much water as the other. How many litres of water did each tank contain originally? 9 A 120-page book has plines to a page. If the number of lines were reduced by three on each page the number of pages would need to be increased by 20 to give the same amount of writing space. How many lines were there on the page originally? 10 Ar ow er travels upstream at 6 km per hour and back to the starting place at 10 km per Example 9 hour. The total journey takes 48 minutes. How far upstream did the rower go?
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 9 11 A shopkeeper buys a crate of eggs at $1.50 per dozen. He buys another crate, containing 3 dozen more than the first crate, at $2.00 per dozen. He sells them all for $2.50 a dozen and makes $15 profit. How many dozens were there in each of the crates? 12 Jess walked for 45 minutes at 3 km/h and then ran for half an hour at xkm/h. At the end of Example 9 that time she was 6 km from the starting point. Find the value of x. 13 A man travels from Ato Bat 4 km/h and from Bto Aat 6 km/h. The total journey takes 45 minutes. Find the distance travelled. 14 Ab oy is 24 years younger than his father. In 2 years time the sum of their ages will be 40. Find the present ages of father and son. 1.3 Simultaneous equations A linear equation that contains two unknowns, e.g. 2 y+ 3x= 10, does not have a single solution. Such an equation actually expresses a relationship between pairs of numbers, xand y, that satisfy the equation. If all the possible pairs of numbers ( x,y)that will satisfy the equation are represented graphically, the result is a straight line. Hence the name linear relation . y 4 3 2 1 0–1 –2 –3 –4 –3 –2 –1 123 2x – y = 4 x + 2y = –3 (1, –2) x If the graphs of two such equations are drawn on the same set of axes, and they are non-parallel, the lines will intersect at one point only. Hence there is one pair of numbers that will satisfy both equations simultaneously. Finding the intersection of two straight lines can be done graphically; however the accuracy of the solution will depend on the accuracy of the graphs. Alternatively this point of intersection may be found algebraically by solving the pair of simultaneous equations. Three techniques for solving simultaneous equations will be considered. Example 10 Solve the equations 2 x− y= 4and x+ 2y=− 3. Solution 1B ysubstitution 2x− y= 4 (1) x+ 2y=− 3 (2) First express one unknown from either equation in terms of the other unknown. From equation (2) we get x=− 3− 2y. Then substitute this expression into the other equation.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 10 Essential Mathematical Methods Unit s1&2 Equation (1) then becomes 2( −3− 2y)− y= 4 (reducing it to one equation Solving (1) −6− 4y− y= 4 in one unknown) −5y= 10 y=− 2 Substituting the value of yinto (2) x+ 2(−2) =− 3 x= 1 Check in (1) : LHS = 2(1) − (−2) = 4 RHS = 4 Note: This means that the point (1, –2) is the point of intersection of the graphs of the two linear relations. 2B yelimination 2x− y= 4 (1) x+ 2y=− 3 (2) If the coefficient of one of the unknowns in the two equations is the same, we can eliminate that unknown by subtracting one equation from the other. It may be necessary to multiply one of the equations by a constant to make the coefficients of xor ythe same for the two equations. To eliminate x,multiply equation (2) by 2 and subtract the result from equation (1). Equation (2) becomes 2 x+ 4y=− 6(2 ) Then 2 x− y= 4 (1) 2x+ 4y=− 6(2 ) Subtracting (1) − (2): −5y= 10 y=− 2 Now substitute for yin equation (1) to find x,and check as in substitution method. Using a graphics calculator Write each of the equations with yas the subject. Hence: y=− 1 2x− 3 2 and y= 2x− 4 Enter each of these in the Y = screen and GRAPH as shown in the following screens. Select 5: intersect from the CALC menu. Use the arrow keys to take the cursor to a point near the intersection. Select the two lines by pressing ENTER after each of the two first prompts. The guess has been established by the placement of the cursor. Press ENTER to obtain coordinates as shown.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 11 Exercise 1C 1 Solve each of the following pairs of simultaneous equations by the substitution method: a y= 2x+ 1 y= 3x+ 2 b y= 5x− 4 y= 3x+ 6 c y= 2− 3x y= 5x+ 10 2 Solve each of the following pairs of simultaneous equations by the elimination method: a x+ y= 6 x− y= 10 b y− x= 5 x+ y= 3 c x− 2y= 6 x+ 6y= 10 3 Solve each of the following pairs of simultaneous linear equations by either the substitution Example 10 or the elimination method: a 2x–3 y= 7 3x+ y= 5 b 2x–5 y= 10 4x+ 3y= 7 c 2m –n= 1 2n+ m = 8 d 7x–6 y= 20 3x+ 4y= 2 e 3s–t= 1 5s+ 2t= 20 f 4x–3 y= 1 4y–5 x= 2 g 15 x–4 y= 6 9x–2 y= 5 h 2p+ 5q= –3 7p–2 q= 9 i 2x–4 y= –12 2y+ 3x–2 = 0 1.4 Constructing and solving simultaneous linear equations Example 11 The sum of two numbers is 24 and their difference is 96, find the two numbers. Solution Let xand ybe the two numbers. Then x+ y= 24 (1) and x− y= 96 (2) Add equations (1) and (2). This gives 2 x= 120 and hence x= 60 Substitute in equation (1). 60 + y= 24 Hence y=− 36 The two numbers are 60 and −36. Check in (2): 60 − (−36) = 96
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 12 Essential Mathematical Methods Unit s1&2 Example 12 3kilograms of jam and 2 kilograms of butter cost $29 and 6 kilograms of jam and 3kilograms of butter cost $54. Find the cost of 1 kilogram of jam and 1 kilogram of butter. Solution Let the cost of 1 kg of jam = xdollars and the cost of 1 kg of butter = ydollars. Then 3 x+ 2y= 29 (1) and 6 x+ 3y= 54 (2) Multiply (1) by 2 : 6 x+ 4y= 58 (1 ) Subtract (1 )from (2) : −y=− 4 y= 4 Substituting in (2) : 6 x+ 3(4) = 54 6x= 42 x= 7 Jam costs $7 per kg and butter $4 per kg. Check in the original problem: 3kgofjam = $21 and 2 kg of butter = $8 Total = $29 6kgofjam = $42 and 3 kg of butter = $12 Total = $54 Exercise 1D 1 Find two numbers whose sum is 138 and whose difference is 88. 2 Find two numbers whose sum is 36 and whose difference is 9. 3 Six stools and four chairs cost $58, while five stools and two chairs cost $35. Example 11 a How much do ten stools and four chairs cost? b How much do four stools cost? c How much does one stool cost? 4 A belt and a wallet cost $42, while seven belts and four wallets cost $213. a How much do four belts and four wallets cost? b How much do three belts cost? c How much does one belt cost? Use simultaneous equations to solve the following. 5 Find a pair of numbers whose sum is 45 and whose difference is 11.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 13 6 In four years time a mother will be three times as old as her son. Four years ago she was five times as old as her son. Find their present ages. 7 A party was organised for thirty people at which they could have either a hamburger or a pizza. If there were five times as many hamburgers as pizzas calculate the number of each. 8 Tw ochildren had 110 marbles between them. After one child had lost half her marbles and the other had lost 20 they had an equal number. How many marbles did each child start with and how many did they finish with? 9 An investor received $1400 interest per annum from a sum of money, with part of it invested at 10% and the remainder at 7% simple interest. This investor found that if she interchanged the amounts she had invested she could increase her return by $90 per annum. Calculate the total amount invested. 10 A shopkeeper sold his entire stock of shirts and ties in a sale for $10 000. The shirts were Example 12 priced at 3 for $100 and the ties $20 each. If he had sold only half the shirts and two-thirds of the ties he would have received $6000. How many of each did he sell in the sale? 11 Atent manufacturer produces two models, the Outback and the Bush Walker. From earlier sales records it is known that 20 per cent more of the Outback model is sold than the Bush Walker. A profit of $200 is made on each Outback sold, but $350 is made on each Bush Walker. If during the next year a profit of $177 000 is planned, how many of each model must be sold? 12 Oz Jeans has factories in Mydney and Selbourne. At the Mydney factory fixed costs are $28 000 per month and the cost of producing each pair of jeans is $30. At the Selbourne factory, fixed costs are $35 200 per month and each pair of jeans costs $24 to produce. During the next month Oz Jeans must manufacture 6000 pairs of jeans. Calculate the production order for each factory, if the total manufacturing costs for each factory are to be the same. 13 A tea wholesaler blends together three types of tea that normally sell for $10, $11 and $12 per kilogram so as to obtain 100 kilograms of tea worth $11.20 per kilogram. If the same amounts of the two higher priced teas are used, calculate how much of each type must be used in the blend. 1.5 Solving linear inequations An inequation is a mathematical statement that contains an inequality symbol rather than an equals sign, for example 2 x+ 1< 4. To solve linear inequations, proceed exactly as for equations except that: when multiplying or dividing both sides by a negative number, the ‘direction’ of the inequality symbol is reversed .
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 14 Essential Mathematical Methods Unit s1&2 Example 13 Solve the inequation 2 x+ 1< 4. Solution 2x+ 1< 4 2x< 3 Subtract 1 from both sides. x< 3 2 Divide both sides by 2. The solution to a linear inequation is an infinite set of numbers and may be represented on a number line: –1 0 –2 1 2 32 In the number line diagram, the ‘endpoint’ of the interval is indicated with a closed circle if the point is included and an open circle if it is not. Example 14 Solve the inequation 3 − 2x≤ 4. Solution 3− 2x≤ 4 −2x≤ 1 Subtract 3 from both sides. x≥ −1 2 Divide both sides by −2. (Note that inequality symbol is reversed.) The solution can be represented on a real number line: –2 –1 01 2 12– Example 15 Solve the inequality 2x+ 3 5 > 3− 4x 3 + 2. Solution 2x+ 3 5 > 3− 4x 3 + 2 2x+ 3 5 − 3− 4x 3 > 2 Obtain the common denominator on the left-hand side: 3(2 x+ 3) 15 − 5(3 − 4x) 15 > 2 Therefore 3(2 x+ 3) − 5(3 − 4x)> 30 and 6 x+ 9− 15 + 20 x> 30 Therefore 26 x− 6> 30 and x> 36 26 ∴ x> 18 13
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 15 Exercise 1E 1 Solve each of the following inequalities for x: Examples 13, 14 a x+ 3< 4 b x− 5> 8 c 2x≥ 6 d x 3≤ 4 e −x≥ 6 f −2x< −6 g 6− 2x> 10 h −3x 4 ≤ 6 i 4x− 4≤ 2 2 Solve for xin each of the following and show the solutions on a real number line: Example 15 a 4x+ 3< 11 b 3x+ 5< x+ 3 c 1 2(x+ 1) – x> 1 d 1 6(x+ 3) ≥ 1 e 2 3(2x–5) < 2 f 3x− 1 4 − 2x+ 3 2 < −2 g 4x− 3 2 − 3x− 3 3 < 3 h 1− 7x −2 ≥ 10 i 5x− 2 3 − 2− x 3 > −1 3 a Fo rw hich real numbers xis 2 x+ 1a positive number? b Fo rw hich real numbers xis 100 − 50 xapositive number? c Fo rw hich real numbers xis 100 + 20 xapositive number? 4 In a certain country it cost $1 to send a letter weighing less than 20 g. A sheet of paper weighs 3 g. Write a suitable inequation and hence state the number of pages that can be sent for $1. 5 A student scores marks of 66 and 72 on two tests. What is the lowest mark she can obtain on a third test to have an average for the three tests greater than or equal to 75? 1.6 Using and transposing formulae An equation containing symbols that states a relationship between two or more quantities is called a formula .A= lw (Area = length × width) is an example of a formula. The value of A,called the subject of the formula, can be found by substituting in given values of land w. Example 16 Find the area of a rectangle with length ( l)1 0c m and width ( w)4 cm. Solution A= lw A= 10 × 4 A= 40 cm 2 Sometimes we wish to re-write a formula to make a different symbol the subject of the formula. This process is called transposing the formula. The techniques for transposing formulae include those used for solving linear equations detailed in section 1.1.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 16 Essential Mathematical Methods Unit s1&2 Example 17 Transpose the formula v= u+ at,to make athe subject. Solution v= u+ at v− u= at Subtract ufrom both sides. v− u t = a Divide both sides by t. If we wish to evaluate an unknown that is not the subject of the formula, we can either substitute the values for the given variables and solve the resulting equation or we can first transpose the formula and then substitute the given values. Example 18 Evaluate pif 2( p+ q)– r= z,and q= 2, r=− 3and z= 11. Solution 1 Substituting then solving 2(p+ 2) − (−3) = 11 2p+ 4+ 3= 11 2p= 4 p= 2 2T ransposing then substituting 2(p+ q)− r= z 2(p+ q)= z+ r p+ q= z+ r 2 p= z+ r 2 − q ∴ p= 11 + (−3) 2 − 2 p= 2 Example 19 A path xmetres wide surrounds a rectangular lawn. The lawn is lmetres long and bmetres wide. The total area of the path is Am2. a Find Ain terms of l,band x. b Find bin terms of l,Aand x.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 17 Solution a b + 2 x m b l l + 2 x m The area of the path A= (b+ 2x)(l+ 2x)− bl = bl + 2xl + 2xb + 4x2− bl A= 2xl + 2xb + 4x2 b A− (2xl + 4x2)= 2xb Therefore b= A− (2xl + 4x2) 2x Example 20 Fo reach of the following make cthe subject of the formula. a e= √3c− 7a b 1 a− 1 b= 1 c− 2 Solution a e= √3c− 7a Square both sides of the equation. e2= 3c− 7a Therefore 3 c= e2+ 7a and c= e2+ 7a 3 b 1 a− 1 b= 1 c− 2 Establish common denominator on left-hand side of the equation. b− a ab = 1 c− 2 Ta ke the reciprocal of both sides. ab b− a= c− 2 Therefore c= ab b− a+ 2
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 18 Essential Mathematical Methods Unit s1&2 Exercise 1F 1 Fo reach of the following find the value of the letter in parentheses: a c= ab ,a= 6, b= 3( c) b r= p+ q,p= 12, q=− 3( r) c c= ab ,a= 6, c= 18 ( b) d r= p+ q,p= 15, r=− 3( q) e c= √a,a= 9( c) f c= √a,c= 9( a) g p= u v,u= 10, v= 2( p) h p= u v,p= 10, v= 2( u) 2 Fo reach of the following construct a formula using the given symbols: a S,the sum of three numbers a,band c b P,the product of two numbers xand y c the cost, $ C,o ffi ve CDs which each cost $ p d the total cost, $ T,of dchairs which cost $ peach and ctables which cost $ qeach e the time, T,inminutes, of a train journey that takes ahours and bminutes 3 Find the values of the following: a E= IR,w hen I= 5and R= 3 b C = pd ,w hen p= 3.14 and d= 10 c P = RT V ,w hen R= 60, T= 150 and V= 9 d I= E R,w hen E= 240 and R= 20 e A= rl,w hen = 3.14, r= 5and l= 20 f S= 90(2 n–4), when n= 6 4 Fo reach of the following make the symbol indicated the subject of the formula: Example 17 a PV = c;V b F= ma ;a c I= Prt ;P d w= H + Cr ;r e S= P(1 + rt);t f V = 2R R− r;r 5 Find the value of the unknown symbol in each of the following: Example 18 a D = T+ 2 P ,w hen D = 10, P= 5 b A= 1 2bh ,when A= 40, h= 10 c V = 1 3r2h,w hen = 3.14, V= 100, r= 5 d A= 1 2h(a+ b),when A= 50, h= 5, a= 10 b m b m b m b m b m b m a m a m a m a m 6 The diagram represents the brick wall of ad welling with three windows. Each of the windows is hm high and wm wide. Other dimensions are as shown. a Find the length of the wall. b Find the height of the wall. c Find the total area of the three windows. d Find the total area of brickwork.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Chapte r1—Revie wing Linear Equations 19 7 A lampshade has a metal frame consisting of two circular hoops of radii pcm and qcm joined by four straight struts of length hcm. The total length of metal is Tcm. ai Find an expression for Tin terms of p,qand h. ii Find Twhen p= 20, q= 24 and h= 28. b The area of the material covering the frame is Acm 2, where A= h(p+ q). Find an expression for pin terms of A,h,q. 8 Find the value of the ‘unknown’ symbol in each of the following: a P = T− M D ,P= 6, T= 8, M = 4 b H = a 3+ a b,H = 5and a= 6 c a= 90(2 n− 4) n ,a= 6 d R= r a+ r 3,a= 2and R= 4 9 Right-angled triangles XYZ and ABC Example 19 are similar. XY AB = YZ BC = ZX CA = k IfAB = ccm, AC = bcm, find: A B C Y Z c cm X b cm a the area, D cm 2,o fthe shaded region in terms of c,band k b kin terms of D,band c c the value of kifD = 2, b= 3and c= 4 10 Tw orectangles each with dimensions ccm × bcm are used to form a cross as shown. The arms of the cross are all of equal length. b cm b c c c cm b cm a Find the perimeter, Pcm, of the cross in terms of band c. b Find the area, Acm 2,o fthe cross in terms of cand b. c Find bin terms of Aand c. 11 Fo reach of the following make the symbol in brackets the subject of the formula: a a= √a+ 2b (b) Example 20 b a+ x a− x= b− y b+ y (x) c px = 3q− r2 (r) d x y= 1− v2 u2 (v)
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Review 20 Essential Mathematical Methods Unit s1&2 Chapter summary A linear equation is one in which the variable has exponent 1. It is often helpful to look at how the equation has been constructed so that the steps necessary to ‘undo’ the equation can be identified. It is most important that the steps taken to solve the equation are done in the correct order. An inequation is a mathematical statement that contains an inequality symbol rather than an equals sign, for example 2 x+ 1< 4. To solve linear inequations, proceed exactly as for equations except that, when multiplying or dividing both sides by a negative number, the ‘direction’ of the inequality symbol is reversed. An equation containing symbols that states a relationship between two or more quantities is called a formula. A= lw (area = length × width) is an example of a formula. Ais called the subject of the formula. If we wish to evaluate an unknown that is not the subject of the formula, we can either substitute the values for the given variables and solve the resulting equation or we can first transpose the formula and then substitute the given values. An equation for the variable xin which the coefficients of x,including the constants, are pronumerals is known as a literal equation. Multiple-choice questions 1 The solution of the linear equation 3 x–7 = 11 is A 4 3 B 11 3 C −3 4 D 6 E −6 2 If x 3+ 1 3= 2then x= A 1 3 B 2 3 C 7 3 D 5 E 7 3 The solution of the equation x− 8= 3x− 16 is A x= −8 3 B x= 11 3 C x= 4 D x= 2 E x=− 2 4 In algebraic form, 7 is 11 times 2 less than xcan be written as A 7= 11( x− 2) B 11 x− 2= 7 C 7= 11( x+ 2) D 11 x− 2= 7 E 7 11 − 2= x 5 The solution of the simultaneous equations 2 x− y= 10 and x+ 2y= 0is A x=− 2and y= 3 B x= 2and y=− 3 C x= 4and y=− 2 D x= 6and y= 2 E x= 1and y=− 8
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Review Chapte r1—Revie wing Linear Equations 21 6 Ibought xCDs for $ aand yDVDs for $ b.The average price paid, in dollars, is A x+ y a+ b B a+ b y+ x C xa + yb y+ x D y+ x xa + yb E a x+ b y 7 The solution of the equation x+ 1 4 − 2x− 1 6 = xis A x= 8 5 B x= 5 13 C x= 1 D x=− 1 5 E x=− 1 8 The values of zthat satisfy the inequation 72 + 15 z 3 > 4are A z> 4 B z> −4 C z=− 4 D z< 4 E z< −4 9 IfA= hw + k w then A w= k A− h B w= ht + k A C w= A− 2k 2h D w= 3Ah 2 − k E w= 2 3h(A+ k) 10 Bronwyn walks one lap of an oval at 2.5 km/h and then jogs another 8 laps of the oval at 5km/h. If it takes her 30 minutes in total, how long in metres is each lap? A 200 m B 250 m C 300 m D 350 m E 400 m Short-answer questions (technology-free) 1 Solve each of the following equations for x: a 2x+ 6= 8 b 3− 2x= 6 c 2x+ 5= 3− x d 3− x 5 = 6 e x 3= 4 f 13 x 4 − 1= 10 g 3(2 x+ 1) = 5(1 − 2x) h 3x+ 2 5 + 3− x 2 = 5 2 Solve each of the following for t: a a− t= b b at + b c = d c a(t− c)= d d a− t b− t= c e at + b ct − b= 1 f 1 at + c= d 3 Solve each of the following inequalities for x: a 2− 3x> 0 b 3− 2x 5 ≥ 60 c 3(58 x− 24) + 10 < 70 d 3− 2x 5 − x− 7 6 ≤ 2 4 Make xthe subject of the formula z= 1 2x− 3t.F ind xwhen z= 4and t=− 3.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Review 22 Essential Mathematical Methods Unit s1&2 5 A number dis equal to the square of a number eplus twice a number f. a Find a formula for din terms of eand f. b Make fthe subject of the formula. c Find fwhen d= 10 and e= 3. 6 Solve b− cx a + a− cx b + 2= 0for x. 7 Solve a x+ a+ b x− b= a+ b x+ cfor x. Extended-response questions 1 The formula for converting degrees Celsius, C,tod egrees Fahrenheit, F,is F= 9 5C+ 32. a Convert 30 ◦Ft oCelsius. b Convert 30 ◦Ct oF ahrenheit. c Ifx◦C= x◦Ffind x. d If ( x+ 10) ◦C= x◦Ffind x. e If 2 x◦C= x◦Ffind the value of x. f Ifk◦F= (−3k)◦Cfind k. 2 Fo ra spherical mirror of radius rcm, 1 v+ 1 u= 2 r,w here ucm is the distance from the mirror to the object and vcm is the distance from the mirror to the image. The magnification is given by m = v− r r− u. a Find rin terms of vand ufrom the first formula. b Find m in terms of vand uonly. 3 The diagram shows a section of wire mesh wmetres in width and lmetres in length. w m l m a Find an expression in terms of wand lfor the total length of wire required for the mesh. bi Ifw= 3l,find an expression in terms of wfor the total length of wire required. ii If the length of wire used is 100 m, find the value of wand the value of l. c The total length of wire, Lm, required for another rectangular mesh of dimensions xmby ym, is given by the formula L= 6x+ 8y. i Find yin terms of xand L. ii Find yifL= 200 and x= 4. d A third type of mesh can also be used to cover a rectangular region with dimensions xm and ym. In this case, the type of mesh introduced in part c requires 100 m of wire and so 6 x+ 8y= 100. This third type of mesh requires 80 m and this gives the equation 3x+ 2y= 40. Find the values of xand y.
P1: FXS/ABE P2: FXS 0521609976c01.xml CUAT006-EVANS August 3, 2005 13:19 Review Chapte r1—Revie wing Linear Equations 23 4 To m leaves town Aand travels at a constant speed of ukm/h towards town B.A tthe same time Julie leaves town Band travels towards town Aat a constant speed of vkm/h. Town Bis dkm from town A. a How far has each travelled after thours? b By considering that the sum of their distances travelled must be dkm when they meet, find: i the time it takes for them to meet ii their distance from town Awhen they meet. c Ifu= 30, v= 50 and d= 100, find the time it takes for them to meet and their distance from town A. 5 Xiu travels from town Ato town Bat ukm/h and then returns at vkm/h. Town Aisdkm from town B. a Find the average speed at which Xiu travels for the complete journey, in terms of uand v. Remember, average speed = total distance travelled total time taken . b If it takes Thours to travel from Ato B,find the time taken: i for the return trip from Bto A,inter ms of T,uand v ii the entire trip in terms of T,uand v. 6 A man on a bicycle rides one-third of the way from town Ato town Bat a speed akm/h and the remainder of the way at 2 bkm/h. a If the distance between the two towns is 9 km find the time taken to ride from Ato B. If the man had travelled at a uniform rate of 3 ckm/h, he could have ridden from Ato B and back again in the same time. b Show that 2 c= 1 a+ 1 b. ci Make cthe subject of this formula. ii Find c,w hen a= 10 and b= 20. 7 A man walks 70 km. He walks xkm at 8 km/h and ykm at 10 km/h. a Find the length of time he was walking at 8 km/h in terms of x,and the length of time he wa sw alking at 10 km/h in terms of y. b Find his average speed in terms of xand y. c If the man walks at 10 km/h for the time he was walking at 8 km/h and at 8 km/h for the time he was walking at 10 km/h, he walks 72 km. Find xand y.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 CHAPTER 2 Linear Relations Objectives To calculate the gradient of a straight line. To interpret and use the general equation of a straight line y=mx +c. To solve simultaneous linear equations graphically. To calculate the gradient of a straight line as a tangent . To calculate the product of the gradients of two perpendicular straight lines. To find the distance between two points . To find the midpoint of a straight line. To calculate the angle between two intersecting straight lines. To apply aknowledge of linear relations to solving problems. A relation is defined as a set of ordered pairs in the form ( x,y). Ar ule relating the x-value to the y-value of each ordered pair sometimes exists, such as y= 2x+ 1, and this is a more convenient way of describing the relation. A relation may also be represented graphically on a set of axes. If the resulting graph is a straight line, then the relation is called a linear relation . 2.1 Th e gradient of a straight line run rise Through any two points it is only possible to draw a single straight line. Therefore a straight line is defined by any two points on the line. From previous work, you should be familiar with the concept of the gradient or slope of a line. The symbol used for gradient is m.This may be defined as: Gradient = rise run 24
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 25 Hence given any two points on the line, ( x1,y1)and ( x2,y2), the gradient of the line can be found. y x 0 A (x1, y1) B (x2, y2) run = x2 – x1 rise = y2 – y1 Gradient m = rise run = y2− y1 x2− x1 Example 1 Find the gradient of the given line. y x 2 1 –2 –1 0 Solution Let ( x2,y2)= (0,2) and let ( x1,y1)= (−2,0) Gradient m = y2− y1 x2− x1 = 2− 0 0− (−2) = 2 2 = 1 Example 2 Find the gradient of the given line. 0 y x 1 12 2 3 Solution Let ( x1,y1)= (0,3) and ( x2,y2)= (2,0) m = y2− y1 x2− x1 = 0− 3 2− 0 =− 3 2
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 26 Essential Mathematical Methods Unit s1&2 It should be noted that the gradient of a line that slopes upwards from left to right is positive , as illustrated in Example 1, and the gradient of a line that slopes downwards from left to right isnegative ,asillustrated in Example 2. The gradient of a horizontal (parallel to the x-axis) line is zero, since y2− y1= 0. The gradient of a vertical (parallel to the y-axis) line is undefined, since x2− x1= 0. Example 3 Find the gradient of the line that passes through the points (1, 6) and ( −3, 7). Solution The gradient = y2− y1 x2− x1= 7− 6 −3− 1= 1 −4=− 1 4 Exercise 2A 1 Calculate the gradient of each of the following lines: Examples 1, 2 a x y –2 0 1 2 3 4 1 –1 b x 0 2 4 6 123 y c x 0–1 1 2 123 4 y d x 0 1 2 1 –1 2 3 4 y e x 1 2 123 3 y 0 f x 1 2 –1 –2 –3 1 3 y 0 g x y 0 (6, 10) (–2, 0) h x y 0 (0, 8) (3, 2) i 0 y x 4 5
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 27 j y x 0 –3 4 k 0 x y 3 2 Sketch a graph of a line with gradient 1. 3 Sketch a graph of a line with gradient 0 which passes through the point (1, 6). 4 Fo reach of the following find the gradient of the line that passes through the two points Example 3 with the given coordinates: a (6, 3) (2, 4) b (−3, 4) (1, −6) c (6, 7) (11, −3) d (5, 8) (6, 0) e (6, 0) ( −6, 0) f (0, −6) ( −6, 0) g (3, 9) (4, 16) h (5, 25) (6, 36) i (−5, 25) ( −8, 64) j (1, 1) (10, 100) k (1, 1) (10, 1000) l (5, 125) (4, 64) 5 a Find the gradient of the straight line that passes through the points with coordinates (5a,2 a)and (3 a,6 a). b Find the gradient of the straight line that passes through the points with coordinates (5a,2 a)and (5 b,2 b). 6 a A line has gradient 6 and passes through the points with coordinates ( −1, 6) and (7, a). Find the value of a. b A line has gradient −6and passes through the points with coordinates (1, 6) and (b,7). Find the value of b. 7 a Find the equation of the line that is parallel to the y-axis and passes through the point with coordinates (4, 7). b Find the equation of the line that is parallel to the x-axis and passes through the point with coordinates ( −4, 11). 2.2 Th e general equation of a straight line The general equation of a straight line is y= mx + c,w here m is the gradient of the line. This form, expressing the relation in terms of y,iscalled the gradient form . Let x= 0, then y= m(0) + c and thus y= c i.e. the y-axis intercept is equal to c.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 28 Essential Mathematical Methods Unit s1&2 Example 4 Find the gradient and y-axis intercept of the graph of y= 3x− 4. Solution The value of m is 3 and the value of cis−4. Therefore the gradient of the above line is 3 and the y-axis intercept is −4. If the rule of a straight line is given, the graph can be sketched using the gradient and the y-axis intercept. Example 5 Sketch the graph of y= 3x− 1. x y 2 3 1 1 –1 0 1 –1 Solution Gradient = 3,i.e.rise run = 3 1 y-axis intercept =− 1 Plot the point (0, −1), the y-axis intercept. From there move across 1 (run) and up 3 (rise) to plot the point (1, 2). If the equation for the straight line is not written in gradient form ,touse the above method for sketching a graph, the equation must first be transposed into gradient form . Example 6 Sketch the graph of 3 y+ 6x= 9. x y 2 1 1 0 –1 –1 1 2 3 Solution First rearrange the equation into gradient form: 3y+ 6x= 9 3y= 9− 6x (Subtract 6 xfrom both sides.) y= 9− 6x 3 (Divide both sides by 3.) y= 3− 2x i.e. y=− 2x+ 3 Therefore m =− 2 and c= 3
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 29 Parallel lines 21 x y 0 3 4 y = 2x + 3 y = 2x – 4 –2 –4 –3 12 –1 –2 If the value of m is the same for two rules, then the lines are parallel. Fo re xample, consider the lines with the following rules: y= 2x+ 3 y= 2x− 4 Exercise 2B 1 Sketch the graphs of each of the following: Example 4 a y= x+ 2 b y=− x+ 2 c y= 2x+ 1 d y=− 2x+ 1 2 Sketch the graphs of each of the following using the gradient form, y= mx + c: Example 5 a x+ y= 1 b x− y= 1 c y− x= 1 d −x− y= 1 3 Sketch the graphs of each of the following using y= mx + c: a y= x+ 3 b y= 3x+ 1 c y= 4− 1 2x d y= 3x− 2 e 4y+ 2x= 12 f 3x+ 6y= 12 g 4y− 6x= 24 h 8x− 3y= 24 4 Fo rw hich of the following pairs of equations are the corresponding lines parallel to each other? Sketch graphs to show the non-parallel lines. a 2y= 6x+ 4; y= 3x+ 4 b x= 4− y;2 x+ 2y= 6 c 3y− 2x= 12; y+ 1 3= 2 3x d 4y− 3x= 43 y= 4x− 3 5 Fo rw hich of the following do the lines pass through the origin? a y+ x= 1 b y+ 2x= 2(x+ 1) c x+ y= 0 d x− y= 1 6 Give the gradient for each of the lines in Question 5. 2.3 Finding the equation of a straight line The equation of a straight line may be found if the gradient and y-axis intercept are known. Example 7 Find the equation if m =− 3and c= 10. Solution Equation is y=− 3x+ 10.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 30 Essential Mathematical Methods Unit s1&2 Example 8 Find the equation of the straight line with gradient −3w hich passes through the point with coordinates ( −5, 10). Solution The general equation of lines with gradient −3is y=− 3x+ c. If a point on the line is given the value of ccan be determined. When x=− 5, y= 10 Thus 10 =− 3×− 5+ c. Solving for c:10 − 15 = c and therefore c=− 5. The equation of the line is y=− 3x− 5. Example 9 Find the equation of the straight line with y-axis intercept −3w hich passes through the point with coordinates (1, 10). Solution The general equation of lines with y-axis intercept −3is y= mx − 3. The line passes through the points with coordinates (0, −3) and (1, 10). Therefore the gradient m = 10 −− 3 1− 0 = 13. The equation is y= 13 x− 3. In general the equation of a straight line can be determined by two ‘independent pieces of information’. Two cases are considered below. Case 1 Given any two points A(x1,y1)and B(x2,y2) Using these two points, the gradient of the line AB can be determined: x y 0 A(x1, y1) P(x, y) B(x2, y2) m = y2− y1 x2− x1 Using the general point P(x,y), also on the line, m = y− y1 x− x1 Therefore the equation of the line is y− y1= m(x− x1)w here m = y2− y1 x2− x1
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 31 Example 10 Find the equation of the straight line passing through the points (1, −2) and (3, 2). 1 0 –2 –4 2 2 (1, –2) 34 (3, 2) P(x, y) y x Solution m = y2− y1 x2− x1 = 2− (−2) 3− 1 = 4 2 = 2 ∴ 2= y− (−2) x− 1 2x− 2= y+ 2 ∴ y= 2x− 4 Case 2 Given the gradient m and one other point, A(x1,y1) As the gradient m is already known, the rule can be found using y− y1= m(x− x1) Example 11 Find the equation of the line that passes through the point (3, 2) and has a gradient of −2. 0 123 (3, 2) 4 1 2 8 y x Solution y− 2=− 2(x− 3) y− 2=− 2x+ 6 y=− 2x+ 8 y=− 2x+ 8i sthe equation which could also be expressed as y+ 2x− 8= 0 The equation of a straight line can also be found from the graph by reading off two points and using them to find the equation as outlined above.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 32 Essential Mathematical Methods Unit s1&2 Example 12 Find the equation of the line shown in the graph. 0 12 4 3 2 1 A B y x Solution From the graph it can be seen that the y-axis intercept is (0, 4), i.e. c= 4. As the coordinates of Aand Bare (0, 4) and (2, 0): Gradient m = y2− y1 x2− x1 = 0− 4 2− 0 =− 4 2 =− 2 The equation of the line is y=− 2x+ 4. V ertical and horizontal lines Ifm = 0, then the line is horizontal and the equation is simply y= c,w here cis the y-axis intercept. If the line is vertical ,the gradient is undefined and its rule is given as x= a, where ais the x-axis intercept. Note that the equation of a vertical line is not in the form y= mx + c. Exercise 2C 1 a Find the equation of the straight line with gradient 3 and y-axis intercept 5. b Find the equation of the straight line with gradient −4and y-axis intercept 6. c Find the equation of the straight line with gradient 3 and y-axis intercept −4. Example 7 2 a Find the equation of the straight line with gradient 3 and which passes through the Example 8 point with coordinates (6, 7). b Find the equation of the straight line with gradient −2and which passes through the point with coordinates (1, 7).
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 33 3 Fo rthe straight line with y-axis intercept 6 and passing through the point with coordinates Example 9 (1, 8) find: a the gradient b the equation 4 Fo rthe straight line with y-axis intercept 6 and passing through the point with coordinates (1, 4) find: a the gradient b the equation 5 Find the equation of the straight line that passes through the point with (1, 6) and has gradient: a 2 b −2 6 Find the equation of each of the following lines: ab c –2 –1 0 12 4 y x 3 2 1 3 1 1 0 –1 –2 –3 y x 2 2 1 0 12 –1 –2 –1 –2 y x de f 2 1 0 12 –1 –2 y x 3 –1 2 3 4 1 0 12 –1 –2 y x 2 3 4 1 0 12 –1 –2 y x 7 Write another equation that would give a parallel line for each of those shown in Question 6. Check your answers by sketching graphs. 8 Find the equations of the following straight lines. (Hint: It may help to sketch the graphs.) a Gradient 3 4,passing through ( −6, 5) b Gradient −1 2,passing through (4, −3) c Gradient 0, passing through (0, 3) d Gradient 0, passing through (0, −3) 9 Write, in the form y= mx + c,the equations of lines which have the given gradient and pass through the given point: a 11 3,(0, 3) b −1 2,(0, 3) c −0.7, (1, 6) d 11 2,(4, 3) e −3 4,(4, 3) f −1, (0, 0) 10 Find equations defining the lines which pass through the following pairs of points: Example 10 a (0, 4), (6, 0) b (−3, 0), (0, −6) c (0, 4), (4, 2) d (2, 6), (5, 3)
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 34 Essential Mathematical Methods Unit s1&2 11 Find the equations, in the form y= mx + c,o fthe lines which pass through the following pairs of points: a (0, 4), (3, 6) b (1, 0), (4, 2) c (−3, 0), (3, 3) d (−2, 3), (4, 0) e (−1.5, 2), (4.5, 8) f (−3, 1.75), (4.5, −2) 12 Do the points P(1, −3), Q(2, 1) and R 21 2,3 lie on the same straight line? 13 Find the equations defining each of the three sides of the triangle ABC for which the coordinates of the vertices are A(−2, −1), B(4, 3) and C(6, 0). 2.4 Equation of a straight line in intercept form and sketching graphs Often we encounter a linear relation that is not expressed in the form y= mx + c.An alternative standard notation is ax + by = c This is sometimes referred to as the intercept form . While it is necessary to transpose the equation into gradient form if you wish to find the gradient, it is often convenient to work with linear relations in the intercept form. Sketching graphs in intercept form A convenient way to sketch graphs of straight lines is to plot the two axes intercepts. Example 15 Sketch the graph of 2 x+ 4y= 10. Solution x-axis intercept ( y= 0): 2 x+ 4(0) = 10 x= 5 y-axis intercept ( x= 0): 2(0) + 4y= 10 y= 2.5 2.5 5 14 3 2 y x When finding the equation of a straight line, it is also sometimes more convenient to express it in intercept form. Example 16 Find the equation, in intercept form, of the line passing through the points A(2, 5) and B(6, 8). Solution m = 8− 5 6− 2= 3 4
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 35 Therefore, using the gradient and the point A(2, 5), we have the equation: y− 5= 3 4(x− 2) 4(y− 5) = 3(x− 2) 4y− 20 = 3x− 6 4y− 3x= 14 −3x+ 4y= 14 Straight lines may be sketched by finding the axes intercepts even when the equation is given in the form y= mx + c. Example 17 0 y 3 –6 x Sketch the graph of y= 2x− 6, by first finding the intercepts. Solution When x= 0, y=− 6. Hence the y-axis intercept is −6. When y= 0, 2 x− 6= 0 ∴ 2x= 6 and x= 3 The x-axis intercept is 3. Exercise 2D 1 Fo reach of the following give the coordinates of the axes intercepts: Example 15 a x+ y= 4 b x− y= 4 c −x− y= 6 d y− x= 8 2 Fo reach of the following find the equation of the straight line graph passing through the Example 16 points Aand B: a A(0, 6) and B(3, 0) b A(0, −2) and B(4, 0) c A(2, 2) and B(6, 6) d A(2, 2) and B(−6, 6) 3 Fo reach of the following sketch the graph by first finding the axes intercepts: Example 17 a y= x− 1 b y= x+ 2 c y= 2x− 4 4 Sketch the graphs of each of the following linear relations: a 2x− 3y= 12 b x− 4y= 8 c −3x+ 4y= 24 d −5x+ 2y= 20 e 4x− 3y= 15 f 7x− 2y= 15
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 36 Essential Mathematical Methods Unit s1&2 5 Find the equations of the straight lines passing through the following pairs of points. (Express your answer in intercept form.) a (−1, 4), (2, 3) b (0, 4), (5, −3) c (3, −2), (4, −4) d (5, −2), (8, 9) 6 Transpose from the intercept form to the gradient form and hence state the gradient of each of the following linear relations: a 2x− y= 9 b 3x+ 4y= 10 c −x− 3y= 6 d 5x− 2y= 4 7 Sketch the graphs of each of the following by first determining the axes intercepts: a y= 2x− 10 b y= 3x− 9 c y= 5x+ 10 d y=− 2x+ 10 8 A straight line has equation y= 3x− 4. The points with coordinates (0, a), ( b,0), (1, d) and ( e,10) lie on the line. Find the values of a,b,dand e. 2.5 Linear models In many practical situations a linear function can be used. Example 18 Austcom’s rates for local calls from private telephones consist of a quarterly rental fee of $40 plus 25c for every call. Construct a cost function that describes the quarterly telephone bill. 100 80 60 40 20 50 100 150 200 n C Solution Let C= cost ($) of quarterly telephone bill n= number of calls then: C= 0.25 n+ 40 As the number of calls is counted in whole numbers only, the domain of this function is N∪{0}. Draw the graph of the function C = 0.25 n+ 40 ,n∈ N ∪{ 0} Note: The graph should be a series of discrete points rather than a continuous line because n∈N∪{0}.W ith the scale used it is not practical to show it correctly. Example 19 The tyres on a racing car had lost 3 mm of tread after completing 250 km of a race and 4 mm of tread after completing 1000 km. Assuming that the loss of tread was proportional to the distance covered, find the total loss of tread, dmm, after skm from the start of the race. What would be the tread loss by the end of a 2000 km race?
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 37 Solution d (mm) 4 3 0 250 s increase d increase 1000 s (km) Gradient m = dincrease sincrease = 4− 3 1000 − 250 = 1 750 ∴ d= 1 750 s+ c When s= 250, d= 3 ∴ 3= 1 3+ c ∴ c= 22 3 Total loss of tread after skm ,d= 1 750 s+ 22 3 When s= 2000, d= 1 750 × 2000 + 22 3 = 22 3+ 22 3 = 51 3 ∴the loss of tread at the end of a 2000 km race is 5 1 3mm. An important linear relation is the relation between distance travelled and time taken when an object is travelling with constant speed. If a car travels at 40 km/h, the relationship between distance travelled ( skilometres) and time taken ( thours) is s= 40 t,t≥ 0. The graph of s against tis a straight line graph through the origin. The gradient of the graph is 40. Example 20 A car starts from point Aon a highway 10 kilometres past the Wangaratta post office. The car travels at an average speed of 90 km/h towards picnic stop B,w hich is 120 kilometres further on from A.Let thours be the time after the car leaves point A. a Find an expression for the distance d1of the car from the post office at time thours. b Find an expression for the distance d2of the car from point Bat time thours. c On separate sets of axes sketch the graphs of d1against tand d2against tand state the gradient of each graph. Solution a At time tthe distance of the car from the post office is 10 + 90 tkilometres. b At time tthe distance of the car from Bis 120 − 90 tkilometres.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 38 Essential Mathematical Methods Unit s1&2 c O 10 d1 , 130 43 t 120 d2 O t 43 Gradient = 90 Gradient =− 90 Exercise 2E 1 a A train moves at 50 km/h in a straight line away from town. Give a rule for the distance, dkm, from the town at time thours after leaving the town. b A train has stopped at a siding 5 km from the town and then moves at 40 km/h in a straight line away from the siding. Give a rule for the distance, dkm, from the town at time thours after leaving the siding. 2 a An initially empty container is being filled with water at a rate of 5 litres per minute. Give a rule for the volume, Vlitres, of water in the container at time tminutes after the filling of the container starts. b A container contains 10 litres of water. Water is then poured in at a rate of 5 litres per minute. Give a rule for the volume, Vlitres, of water in the container at time tminutes after the pouring starts. 3 The weekly wage, $ w,o fav acuum cleaner salesperson consists of a fixed sum of $350 Example 18 plus $20 for each cleaner sold. If ncleaners are sold per week, construct a rule that describes the weekly wage of the salesperson. 4 The reservoir feeding an intravenous drip contains 500 mL of a saline solution. The drip releases the solution into a patient at the rate of 2.5 mL/minute. a Construct a rule which relates v,the amount of solution left in the reservoir, to time, tminutes. b State the possible values of tand v. c Sketch the graph of the relation. 5 The cost ($ C)o fhiring a taxi consists of two elements, a fixed flagfall and an amount that Example 18 varies with the number ( n)o fkilometres travelled. If the flagfall is $2.60 and the cost per kilometre is $1.50, determine a rule which gives Cin terms of n. 6 A car rental company charges $85, plus an additional amount of 24c per kilometre. a Write a rule to determine the total charge $ Cfor hiring a car and travelling xkilometres. b What would be the cost to travel 250 kilometres? 7 Tw o t owns Aand Bare 200 km apart. A man leaves town Aand drives at a speed of Example 20 5km/h towards town B.F ind the distance of the man from town Bat time thours after leaving town A.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 39 8 The following table shows the extension of a spring when weights are attached to it. x,extension (cm) 0 123456 w,w eight (g) 50 50.2 50.4 50.6 50.8 51.0 51.2 a Sketch a graph to show the relationship between xand w. b Write a rule that describes the graph. c What will be the extension if w= 52.5 g? 9 A printing firm charges $35 for printing 600 sheets of headed notepaper and $47 for printing 800 sheets. a Find a formula, assuming the relationship is linear, for the charge, $ C,inter ms of number of sheets printed, n. b How much would they charge for printing 1000 sheets? 10 An electronic bankteller registered $775 after it had counted 120 notes and $975 after it had counted 160 notes. a Find a formula for the sum registered ($ C)interms of the number of notes ( n)counted. b Wa sthere a sum already on the register when counting began? c If so, how much? 2.6 Problems involving simultaneous linear models Example 21 There are two possible methods for paying gas bills. Method A:Afi xedcharge of $25 per quarter + 50c per unit of gas used Method B:Afi xedcharge of $50 per quarter + 25c per unit of gas used Determine the number of units which must be used before method Bbecomes cheaper than method A. 0255075 100 125 150 C ($) x (units) C1 = 0.5 x + 25 C2 = 0.25 x + 50 100 25 50 Solution Let C1= charge in $ using method A C2= charge in $ using method B x= number of units of gas used Now C1= 25 + 0.5x C2= 50 + 0.25 x It can be seen from the graph that if the number of units exceeds 100 method B is cheaper.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 40 Essential Mathematical Methods Unit s1&2 The solution can be obtained by solving simultaneous linear equations: C1= C2 25 + 0.5x= 50 + 0.25 x 0.25 x= 25 x= 100 Example 22 Robyn and Cheryl race over 100 metres. Robyn runs so that it takes aseconds to run 1 metre and Cheryl runs so that it takes bseconds to run 1 metre. Cheryl wins the race by 1 second. The next day they again race over 100 metres but Cheryl gives Roby na5 metre start so that Robyn runs 95 metres. Cheryl wins this race by 0.4 second. Find the values of aand band the speed at which Robyn runs. Solution Fo rthe first race: time for Robyn − time for Cheryl = 1s 100 a− 100 b= 1 (1) Fo rthe second race: time for Robyn − time for Cheryl = 0.4 s 95 a− 100 b= 0.4 (2) Subtract (2) from (1). Hence 5 a= 0.6 and a= 0.12. Substitute in (1) to find b= 0.11. Robyn’s speed = 1 0.12 = 25 3 metres per second. Exercise 2F 1 Tw obicycle hire companies have different charges. Company A charges $ C,according to the rule C= 10 t+ 20, where tis the time in hours for which a bicycle is hired. Company B charges $ C,according to the rule C= 8t+ 30. a Sketch each of the graphs on the same set of axes. b Find the time, t,for which the charge of both companies is the same. 2 The distances, dAkm and dBkm, of cyclists A and B travelling along a straight road from a town hall step are given respectively by dA= 10 t+ 15 and dB= 20 t+ 5, where tis the time in hours after 1.00 pm. a Sketch each of the graphs on the one set of axes. b Find the time in hours at which the two cyclists are at the same distance from the town hall step. 3 A helicopter can be hired for $210 per day plus a distance charge of $1.60 per km or, alternatively, at a fixed charge of $330 per day for an unlimited distance.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 41 a Fo reach of the methods of hiring, find an expression for cost, $ C,inter ms of xkm, the distance travelled. b On one set of axes, draw the graph of cost versus distance travelled for each of the methods. c Determine for what distances the fixed-charge method is cheaper. Example 21 4 Three power boats in a 500 km handicap race leave at 5 hourly intervals. Boat A leaves first and has a speed for the race of 20 km/h. Boat B leaves 5 hours later and travels at an average speed of 25 km/h. Boat C leaves last, 5 hours after B, and completes the race at a speed of 40 km/h. a Draw a graph of each boat’s journey on the same set of axes. b Use your graphs to find the winner of the race. c Check your answer algebraically. d Write a short description of what happened to each boat in the race. 5 If the line OT has the equation y=− 3 4xand the line HT has the equation y= 3 2x− 12, determine the point over which both craft would pass. 10 –10 O T Trawler H Hovercraft y x Scale: 1 unit = 1 nautical mile 6 A school wishes to take some of its students on an excursion. If they travel by tram it will cost the school $2.80 per student. Alternatively, the school can hire a bus at a cost of $54 for the day plus a charge of $1 per student. a Fo reach mode of transport, write an expression for the cost ($ C)o ftransport in terms of the number of students ( x). b On one set of axes, draw the graph of cost, $ C,v ersus number of students, x,for each mode of transport. c Determine for how many students it will be more economical to hire the bus. Example 21 7 Anne and Maureen live in towns that are 57 km apart. Anne sets out at 9.00 am one day to ride her bike to Maureen’s town at a constant speed of 20 km/h. At the same time Maureen sets out to ride to Anne’s town at a constant speed of 18 km/h. a Write down a rule for the distance, dkm, that each of them is from Anne’s place at a time tminutes after 9.00 am. b On the same set of axes, draw graphs of the distance, dkm, versus time, tminutes after 9.00 am, for each cyclist.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 42 Essential Mathematical Methods Unit s1&2 c Find the time at which they will meet. d How far has each of them travelled when they meet? 8 John and Michael race over 50 metres. John runs so that it takes at aseconds to run 1 metre Example 22 and Michael runs so that it takes bseconds to run 1 metre. Michael wins the race by 1second. The next day they again race over 50 metres but Michael gives John a 3 metre start so that John runs 47 metres. Michael wins this race by 0.1 second. Find the values of aand band the speed at which Michael runs. 2.7 Th e tangent of the angle of slope and perpendicular lines adjacent opposite θ From year 10 you will be familiar with the trigonometric ratio tan = opposite adjacent The gradient, m,o fa straight line is given by m = y2− y1 x2− x1 (x2= x1) It therefore follows that m = tan ,w here is the angle that the line makes with the positive direction of the x-axis. y x y x B(x2, y2) B(x2, y2) A(x1, y1) A(x1, y1) x2 – x1 y2 – y1 θ x2 – x1 y2 – y1 0 0 θ θ Ifm is positive, will be an acute angle. If m is negative, will be an obtuse angle. It follows that the value of m in the general equation of a straight line y= mx + cgives the value of the tangent of the angle made by the line and the x-axis. Example 23 Determine the gradient of the line passing through the given points and the angle the line makes with the positive direction of the x-axis for: a (3, 2) and (5, 7) b (5, −3) and ( −1, 5)
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 43 Solution a m = 7− 2 5− 3 = 2.5 Tangent of angle = 2.5 ∴ Angle = 68 .20 ◦ b m = 5− (−3) −1− 5 =− 8 6 Tangent of angle =− 4 3 This implies the angle is obtuse ∴ angle = 180 ◦− (53 .130 ... )◦ = 126 .87 ◦ Example 24 Find the magnitude of the angle each of the following make with the positive direction of the x-axis: a y= 2x+ 3 b 3y= 3x− 6 c y=− 0.3 x+ 1.5 Solution a y= 2x+ 3 Gradient = 2 ∴ Tangent of angle = 2 giving an angle of 63 .43 ◦(63 ◦26 ) b 3y= 3x− 6 y= x− 2 Gradient = 1 ∴ Tangent of angle = 1 giving an angle of 45 ◦ c y=− 0.3x+ 1.5 Gradient =− 0.3 ∴ Tangent of angle =− 0.3 giving an angle of (180 − 16 .7)◦with the positive direction of the x-axis = 163 .3◦(163 ◦18 ) Perpendicular lines y x 0 Eb F c a D (1) (2) α α θ θ Multiplying the gradients of two perpendicular straight lines leads to a useful result. Gradient of (1) = tan = a b = m1 Gradient of (2) = m2 =− a c =− tan =− b a (from DEF ) =− 1 m1 ∴ m1m2=− 1
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 44 Essential Mathematical Methods Unit s1&2 If two straight lines are perpendicular, the product of their gradients is −1. Conversely, if the product of the gradients of two lines is −1, then the two lines are perpendicular. Note: This result holds if one of the two lines is not parallel to an axis. Example 25 Find the equation of the straight line which passes through (1, 2) and is: a parallel to the line with equation 2 x− y= 4 b perpendicular to the line with equation 2 x− y= 4. Solution The equation 2 x− y= 4can be rearranged to y= 2x− 4. Hence the gradient of the line can be seen to be 2. a Therefore the gradient of any line parallel to this line is 2. The equation of the straight line with this gradient and passing through the point with coordinates (1, 2) is y− 2= 2(x− 1) Therefore y= 2xis the equation of the line which passes through (1, 2) and is parallel to the line with equation 2 x− y= 4. b The gradient of any line perpendicular to the line with equation y= 2x− 4is −1 2. The equation of the straight line with this gradient and passing through the point with coordinates (1, 2) is y− 2=− 1 2(x− 1) Therefore 2 y− 4=− x+ 1and equivalently 2 y+ x= 5. Therefore 2 y+ x= 5i sthe equation of the line which passes through (1, 2) and is perpendicular to the line with equation 2 x−y= 4. Example 26 The coordinates of the vertices of a triangle ABC are A(0, −1), B(2, 3) and C 3, −21 2 .Show that the side AB is perpendicular to side AC . Solution Note: mAB = the gradient of the line AB . mAB = 3− (−1) 2− 0 = 4 2 = 2
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 45 mAC = −21 2− (−1) 3− 0 = −11 2 3 =− 1 2 Since mAB × mAC = 2×− 1 2=− 1, the lines AB and AC are perpendicular to each other. Exercise 2G 1 Find the angle that the lines joining the given points make with the positive direction of the x-axis: a (0, 3), (3, 0) b (0, −4), (4, 0) c (0, 2), ( −4, 0) d (0, −5), ( −5, 0) 2 Find the magnitude of the angle made by each of the following with the positive direction of the x-axis: a y= x b y=− x c y= x+ 1 d x+ y= 1 e y= 2x f y=− 2x 3 Find the angle that the lines joining the given points make with the positive direction of the Example 23 x-axis: a (−4, −2), (6, 8) b (2, 6), ( −2, 4) c (−3, 4), (6, 1) d (−4, −3), (2, 4) e (3b, a ), (3 a, b ) f (c, b ), ( b, c ) 4 Find the magnitude of the angle made by each of the following with the positive direction Example 24 of the x-axis: a y= 3x+ 2 b 2y=− 2x+ 1 c 2y− 2x= 6 d 3y+ x= 7 5 If the points A, B and Chave the coordinates A(5, 2), B(2, −3) and C(−8, 3), show that the triangle ABC is a right-angled triangle. 6 Show that RSTU is a rectangle if the coordinates of the vertices are respectively R(2, 6), S(6, 4), T(2, −4) and U(−2, −2). 7 Find the equation of the straight line which passes through the point (1, 4) and is Example 25 perpendicular to the line with equation y=− 1 2x+ 6. 8 Find the equation of the straight line which passes through (4, −2) and is: a parallel to the line with equation 2 x− 3y= 4 b perpendicular to the line with equation 2 x− 3y= 4.
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 46 Essential Mathematical Methods Unit s1&2 9 Given that the lines 4 x− 3y= 10 and 4 x− ly = m are perpendicular and intersect at the point (4, 2), find the values of land m. 2.8 Th e distance between two points y x B(x2, y2) A(x1, y1) x2 – x1 y2 – y1 0 C The distance between the given points A(x1,y1)and B(x2,y2)can be found by applying the Theorem of Pythagoras to triangle ABC : AB 2= AC 2+ BC 2 = (x2− x1)2+ (y2− y1)2 ∴The distance between the two points Aand Bis AB = (x2− x1)2+ (y2− y1)2 Example 27 Calculate the distance EF ifEis ( −3, 2) and Fis (4, −2). x y 1 0–1 –2 –1 –2 –3 E 123 4 F 2 Solution EF = (x2− x1)2+ (y2− y1)2 = (4 − (−3)) 2+ (−2− 2)2 = 72+ (−4)2 = √65 = 8.06 (to 2 decimal places) Exercise 2H 1 Find the distance between each of the following (correct to 2 decimal places): Example 27 a (3, 6) and ( −4, 5) b (4, 1) and (5, −3) c (−2, −3) and ( −5, −8) d (6, 4) and ( −7, 4) 2 Calculate the perimeter of a triangle with vertices ( −3, −4), (1, 5) and (7, −2). 3 There is an off-shore oil drilling platform in Bass Strait situated at D(0, 6), where 1unit = 5km. Pipes for this oil drill come ashore at M(−6, 1) and N(3, −1). Assuming the pipelines are straight, which is the shorter DM or DN ?
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 47 2.9 Midpoint of a line segment y x C D P(x, y) B(x2, y2) A(x1, y1) x – x1 y – y1 y2 – y 0 x2 – x Let P(x, y )b ethe midpoint of the line segment joining A(x1,y1)and B(x2,y2). The triangles APC and PBD are congruent. AC =PD and PC =BD ∴ x− x1= x2− xy − y1= y2− y 2x= x1+ x2 2y= y1+ y2 x= x1+ x2 2 y= y1+ y2 2 To find the midpoint ( x, y ) x= x1+ x2 2 and y= y1+ y2 2 Example 28 Find the midpoint of the line segment joining A(2, 6) with B(−3, −4). Solution Midpoint of line segment AB has coordinates 2+− 3 2 ,6+− 4 2 = −1 2,1 . Exercise 2I 1 Find the coordinates of M,the midpoint of AB ,w here Aand Bhave the following Example 28 coordinates: a A(2, 12), B(8, 4) b A(−3, 5), B(4, −4) c A(−1.6, 3.4), B(4.8, −2) d A(3.6, −2.8), B(−5, 4.5) 2 Find the midpoints of each of the sides of a triangle ABC ,w here Ais (1, 1), Bis (5, 5) and Cis (11, 2). 3 The secretary of a motorcross club wants to organise two meetings on the same weekend. One is a hill climb starting from point A(3.1, 7.1) and the other is a circuit event with the start at B(8.9, 10.5), as shown on the map. Only one ambulance can be provided. The ambulance can be called up by radio, so it is decided to keep it at C,halfway between A and B.What are the coordinates of C? A C B N Scale1cm = 2 km + km 14 12 10 8 6 4 2 2 D 46 8 EF 10 km
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 48 Essential Mathematical Methods Unit s1&2 4 The diagram shows the four points A(6, 6), B(10, 2), C(−1, 5) and D(−7, 1). a If the midpoint of AB isP and the midpoint of CD is M,calculate the distance PM . b Does the line joining these midpoints pass through 0,31 4 ? 8 x D C A –8 –4 0 4 8 4 B y 5 IfM is the midpoint of XY ,find the coordinates of Ywhen Xand M have the following values: a X(−4, 2), M(0, 3) b X(−1, −3), M(0.5, −1.6) c X(6, −3), M(2, 1) d X(4, −3), M(0, −3) 6 Find the coordinates of the midpoint of the line joining (1, 4) and ( a, b ), in terms of aand b.If(5, −1) is the midpoint find the values of aand b. 2.10 Angle between intersecting lines It is possible to use coordinate geometry to find the angle between two intersecting lines. Let = angle between the intersecting lines AB and CD . 1= angle between the positive direction of the x-axis and the line AB . 2= angle between the positive direction of the x-axis and the line CD . x y 0 θ1 θ2 α D C B A Then 1+ = 2.(The exterior angle of a triangle equals the sum of the two opposite interior angles.) ∴ = 2− 1
P1: FXS/ABE P2: FXS 0521609976c02.xml CUAT006-EVANS August 3, 2005 13:22 Chapte r2— Linear Relations 49 Example 29 Find the two angles between the intersecting lines 3 y+ 2x= 6and y= x+ 1. Sketch each line and label the angles. Solution x y C B D A –1 0 1 2 12 3 β α θ1 θ2 To find 1 y= x+ 1 ∴ m = 1 tan 1= 1 1= 45 ◦ To find 2 3y+ 2x= 6 3y=− 2x+ 6 y=− 2 3x+ 2 ∴ m =− 2 3 =− 0.6667 ∴ tan 2=− 0.6667 2= 180 ◦− 33 ◦41 = 146 ◦19 ∴ = 2− 1 = 146 ◦19 − 45 ◦ = 101 ◦19 = 180 − 101 ◦19 = 78 ◦41 Exercise 2J 1 Find the acute angle between each of the following pairs of straight lines: Example 29 a With gradients 3 and 3 4 b With gradients –2 and 3 c With gradients 2 3and −3 2 d With equations 2 y= 8x+ 10 and 3 x− 6y= 22 e With equations 4 x− 3y= 5and 2 x− 4y= 9
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