=> 18h+ 3d= 120 => 4h+3d = 36

Transcription

1 Linear simultaneous equations EAMPLE 1.34 At a poultr farm, six hens and one duk ost $4, while four hens and three duks ost $36. What is the ost of eah tpe of bird? Let the ost of one hen be $hand the ost of one duk be $d. Then the information given an be written as: 6h+ d = 4 4h+3d = There are several methods of solving this pair of equations. Method 1: Elimination Multipling equation b 3 Leaving Subtrating Dividing both sides b 14 Substituting h = 6 in equation gives => Therefore a hen osts $6 and a duk $4. => 18h+ 3d= 12 => 4h+3d = 36 => 14h 84 => h d = 4 d = 4 Note 1 The first step was to multipl equation b 3 so that there would be a term 3d in both equations. This meant that when subtrated, the variable dwas eliminated and so it was possible to find the value of h. 2 The value h = 6 was substituted in equation but it ould equall well have been substituted in the other equation. Chek for ourself that this too gives the answer d= 4. Before looking at other methods for solving this pair of equations, here is another example. EAMPLE 1.35 Solve 3x+ 5= 12 2x- 6= -2 Adding Giving Substituting x = -1 in equation Adding 3 to eah side Dividing b 5 Therefore x= -1, = 3. => => => => => 18x+ 3 = 72 1x- 3 = -1 28x = 12 5 = 15 = 3

2 Note In this example, both equations were multiplied, the first b 6 to give +3 and the seond b 5 to give -3. Beause one of these terms was positive and the other negative, it was neessar to add rather than subtrat in order to eliminate. Returning now to the pair of equations giving the pries of hens and duks, 6h+ d = 4 4h+3d = here are two alternative methods of solving them. Method 2: Substitution The equation 6h + d = 4 is rearranged to make d its subjet: 3" ;:; I» :I Cll J:l I» ; :I Cll d=4-6h. This expression for d is now substituted in the other equation, 4h + 3d= 36, giving 4h+ 3(4-6h) = 36 => 4h h = 36 => -l4h = -84 => h = 6 Substituting for h in d = 4-6h gives d = 4-36 = 4. Therefore a hen osts $6 and a duk $4 (the same answer as before, of ourse). Method 3: Intersetion of the graphs of the equations Figure 1.13 shows the graphs of the two equations, 6h + d = 4 and 4h +3d= 36. As ou an see, the interset at the solution, h = 6 and d = 4. d Figure \ '\ r\ r\. 3d 36 \ - t -f--- - r r--- I -- I 1 I -r-. --f \ -- \ - '--- - [\ ±- - r h

3 Non-linear simultaneous equations The simultaneous equations in the examples so far have all been linear, that is their graphs have been straight lines. A linear equation in, sa, x and ontains onl terms in x and and a onstant term. So 7x + 2 = 11 is linear but 7x 2 + 2= 11 is not linear, sine it ontains a term in x 2 You an solve a pair of simultaneous equations, one of whih is linear and the other not, using the substitution method. This is shown in the next example. EAMPLE 1.36 Solve x+ 2= 7 x = Rearranging equation gives x= 7-2. Substituting for x in equation@: (7-2)2 + 2 = 1 Multipling out the (7-2) x (7-2) gives = , so the equation is This is rearranged to give :::::} s2-1s :::::} S(- 3) -13(- 3) :::::} (S- 13)(- 3) A quadrati in whih ou an now solve using fatorisation or the formula. Either S-13= :::::} = 2.6 Or -3= :::::} = 3 Substituting in equation, x+ 2= 7: = 2.6 :::::} = 1.8 = 3 :::::} x= 1 The solution is either x = 1.8, = 2.6 or x = 1, = 3. A Alwas substitute into the linear equation. Substituting in the quadrati will give ou extra answers whih are not orret.

4 EERCISE 1G 1 Solve the following pairs of simultaneous equations. (i) 2x+ 3= 8 (ii) x+4= 16 (iiil 7x+ = 15 3x+ 2= 7 3x+ 5=2 4x+ 3= 11 (iv) Sx- 2= 3 (v) 8x- 3= 21 (vi) 8x+ = 32 x+ 4= 5 5x+ = 16 7x- 9=28 (viil 4x+ 3= 5 (viii) 3u-2v= 17 (ixl 41-3m= 2 2x- 6=-5 5u-3v=28 5l-7m=9 2 A student wishes to spend exatl $1 at a seond-hand bookshop. All the paperbaks are one prie, all the hardbaks another. She an bu five paperbaks and eight hardbaks. Alternativel she an bu ten paperbaks and six hardbaks. m Write this information as a pair of simultaneous equations. (iil Solve our equations to find the ost of eah tpe of book. 3 The ost of a pear is Se greater than that of an apple. Eight apples and nine pears ost $1.64. (i) Write this information as a pair of simultaneous equations. (iil Solve our equations to find the ost of eah tpe of fruit. 4 A ar journe of 38 km lasts 4 hours. Part of this is on a motorwa at an average speed of 11 kmh- 1, the rest on ountr roads at an average speed of 7 kmh- 1. m Write this information as a pair of simultaneous equations. (ii) Solve our equations to find how man kilometres of the journe is spent on eah tpe of road. 5 Solve the following pairs of simultaneous equations. m x2+2=1 x+=4 (iil x =8 x+ 2= 8 (iiil 2x = 12 x- =-1 (iv) k2 + km = 8 m=k-6 (v) t 2 - t 2 = tl = 2t2 (vi) p + q + 5 = p2 = q2 + 5 (viil k(k- m) = 12 k(k+ m)= 6 tviiil P/ - P/ = PI+ p2 = 2

5 6 The diagram shows the net of a lindrial ontainer of radius r m and height hem. The full width of the metal sheet from whih the ontainer is made is 1 m, and the shaded area is waste. The surfae area of the ontainer is 14n: m 2 h (i) Write down a pair of simultaneous equations for rand h. (ii) Find the volume of the ontainer, giving our answers in terms of n:. (There are two possible answers.) 7 A large window onsists of six square panes of glass as shown. Eah pane is x m b x m, and all the dividing wood is m wide.!m (i) Write down the total area of the whole window in terms of x and. (iil Show that the total area of the dividing wood is 7 x (iii) The total area of glass is 1.5 m 2, and the total area of dividing wood is 1m 2 Find x, and hene find an equation for and solve it. [MEI] Inequalities Not all algebrai statements involve the equals sign and it is just as important to be able to handle algebrai inequalities as it is to solve algebrai equations. The solution to an inequalit is a range of possible values, not speifi value(s) as in the ase of an equation.

6 Linear inequalities A The methods for linear inequalities are muh the same as those for equations but ou must be areful when multipling or dividing through an inequalit b a negative number. Take for example the following statement: :i.q!. ;::;: ; Multipl both sides b -1 5 > 3 is true -5 > -3 is false. A It is atuall the ase that multipling or dividing b a negative number reverses the inequalit, but ou ma prefer to avoid the diffiult, as shown in the examples below. EAMPLE 1.37 Solve Sx- 3 o% 2x- 15. Add 3 to, and subtrat 2x from, both sides => Sx- 2x,;;;: Tid up => 3x o% -12 Divide both sides b 3 o% -4 Note Sine there was no need to multipl or divide both sides b a negative number, no problems arose in this example. EAMPLE 1.38 Solve 2+ 6 > Subtrat 6 and 7 from both sides Tid up :::::} 2-7 > 11-6 :::::} Add 5 to both sides and subtrat 5 :::::} Divide both sides b +5-1 > Note that logiall -1 > is the same as < -1, so the solution is < -1.

7 EAMPLE 1.39 Quadrati inequalities Solve (il x 2-4x+ 3 > liil x 2-4x+ 3 :os;. The graph of = x 2-4x + 3 is shown in figure 1.14 with the green parts of the x axis orresponding to the solutions to the two parts of the question. (i) You want the values of x for whih >, whih that is where the urve is above the x axis. (ii) You want the values of x for :os;, that is where the urve rosses or is below the x axis. Here the end points are not inluded in the inequalit so ou draw open irles: I r--2-i\ I I f--l- t--1 I I V Figure 1.14 The solution is x < 1 or x > 3. The solution is x 1 and x :os; 3, usuall witten 1 :os; x :os; 3. EAMPLE 1.4 Find the set of values of k for whih x 2 + kx + 4 = has real roots. A quadrati equation, ax2 + bx + =, has real roots if b 2-4a. So x 2 + kx + 4 = has real roots if k 2-4 x 4 x 1. So the set of values is k 4 and k -4. Take are: (- 5) 2 = 25 and (-3)2 = 9, so k must be less than or equal to -4.

8 EERCISE 1H 1 Solve the following inequalities. m Sa+ 6 > 2a + 24 (iiil 4(-l) > 3(- 2) (v)!e + 3! < e 2 2 (vii) 5(2-3g) + g 8(2g- 4) (ii) 3b- 5,::;; b-1 (iv) d- 3(d + 2) 2(1 + 2d) (vi) -J- 2f- 3 < 4(1 +f) (viii) 3(h + 2)- 2(h- 4) > 7(h + 2) 2 Solve the following inequalities b skething the urves of the funtions involved. (i) p 2-5p+4 < (ii) p 2-5p+4 (iii) x 2 + 3x+ 2 (iv) x 2 + 3x > -2 (v) > (vi) z(z-1) 2 (vii) q 2-4q + 4 > (viii) (- 2) > 8 (ix) 3x 2 + 5x- 2 < (x) ll- 6 (xi) 4x- 3 x 2 (xii) 1 2 > Find the set of values of k for whih eah of these equations has two real roots.. ::t (i) 2x 2-3x + k = (iiil 5x 2 + kx+ 5 = (iil = (iv) 3x 2 + 2kx+ k= 4 Find the set of values of k for whih eah of these equations has no real roots. (i) x 2-6x + k = liiil 4x 2 - kx+ 4 = (iil (iv) I<EY POINTS 1 The quadrati formula for solving ax 2 + bx + = is -b ±.Jb 2-4a = -=--=--'-=-----="'- 2a where b 2-4a is alled the disriminant. If b 2 - If b 2 - If b 2-4a >, the equation has two real roots. 4a =, the equation has one repeated root. 4a <, the equation has no real roots. 2 To solve a pair of simultaneous equations where one equation is non-linear: first make x or the subjet of the linear equation then substitute this rearranged equation for x or in the non-linear equation solve to find or x substitute bak into the linear equation to find pairs of solutions. 3 Linear inequalities are dealt with like equations but if ou multipl or divide b a negative number ou must reverse the inequalit sign. 4 When solving a quadrati inequalit it is advisable to sketh the graph.

9 E Cll : (J Co-ordinate geometr A plae for everthing, and everthing in its plae Samuel Smiles Co-ordinates Co-ordinates are a means of desribing a position relative to some fixed point, or origin. In two dimensions ou need two piees of information; in three dimensions, ou need three piees of information. In the Cartesian sstem (named after Rene Desartes), position is given in perpendiular diretions: x, in two dimensions; x,, z in three dimensions (see figure 2.1). This hapter onentrates exlusivel on two dimensions. z 5 4 (3, 4, 5) 3 3 (3, 2) Figure 2.1

10 Plotting, skething and drawing In two dimensions, the o-ordinates of points are often marked on paper and joined up to form lines or urves. A number of words are used to desribe this proess. Plot (a line or urve) means mark the points and join them up as auratel as ou an. You would expet to do this on graph paper and be prepared to read information from the graph. Sketh means mark points in approximatel the right positions and join them up in the right general shape. You would not expet to use graph paper for a sketh and would not read preise information from one. You would however mark on the o-ordinates of important points, like intersetions with the x and axes and points at whih the urve hanges diretion. -l :::r Cll ii! Cl. Cij" :I I» :; Cll Draw means that ou are to use a level of aura appropriate to the irumstanes, and this ould be anthing between a rough sketh and a ver auratel plotted graph. The gradient of a line In everda English, the word line is used to mean a straight line or a urve. In mathematis, it is usuall understood to mean a straight line. If ou know the o-ordinates of an two points on a line, then ou an draw the line. The slope of a line is measured b its gradient. It is often denoted b the letter m. In figure 2.2, A and B are two points on the line. The gradient of the line AB is given b the inrease in the o-ordinate from A to B divided b the inrease in the x o-ordinate from A to B. AL (2, 4) B(6, 7) Figure 2.2

11 Q) E Q) Cl Q) ea :s (.) In general, when A is the point (x 1, 1 ) and B is the point (x 2, z), the gradient is Yz- Yt m= x 1 When the same sale is used on both axes, m= tan e (see figure 2.2). Figure 2.3 shows four lines. Looking at eah one from left to right: line A goes uphill and its gradient is positive; line B goes downhill and its gradient is negative. Line C is horizontal and its gradient is ; the vertial line D has an infinite gradient. Figure ACTIVITY 2.1 On eah line in figure 2.3, take an two points and all them (xl' 1 ) and (x 2, 2 ). Substitute the values of xl' l' x 2 and 2 in the formula m= Yz - Y1 z- I and so find the gradient. G Does it matter whih point ou all (x 1, 1 ) and whih (x 2, 2 )? Parallel and perpendiular lines If ou know the gradients m 1 and m 2 of two lines, ou an tell at one if the are either parallel or perpendiular- see figure 2.4. Figure 2.4 parallel lines: m 1 =m 2 perpendiular lines: m 1m 2 = -I

12 I ACTIVITY 2.2 ACTIVITY 2.3 Lines whih are parallel have the same slope and so m 1 = m 2. If the lines are perpendiular, m 1 m 2 = -1. You an see wh this is so in the ativities below. Draw the line L 1 joining (, 2) to (4, 4), and draw another line l 2 perpendiular to L 1 Find the gradients m 1 and m 2 of these two lines and show that m 1 m 2 = -1. The lines AB and BC in figure 2.5 are equal in length and perpendiular. B showing that triangles ABE and B are ongruent prove that the gradients m 1 and m 2 must satisf m 1 m 2 = -1. gradient m 1 B \.:::::_ B gradient m 2-4 :r a.!' :I n er ::!: :I ::!: "Cl :; Ul A B I E Dl Figure 2.5 A Lines for whih m 1 m 2 = -1 will onl look perpendiular if the same sale has been used for both axes. The distane between two points When the o-ordinates of two points are known, the distane between them an be alulated using Pthagoras' theorem, as shown in figure 2.6. ALJC (2, 4) Figure 2.6

13 This method an be generalised to find the distane between an two points, A(xl' 1 ) and B(x 2, 2 ), as in figure 2.7. Ql E Q).,! u Figure 2.7 The mid-po nt of a line joining two oin s Look at the line joining the points A(2, 1) and B(8, 5) in figure 2.8. The point M(5, 3) is the mid-point of AB Figure Notie that the o-ordinates of M are the means of the o-ordinates of A and B. 5 = i(2 + 8); 3 = io + 5). This result an be generalised as follows. For an two points A(xl' 1 ) and B(x 2, 2 ), the o-ordinates of the mid-point of AB are the means of the o-ordinates of A and B so the mid-point is i+ z Y1 + Yz) ( 2 ' 2.

14 EAMPLE 2.1 A and Bare the points (2, 5) and (6, 3) respetivel (see figure 2.9). Find: (i) the gradient of AB (iil the length of AB (iiil the mid-point of AB (ivl the gradient of a line perpendiular to AB. Taking A(2, 5) as the point (x, 1 ), 1 and B(6, 3) as the point (x 2, 2 ) gives x 1 = 2, 1 =5,x 2 =6, 2 =3. (i) Gradient= 2 -, _3-5_ liil Length AB= x,)l + ( 2 - Yl = = 4 = J2o.... (xl + x2 Y1 + Y2) hi l M1d-pomt = - 2 -, = ( 2; 6) 5 ; 3) = ( 4, 4) A(2, 5) Figure 2.9 B(6, 3) ::r 3 a:,; :; :r o :; :; 'C :; (ivl Gradient of AB= m 1 = If m 2 is the gradient of a line perpendiular to AB, then m 1 m 2 = -1 1 ::::::} = -1 m 2 =2. EAMPLE 2.2 Using two different methods, show that the lines joining P(2, 7), Q(3, 2) and R(O, 5) form a right-angled triangle (see figure 2.1). Method Gradient of RP Gradient ofrq = 3 _ = -1 ::::::} Produt of gradients= 1 x ( -1) = -1 ::::::} Sides RP and RQ are at right angles. 6 R(O, 5) " [ Figure 2.1

15 ; E 1 : (J Method2 Pthagoras' theorem states that for a right-angled triangle whose hpotenuse has length a and whose other sides have lengths band, a 2 = b Conversel, if ou an show that a 2 = b for a triangle with sides of lengths a, b, and, then the triangle has a right angle and the side oflength a is the hpotenuse. This is the basis for the alternative proof, in whih ou use length 2 = (x 2 - x 1 ) 2 + ( 2 - Y 1 V PQ 2 = (3-2) 2 + (2-7) 2 = = 26 RP 2 = (2- ) 2 + (7-5) 2 = = 8 RQ 2 = (3- ) 2 + (2-5) 2 = = 18 Sine 26 = , PQ2 = RP2 + RQ2 :::::} Sides RP and RQ are at right angles. EERCISE 2A 1 For the following pairs of points A and B, alulate: (a) the gradient of the line AB (bl the mid-point of the line joining A to B (l the distane AB (d) the gradient of the line perpendiular to AB. m A(O, 1)!iiil A( -6, 3) (v) A(4, 3) B(2, -3) B(6,3) B(2,)!iil A(3,2) B(4,-1)!ivl A(5, 2) B(2, -8)!vil A(l, 4) B(l, -2) 2 The line joining the point P(3, -4) to Q(q, ) has a gradient of2. Find the value of q. 3 The three points (2, -1), Y(8, ) and Z(ll, 2) are ollinear (i.e. the lie on the same straight line). Find the value of. 4 The points A, B, C and D have o-ordinates (1, 2), (7, 5), (9, 8) and (3, 5). m Find the gradients of the lines AB, BC, and DA. (ii) What do these gradients tell ou about the quadrilateral AB? (iiil Draw a diagram to hek our answer to part (ii). 5 The points A, Band C have o-ordinates (2, 1), (b, 3) and (5, 5), where b > 3 and LABC = 9. Find: m the value of b (iil the lengths of AB and BC (iiil the area of triangle ABC.

16 6 The triangle PQR has verties P(8, 6), Q(O, 2) and R(2, r). Find the values of r when the triangle: lil has a right angle at P liil has a right angle at Q liiil has a right angle at R livl is isoseles with RQ =RP. 7 The points A, B, and C have o-ordinates ( -4, 2), (7, 4) and ( -3, -1). m Draw the triangle ABC. liil Show b alulation that the triangle ABC is isoseles and name the two equal sides. liiil Find the mid-point of the third side. (ivl B alulating appropriate lengths, alulate the area of the triangle ABC. 8 For the points P(.x, ), and Q(3.x, S), find in terms of x and : lil the gradient of the line PQ liil the mid-point of the line PQ liiil the length of the line PQ. 9 A quadrilateral has verties A(O, ), B(O, 3), C(6, 6) and D(12, 6). lil Draw the quadrilateral. liil Show b alulation that it is a trapezium. liiil Find the o-ordinates of E when EB is a parallelogram. 1 Three points A, Band C have o-ordinates (1, 3), (3, 5) and ( -1, ). Find the values of when: 1i1 AB=AC liil AC= BC liiil AB is perpendiular to BC livl A, B and C are ollinear. 11 The diagonals of a rhombus biset eah other at 9, and onversel, when two lines biset eah other at 9, the quadrilateral formed b joining the end points of the lines is a rhombus. Use the onverse result to show that the points with o-ordinates ( 1, 2), (8, -2), (7, 6) and (, 1) are the verties of a rhombus, and find its area.

17 The equation of a straight line The word straight means going in a onstant diretion, that is with fixed gradient. This fat allows ou to find the equation of a straight line from first priniples. ti E : (.) EAMPLE 2.3 Find the equation of the straight line with gradient 2 through the point (, -5). Figure 2.11 Take a general point (.x, ) on the line, as shown in figure The gradient of the line joining (, -5) to (.x, ) is given b. -(-5) +5 gradient= = --. x- x Sine we are told that the gradient of the line is 2, this gives +5=2 =2x-5. Sine (.x, ) is a general point on the line, this holds for an point on the line and is therefore the equation of the line. The example above an easil be generalised (see page 5) to give the result that the equation of the line with gradient m utting the axis at the point (, ) is = mx+. (In the example above, m is 2 and is -5.) This is a well-known standard form for the equation of a straight line.

18 Drawing a line, given its equation There are several standard forms for the equation of a straight line, as shown in figure When ou need to draw the graph of a straight line, given its equation, the first thing to do is to look arefull at the form of the equation and see if ou an reognise it. (a) Equations of the form x= a x= 3 (b) Equations of the form = b (, 2) l = 2-4 ::r J:l a :I = ::r :I (3, ) () Equations of the form = mx (d) Equations of the form = mx + (, I) (, - J) =-tx+ I (e) Equations of the form px + q + r = Figure 2.12 (a), {b): Lines parallel to the axes Lines parallel to the x axis have the form = onstant, those parallel to the axis the form x = onstant. Suh lines are easil reognised and drawn.

19 (), fd): Equations of the form = mx + ; E en «< u The line = mx + rosses the axis at the point (, ) and has gradient m. If =, it goes through the origin. In either ase ou know one point and an omplete the line either b fmding one more point, for example b substituting x = 1, or b following the gradient (e.g. 1 along and 2 up for gradient 2). (e): Equations of the form px + q + r = In the ase of a line given in this form, like 2x =, ou an either rearrange it in the form = mx+ (in this example = + 2), or ou an find the o-ordinates of two points that lie on it. Putting x = gives the point where it rosses the axis, (, 2), and putting = gives its intersetion with the x axis, (3, ). EAMPLE 2.4 Sketh the lines x = 5, = and = x on the same axes. Desribe the triangle formed b these lines. The line x= 5 is parallel to the axis and passes through (5, ). The line = is the x axis. The line = x has gradient 1 and goes through the origin. Figure 2.13 The triangle obtained is an isoseles right-angled triangle, sine OA =AB= 5 units, and LOAB = 9. EA PLE 2.5 Draw = x- 1 and 3x + 4 = 24 on the same axes. The line = x- 1 has gradient 1 and passes through the point (, -1). Substituting = gives x= 1, so the line also passes through (1, ). Find two points on the line 3x + 4 = 24. Substituting x = gives 4 = 24 Substituting = g1ves 3x = 24 so so =6. x=8.

20 The line passes through (, 6) and (8, ). Figure 2.14 "11 :; Q, i:i' ':1'.Cl I» s :I. I» :; EERCISE 28 1 Sketh the following lines. (i) (iv) =-2 =-3x (viil =x+4 (x) =-4x+8 (., 1 2 III Y = - 2 x- (xvil 2x+ 5= 1 (xix) x+ 3-6 = (ii) = 5 (v) = 3x+ 5 (viii) = 2 (xi) = 4x- 8 (xiv) = 1-2x (xvii) 2x+ - 3 = (xx) = 2 -x (iiil (vi) =2x =x-4 (ix) = 2x + (xii) = -x+ 1 (xv) 3x- 2= 6 (xviii) 2= 5x- 4 2 B alulating the gradients of the following pairs of lines, state whether the are parallel, perpendiular or neither. (i) =-4 x=2 (ii) =3x x=3 (iii) 2x+ = 1 x- 2= 1 (iv) = 2x+ 3 4x- + 1 = (v) 3x- + 2 = 3x+ = (vi) 2x+ 3= 4 2= 3x- 2 (vii) x+2-1= x+2+ 1 = (viii) = 2x-1 2x- +3= (ix) =x-2 x+=6 (x) =4-2x x+ 2= 8 (xi) x+3-2= = 3x+ 2 (xii) =2x 4x+ 2= 5 Finding the equation of a line The simplest wa to find the equation of a straight line depends on what information ou have been given.

21 (i) Given the gradient, m, and the o-ordinates (x 1, 1 ) of one point on the line Cl) E Cl) en Cl) : (J Take a general point (x, ) on the line, as shown in figure Figure 2.15 The gradient, m, of the line joining (xl' 1 ) to (x, ) is given b - m=--1 x - x 1 => - 1 =m (x-x 1 ). This is a ver useful form of the equation of a straight line. Two positions of the point (x 1, 1 ) lead to partiularl important forms of the equation. (a) When the given point (xl' 1 ) is the point (, ), where the line rosses the axis, the equation takes the familiar form = mx+ as shown in figure (b) When the given point (xl' 1 ) is the origin, the equation takes the form =mx as shown in figure Figure 2.16 Figure 2.17

22 EAMPLE 2.6 Find the equation of the line with gradient 3 whih passes through the point (2, -4). Using - 1 = m(x- x 1 ) :::::} - (-4) = 3(x- 2) :::::} + 4 = 3x- 6 :::::} =3x-1. (ii) Given two points, (x 1, 1 ) and (x 2, 2 ) The two points are used to find the gradient:, :; Cl. :r ::r.cl m o :I. m :;- This value of m is then substituted in the equation - 1 =m (x- x 1 ). This gives - YI = (Y2- YI)(x- xj x 2 - x 1 Rearranging the equation gives Figure 2.18 Y- Y1 = x- 1 or Y- Y1 = Y2- Y1 EAMPLE 2.7 Find the equation of the line joining (2, 4) to (5, 3). Taking (xi' 1 ) to be (2, 4) and (x 2, 2 ) to be (5, 3), and substituting the values in - YJ = - ] Y2- Y1 x x-2 gives 3-4 = 5-2. This an be simplified to x =. f) Show that the equation of the line in figure 2.19 an be written Figure 2.19

23 Different tehniques to solve problems The following examples illustrate the different tehniques and show how these an be used to solve a problem. ; E Gl Cl Gl u EAMPLE 2.8 Find the equations of the lines (a)- (e) in figure 2.2. I -v 4-_ 1_ (b) -- -() 1\ V "- I \ V v J I ') V, i / 1 \ I v 2-4 l\ t 8 I I I I - I " 1\ >1\ I, (._. I \! I I I I I. Figure 2.2 Line (a) passes through (, 2) and has gradient 1 :::::} equation of (a) is = x+ 2. Line (b) is parallel to the x axis and passes through (, 4) :::::} equation of (b) is = 4. Line () is parallel to the axis and passes through ( -3, ) :::::} equation of () is x = -3. Line (d) passes through (, ) and has gradient -2 :::::} equation of (d) is = -2x. Line (e) passes through (, -1) and has :::::} equation of (e) is = -1. This an be rearranged to give x =. EAMPLE 2.9 Two sides of a parallelogram are the lines 2 = x + 12 and = 4x- 1. Sketh these lines on the same diagram. The origin is a vertex of the parallelogram. Complete the sketh of the parallelogram and find the equations of the other two sides.

24 The line 2= x+ 12 has and passes through the point (, 6) (sine dividing b 2 gives = + 6). The line = 4x- 1 has gradient 4 and passes through the point (, -1).., ; a. :r IQ ::r J:l o = ; Figure 2.21 The other two sides are lines with and 4 whih pass through (, ), i.e. = and = 4x. EAMPLE 2.1 Find the equation of the perpendiular bisetor of the line joining P( -4, 5) to Q(2, 3). Figure 2.22

25 . Q) E Q) D) Cll : The gradient of the line PQ is (-4) 6 = -3 and so the gradient of the perpendiular bisetor is +3. The perpendiular bisetor passes throught the mid-point, R, of the line PQ. The o-ordinates of R are 2+(-4) 3+5). (-1 4) ( 2 ' 2 1.e. ' Using - 1 = m(x- x 1 ), the equation of the perpendiular bisetor is -4=3(x-(-1)) - 4 = 3x+ 3 =3x+7. EERCISE 2C 1 Find the equations of the lines (i) - (x) in the diagrams below.

26 2 Find the equations of the following lines. (i) parallel to= 2x and passing through (1, 5) (iil parallel to= 3x- 1 and passing through (, ) (iiil parallel to 2x = and passing through (-4, 5) (ivl parallel to 3x- - 1 = and passing through (4, -2) (vi parallel to 2x+ 3= 4 and passing through (2, 2) (vi) parallel to 2x- - 8 = and passing through (-1, -5) 3 Find the equations of the following lines. m >C () iii' N n (i) perpendiular to= 3x and passing through (, ) (iil perpendiular to = 2x + 3 and passing through ( 2, -1) (iiil perpendiular to 2x+ = 4 and passing through (3, 1) (ivl perpendiular to 2 = x + 5 and passing through (-1, 4) (vi perpendiular to 2x + 3 = 4 and passing through (5, -1) (vil perpendiular to 4x- + 1 = and passing through (, 6) 4 Find the equations of the line AB in eah of the following ases. lil A(O, ) liiil A(2, 7) (v) A(-2, 4) B(4,3) B(2, -3) B(5, 3) liil A(2, -1) (iv) A(3, 5) (vi) A(-4, -2) B(3, ) B(5, -1) B(3, -2) 5 Triangle ABC has an angle of 9 at B. Point A is on the axis, AB is part of the line x = and C is the point (6, 2). (i) Sketh the triangle. (iil Find the equations of AC and BC. (iiil Find the lengths of AB and BC and hene find the area of the triangle. (iv) Using our answer to part (iiil, find the length of the perpendiular from B toac. 6 A median of a triangle is a line joining one of the verties to the mid-point of the opposite side. In a triangle OAB, is at the origin, A is the point (, 6) and B is the point (6, ). (i) Sketh the triangle. (iil Find the equations of the three medians of the triangle. (iiil Show that the point (2, 2) lies on all three medians. (This shows that the medians of this triangle are onurrent.) 7 A quadrilateral AB has its verties at the points (, ), (12, 5), (, 1) and ( -6, 8) respetivel. (i) Sketh the quadrilateral. (iil Find the gradient of eah side. (iiil Find the length of eah side. (iv) Find the equation of eah side. (vi Find the area of the quadrilateral.

27 The intersetion of two lines ti E Ql Cl Ql ea (J EAMPLE 2.11 The intersetion of an two urves (or lines) an be found b solving their equations simultaneousl. In the ase of two distint lines, there are two possibilities: (il the are parallel (ii) the interset at a single point. Sketh the lines x+ 2= 1 and 2x+ 3= 4 on the same axes, and find the o-ordinates of the point where the interset. The line x+ 2= 1 passes through (, (1, ). The line 2x+ 3= 4 passes through (, j) and (2, ). Figure 2.23 : x+ 2= 1 (i): 2x+ 3= 4 Substituting=-2 in : :x 2: 2x+ 4= 2 (i): 2x+3=4 Subtrat: = -2. x-4=1 ::::} = 5. The o-ordinates of the point of intersetion are ( 5, -2).

28 EAMPLE 2.12 Find the o-ordinates of the verties of the triangle whose sides have the equations x+ = 4, 2x- = 8 and x+ 2= -1. A sketh will be helpful, so first find where eah line rosses the axes. x+ = 4 rosses the axes at (, 4) and (4, 2x- = 8 rosses the axes at (, -8) and ( 4, ). x + 2 = -1 rosses the axes at (, and (-1, ). :r (I) -t ::r (I) (I)!1 o :I :e :r (I) Cll Figure 2.24 Sine two lines pass through the point ( 4, ) this is learl one of the verties. It has been labelled A on figure Point B is found b : Add 4x- 2= 16 x+2=-1 5x =15 so x=3. Substituting x= 3 in@ gives = -2, sob is the point (3, -2). Point C is found b solving and simultaneousl: : : Subtrat x+ =4 x+2=-1 - = 5 so = -5. Substituting = -5 in gives x = 9, so C is the point (9, -5).

29 The line l has equation 2x- = 4 and the line m has equation = 2x- 3. What an ou sa about the intersetion of these two lines? ; E Gl en Gl :s u Historial note Rene Desartes was born near Tours in Frane in At the age of eight he was sent to a Jesuit boarding shool where, beause of his frail health, he was allowed to sta in bed until late in the morning. This habit staed with him for the rest of his life and he laimed that he was at his most produtive before getting up. After leaving shool he studied mathematis in Paris before beoming in turn a soldier, traveller and optial instrument maker. Eventuall he settled in Holland where he devoted his time to mathematis, siene and philosoph, and wrote a number of books on these subjets. In an appendix, entitled La Geometrie, to one of his books, Desartes made the ontribution to o-ordinate geometr for whih he is partiularl remembered. In 1649 he left Holland for Sweden at the invitation of Queen Christina but died there, of a lung infetion, the following ear. EERCISE 2 1 m Find the verties of the triangle ABC whose sides are given b the lines AB: x- 2= -1, BC: 7x+ 6= 53 and AC: 9x+ 2= 11. liil Show that the triangle is isoseles. 2 Two sides of a parallelogram are formed b parts of the lines 2x- = -9 and x- 2=-9. (i) Show these two lines on a graph. (iil Find the o-ordinates of the vertex where the interset. Another vertex of the parallelogram is the point (2, 1). liiil Find the equations of the other two sides of the parallelogram. (ivl Find the o-ordinates of the other two verties. 3 A(O, 1), B(l, 4), C(4, 3) and D(3, ) are the verties of a quadrilateral AB. (i) Find the equations of the diagonals AC and BD. (iil Show that the diagonals AC and BD biset eah other at right angles. (iiil Find the lengths of AC and BD. (iv) What tpe of quadrilateral is AB? 4 The line with equation Sx + = 2 meets the x axis at A and the line with equation x + 2 = 22 meets the axis at B. The two lines interset at a point C. m Sketh the two lines on the same diagram. (iil Calulate the o-ordinates of A, B and C. (iiil Calulate the area of triangle OBC where is the origin. (iv) Find the o-ordinates of the point E suh that ABEC is a parallelogram.

30 5 A median of a triangle is a line joining a vertex to the mid-point of the opposite side. In an triangle, the three medians meet at a point. The entroid of a triangle is at the point of intersetion of the medians. Find the o-ordinates of the entroid for eah triangle shown. (i) (, 12) (iil (, 9) m > et n iij' et N 6 You are given the o-ordinates of the four points A(6, 2), B(2, 4), C( -6, -2) and D( -2, -4). (i) Calulate the gradients of the lines AB, CB, DC and DA. Hene desribe the shape of the figure AB. (ii) Show that the equation of the line DAis 4-3x = -1 and find the length DA. (iiil Calulate the gradient of a line whih is perpendiular to DA and hene find the equation of the line l through B whih is perpendiular to DA. (ivl Calulate the o-ordinates of the point P where l meets DA. (vi Calulate the area of the figure AB. 7 The diagram shows a triangle whose verties are A( -2, 1), B(l, 7) and C(3, 1). The point L is the foot of the perpendiular from A to BC, and M is the foot of the perpendiular from B to A C. [MEI) (il Find the gradient of the line BC. (iil Find the equation of the line AL. (iiil Write down the equation of the line BM. B(l, 7) A (- 2, I)

31 ti E Cl ea : (.) The lines AL and BM meet at H. livl Find the o-ordinates of H. (v) Show that CH is perpendiular to AB. (vi) Find the area of the triangle BLH. 8 The diagram shows a retangle AB. The point A is (, -2) and C is ( 12, 14). The diagonal BD is parallel to the x axis. C(l2, 14) [MEI] lil Explain wh the o-ordinate of D is 6. The x o-ordinate of D is h. liil Express the gradients of AD and in terms of h. liiil Calulate the x o-ordinates of D and B. (ivl Calulate the area of the retangle AB. [Cambridge AS & A Level Mathematis 979, Paper 12 Q9 November 29] 9 The diagram shows a rhombus AB. The points Band D have o-ordinates (2, 1) and (6, 2) respetivel, and A lies on the x axis. The mid-point of BD is M. Find, b alulation, the o-ordinates of eah of M, A and C. A [Cambridge AS & A Level Mathematis 979, Paper 1 QS June 25]

32 1 Three points have o-ordinates A(2, 6), B(8, 1) and C(6, ). The perpendiular bisetor of AB meets the line BC at D. Find (i) the equation of the perpendiular bisetor of AB in the form ax + b= lii) the o-ordinates of D. [Cambridge AS & A Level Mathematis 979, Paper 1 Q7 November 25] 11 The diagram shows a retangle AB. The point A is (2, 14), B is (-2, 8) and C lies on the x axis. m ; iii' N A(2, 14) B(- 2, 8) D Find (i) the equation of BC. liil the o-ordinates of C and D. [Cambridge AS & A Level Mathematis 979, Paper 1 Q6 June 27] 12 The three points A(3, 8), B(6, 2) and C(lO, 2) are shown in the diagram. The point D is suh that the line DA is perpendiular to AB and DC is parallel to AB. Calulate the o-ordinates of D. D A(3, 8) B(6, 2) C(IO, 2) [Cambridge AS & A Level Mathematis 979, Paper 1 Q6 November 27]

33 Q) E Q) en Q) «< : 13 In the diagram, the points A and C lie on the x and axes respetivel and the equation of AC is 2+ x= 16. The point B has o-ordinates (2, 2). The perpendiular from B to AC meets AC at the point. A lil Find the o-ordinates of. The point D is suh that the quadrilateral AB has AC as a line of smmetr. (ii) Find the o-ordinates of D. (iiil Find, orret to 1 deimal plae, the perimeter of AB. [Cambridge AS & A Level Mathematis 979, Paper 1 Q 11 June 28] 14 The diagram shows points A, Band C ling on the line 2= x+ 4. The point A lies on the axis and AB= BC. The line from D(lO, -3) to B is perpendiular to A C. Calulate the o-ordinates of B and C. [Cambridge AS & A Level Mathematis 979, Paper 1 QS June 29]

34 e rawing urves You an alwas plot a urve, point b point, if ou know its equation. Often, however, all ou need is a general idea of its shape and a sketh is quite suffiient. Figures 2.25 and 2.26 show some ommon urves of the form = x" for n = 1, 2, 3 and 4 and = _l_ for n = 1 and 2. x" Curves of the form = x" for n = 1, 2, 3 and 4 ill :E :r ea () =x (al n = l,=x (bl n = 2, = x 2 (l n = 3,=x 3 (dl n = 4, = x 4 Figure 2.25 G How are the urves for even values of n different from those for odd values of n? Stationar points A turning point is a plae where a urve hanges from inreasing (urve going up) to dereasing (urve going down), or vie versa. A turning point ma be desribed as a maximum (hange from inreasing to dereasing) or a minimum (hange from dereasing to inreasing). Turning points are examples of stationar points, where the gradient is zero. In general, the urve of a polnomial of order n has up to n- 1 turning points as shown in figure 2.26.

35 a maximum point Gl E C) Gl Cl Gl ea : :s C) stationar point. a minimum point \ Figure , There are some polnomials for whih not all the stationar points materialise, as in the ase of = x 4-4x 3 + 5x 2 (whose urve is shown in figure 2.27). To be aurate, ou sa that the urve of a polnomial of order n has at most n- 1 stationar points. Figure 2.27

36 Behaviour for large x (positive and negative) What an ou sa about the value of a polnomial for large positive values and large negative values of x? As an example, look at f(x) = x 3 + 2x 2 + 3x+ 9, and take 1 as a large number. Similarl, f(looo) = = f( -1) = = Note 1 The term x 3 makes b far the largest ontribution to the answers. lt is the dominant term. For a polnomial of order n, the term in xn is dominant as ---7 ± oo. 2 In both ases the answers are extremel large numbers. You will probabl have notied alread that awa from their turning points, polnomial urves quikl disappear off the top or bottom of the page. For all polnomials as x ---7 ±oo, either f(x) oo or f(x) oo. When investigating the behaviour of a polnomial of order n as x -7 ± oo, ou need to look at the term in xn and ask two questions. m Is n even or odd? (ii) Is the oeffiient of x" positive or negative? Aording to the answers, the urve will have one of the four tpes of shape illustrated in figure Intersetions with the x and axes The onstant term in the polnomial gives the value of where the urve intersets the axis. So = x 8 + 5x x rosses the axis at the point (, 23). Similarl, = x 3 + x rosses the axis at (, ), the origin, sine the onstant term is zero. When the polnomial is given, or known, in fatorised form ou an see at one where it rosses the x axis. The urve = ( x- 2 x- 8 x- 9), for example, rosses the x axis at x = 2, x = 8 and x = 9. Eah of these values makes one of the brakets equal to zero, and so=.

37 n even n odd. E r::n. ea :s u oeffiient of x" positive oeffiient of x" negative, - \,, \, \ -,. - \,-,,,_/ '.,,/ I Figure 2.28 EAMPLE 2.13 Sketh the urve = x 3-3x 2 - x+ 3 = (x+ 1) (x-1) (x- 3). -, J I \ I '" - 2 =il- 3x2+x+ 3 4 Figure 2.29 Sine the polnomial is of order 3, the urve has up to two stationar points. The term in x 3 has a positive oeffiient ( +1) and 3 is an odd number, so the general shape is as shown on the left of figure The atual equation = x 3-3x 2 - x+ 3 = (x+ 1) (x- 1) (x-3) tells ou that the urve: -rosses the axis at (, 3) -rosses the x axis at ( -1, ), (1, O) and (3, ). This is enough information to sketh the urve (see the right of figure 2.29).