MODULE 3LINEAR LAWCHAPTER 2 : LINEAR LAW Page Contents3.0 CONCEPT MAP 23.1UNDERSTAND AND USE THE CONCEPT OF LINES OF BEST FIT Draw lines of best fit by inspection of given data Exercises 1 Write equations for lines of best fit Exercises 2 Determine values of variables from a) lines of best fit b) equations of lines of best fit Exercises 333.1.13 43.1.23.1.3563.2 3.2.1APPLY LINEAR LAW TO NON- LINEAR RELATIONS Reduce non-linear relations to linear form Exercises 4 Determine values of constants of non-lineas relations given Exercises 57 73.2.283.2.3Steps to plot a straight lineExercises 6103.3 SPM Questions 3.4 Assessment test 3.5 Answers17 27 301CHAPTER 2 : LINEAR LAWLINEAR LAWLinear RelationNon Linear RelationReducing non-linear relation to linear formLines of best fitDetermining values of constants Formation of equationObtaining information2UNDERSTAND AND USE THE CONCEPT OF LINES OF BEST FITTHE CONCEPT OF LINES OF BEST FIT.The properties of the line of best fit. The straight line is drawn such away that it passes through as many points as possible. The number of points that do not lie on the straight line drawn should be more or less the same both sides of the straight line.Example. Draw the line of best fit for the given graph below. y x EXERCISE 1 1. Draw the line of best fit for the following diagrams. xExercises 1(a) y x (b) y x3Exercises 2Example 1 Example 2 yy x (3 , 8) 2 0 1 x (1, 3)x (3, 7)xThe equation that can be formed is: Y = mX + c 82 =2Gradient , m = 30 Y intercept = 2 Therefore, y = 2x + 2Linear form0x The equation that can be formed is: Y = mX + c 73 =2Gradient , m = 3 1 The straight line also passes through the Point (1, 3) xY Another equation can be formed is, 3 = 2(1) + c .c = 1 Therefore, y = 2x + 1 Linear formy a) 9 .x (8, 1) 0 x b)y x (3, 6)01x4y c) x ( 5 , 2) d) x (8, 8)x ( 6,4) 0 7 x 0 x3.1.3 Determine values of variables from:a)lines of best fity X (7, k)b) Equations of lines of best fitExample 1Example 1y x (3, k) y = 3x + 4x (2, 4) 2 42 Gradient = =1 20 k4 Therefore,=1 72 .k 2 = 5 .k = 7 x 0 4xThe straight line also passes through the point (3, k) Another equation that can be formed is .k = 3(3) + 4 .k = 9 + 4 = 135a) lines of best fiti) y X (6, k)b) Equations of lines of best fiti) y Xx (3, 3) 1 0 x (k, 3) x 0.y = 2x - 3xii)y X (k, 6)ii)y y= 6 - xx (3, k) x (5, 2) 0 0 7 x xiii)y X (8, 8)iii) y X (4, k)x (6, 4) .y = x + 2 x 0 (k, 2) 0 x x6APPLY LINEAR LAW TO NON- LINEAR RELATIONS3.2.1 Reduce non- linear relations to linear form Example 12Y = mX + cExample 1x y = ax + bNon- lineara) y = pxq log10 y = log10 (pxq) log10 y = log10 p + log10 xqNon- linear Take log of Both sidersyaxx2 =+ xxxb y1 =a+x xb Y = c + XmDivided by xLinear formlog10 y = log10 p + qlog10 x Y = c mXLinear formExercises 4Equation a) y2 = ax + b Linear form Y X m cb) y = ax2 + bxc)ab =+ 1 yxd) y2 = 5x2 + 3xe) y = 3 x +5 xf) y = abxg) y =4 (x + b)2 a27Exercises 5xy Example 1 X (11, 6)a)y x 10x (1, 2) x2 The above figure shows part of a straight line graph drawn to represent the equation . xy = ax2 + b Find the value of a and b.x (4, 2) 0x The above figure shows part of a straight line graph drawn to represent the equation . y = ax2 + bx Find the value of a and b,62 Gradient , a = 11 1 2 A= 5 22 Therefore, xy = x +b 5 Another equation that can be formed is 2 .(1)(2) = (1) 2 + b 5 28 b, = 2 -= 55 2 8 Hence a = 5 , b= 58y b) x (1, 9) c) x (3, 6)1 .lg x 0 The above figure shows part of a straight line graph drawn to represent the equation .of y = axb Find the value of a and b,1 x The above figure shows part of a straight line graph drawn to represent the equation 0 1 .of xy = a + bx Find the value of a and b,9STEPS TO PLOT A STRAIGHT LINE 3.3.1. Using a graph paper.Change the non-linear function with variables x and y to a linear form Y = mX + cConstruct a tableChoose a suitable scale Label both axesLine of best fitDetermine : gradient m Y-intercept cQUESTIONS X Y 2 2 3 9 4 20 5 35 6 54TableThe above table shows the experimental values of two variables, x and y. It is know that x and y are related by the equation y = px2 + qxNon- lineara) Draw the line of best fit for b) From your graph, find, i)the initial velocity ii)the accelerationy against x x10STEP 1 y = px2 + qx ypx = xx y = px + q x Y = mX + c2Reduce the non-linear To the linear form The equation is divided throughout dy x To create a constant that is free from x On the right-hand side i.e, qqx + xLinear form Y = mX + cSTEP 2construct table.x23456.y y x2920355413559STEP 3Using graph paper, - Choose a suitable scale so that the graph drawn is as big as possible. - Label both axis - Plot the graph of Y against X and draw the line of best fit111210 x 8 x 6 x 4 x 2 x123456x-2-4STEP 4From the graph, find m and c9 1 Gradient , m ==2 62Construct a right-angled triangle, So that two vertices are on the line of best fit, calculate the gradient, mY- intercept = c.= -3Determine the Y-intercept, c .from the straight line graph121. The table below shows some experimental data of two related variable x and y It is know that x and y are related by an equation in the form y .x = ax + bx2 ,1where. a and b are constants 2 3 4 .y 7 16 24 24 . y a) Draw the straight line graphagainst x of x b) Hence, use the graph to find the values of a and b 5 16 6 0 7 -241314.x y0 1.672 1.94 2.216 2.418 2.6510 2.79It is known that x and y are related by an equation in the form axb .y =+, where a and b are constants. yy a) Draw the straight line graph y2 against x b) Hence, use the graph to find the values of a and b15161. SPM 2003( paper 1, question no 10) x and y are related by equation y = px2 + qx , where p and q are constants. A straight y against x, as show in Diagram belowline is obtained by plotting x y x x (2, 9)x (6, 1) 0 Calculate the values af p and q. x17Diagram below shows a straight line graph of y xy against x xx (2, k)x (h, 3) a. x Gigen that y = 6x x2 , calculate the value of k and of h18SPM 2005( paper 1, question no 13) The variable x and y are related by the equation y = kx4 where k is a constant. a) Convert the equation y = kx4 to linear form b) Diagram below shows the straight line obtained by plotting log10 y against log10 x log10 y. x (2, h) (0, 3) x log10 y 0 Find the value of i) log10 k ii) h.19SPM 2003( paper 2, question no 7) Table 1 shows the values of two variables, x and y, obtained from on experiment. It is x 2 , where p and k are constants.known that x and y are related by the equation y = pk x y 1.5 1.59 2.0 1.86 2.5 2.40 3.0 3.17 3.5 4.36 4.0 6.76(a) Plot log y against x2. Hence, draw the line of best fit (b) Use the graph in (a) to find value of (i )p ( ii ) k2021Table 1 shows the values of two variables, x and y, obtained from on experiment. Variables x and y are related by the equation y = pkx , where p and k are constants. x y 2 3.16 4 5.50 6 9.12 8 16.22 10 28.84 12 46.77(a) Plot log10 y against x by using scala of 2 cm to 2 units on the x-axis and 2 cm to 0.2 unit on the log 10-axis. Hence, draw the line of best fit (b) Use the graph in (a) to find value of (i )p ( ii ) k2223Table 1 shows the values of two variables, x and y, obtained from on experiment. r ,The variables x and y are related by the equation y = px + x 1.0 y 5.5 2.0 4.7 px 3.0 5.0 4.0 6.5 5.0 7.7 5.5 8.4(b)where p against x2constants.scala of 2 cm to 5 units on both axes. Plot xy and r are by using Hence, draw the line of best fit (b) Use the graph in (a) to find value of (i )p ( ii ) r24251.xy(0, h).x 0(8, k) xThe above figure shows part of a straight-line graph drawn to represent the equation 61 -Find the value of h and of ky = x226.x Y1 - 1.52 93 43.54 1145 232.5It is know that x and y are related by an equation in the from .y = ax3 - bx , where a and b are constants. a) Change the equation to the linear form and hence draw the straight line graph for values of x and y b) From your graph, I).determine the values of a and b II).find the value of y when x = 3.52728Exercises 2 a) b) c) d) y y y y = = = = -x+9 3x 3 -x+7 2x - 8 a)Exercises 3 i) k = 5 ii) k = 1 iii) k = 5 b) i) k = 3 ii) k = 3 iii) k = 6Exercises 4Exercises 5Ya) b) c) d) .y2Xx xma acb ba) a = 1, b = 2 b) a = 11, b = 8 c) a = 3 , b = 3y x 1 y y2 x1 xx x xb a5 3 lg b1 a3 5 lg aExercises 51) a = - 1 , b = 10 2) a = 0.5 , b = 2.8e) y x f) lg y g)yx2 a2b aSPM Questions 1) p = -2 , q = 13 2) h = 3 , k = 4 3) a) lg y = 4lg x + lg k b) k = 100, h = 11 4) p = 1.259 , k = 1.109 5) p = 1.82, k = 1.307 6) P = 1.37, r = 5.48Assessment test1) h = 6 ,k=22)a)y= ax2 b x b) i. a = 2 , b = 3.6 ii. y = 73.529