YEARS. The Improving Mathematics Education in Schools (TIMES) Project INTRODUCTION TO COORDINATE GEOMETRY. NUMBER AND ALGEBRA Module 29

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1 9 1 YEARS The Improving Mathematics Education in Schools (TIMES) Project INTRODUCTION TO COORDINATE GEOMETRY A guide for teachers - Years 9 1 June 11 NUMBER AND ALGEBRA Module 9

2 Introduction to Coordinate Geometr (Number and Algebra : Module 9) For teachers of Primar and Secondar Mathematics 51 Cover design, Laout design and Tpesetting b Claire Ho The Improving Mathematics Education in Schools (TIMES) Project 9 11 was funded b the Australian Government Department of Education, Emploment and Workplace Relations. The views epressed here are those of the author and do not necessaril represent the views of the Australian Government Department of Education, Emploment and Workplace Relations. The Universit of Melbourne on behalf of the International Centre of Ecellence for Education in Mathematics (ICE EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 1 (ecept where otherwise indicated). This work is licensed under the Creative Commons Attribution- NonCommercial-NoDerivs 3. Unported License

3 The Improving Mathematics Education in Schools (TIMES) Project INTRODUCTION TO COORDINATE GEOMETRY A guide for teachers - Years 9 1 June 11 NUMBER AND ALGEBRA Module 9 Peter Brown Michael Evans David Hunt Janine McIntosh Bill Pender Jacqui Ramagge 9 1 YEARS

4 {1} A guide for teachers INTRODUCTION TO COORDINATE GEOMETRY ASSUMED KNOWLEDGE Fluenc with the arithmetic of the rational numbers Knowledge of ratios Congruent and similar triangles Basic algebraic notation Fluenc with algebraic epressions and equations Basic plotting points in the Cartesian plane including plotting points from a table of values. MOTIVATION Coordinate geometr is one of the most important and eciting ideas of mathematics. In particular it is central to the mathematics students meet at school. It provides a connection between algebra and geometr through graphs of lines and curves. This enables geometric problems to be solved algebraicall and provides geometric insights into algebra. The invention of calculus was an etremel important development in mathematics that enabled mathematicians and phsicists to model the real world in was that was previousl impossible. It brought together nearl all of algebra and geometr using the coordinate plane. The invention of calculus depended on the development of coordinate geometr. CONTENT It is epected that students have met plotting points on the plane and have plotted points from tables of values of both linear and non linear functions. The number plane (Cartesian plane) is divided into four quadrants b two perpendicular aes called the -ais (horizontal line) and the -ais (vertical line). These aes intersect at a point called the origin. The position of an point in the plane can be represented b an ordered pair of numbers (, ). These ordered pairs are called the coordinates of the point.

5 The Improving Mathematics Education in Schools (TIMES) Project {} The point with coordinates (4, ) has been plotted on the Cartesian plane shown. The coordinates of the origin are (, ). 4 -coordinate 3 (4, ) Once the coordinates of two points are known the distance between the two points and midpoint of the interval joining the points can be found coordinate 4 DISTANCE BETWEEN TWO POINTS Distances are alwas positive, or zero if the points coincide. The distance from A to B is the same as the distance from B to A. We first find the distance between two points that are either verticall or horizontall aligned. Find the distance between the following pairs of points. a A(1, ) and B(4, ) b A(1, ) and B(1, 3) a The distance AB = 4 1 = 3 Note: The distance AB is obtained from the difference of the -coordinates of the two points. A(1, ) B(4, ) b The distance AB = 3 ( ) = 5 B(1, 3) Note: The distance AB is obtained from the difference of the -coordinates of the two points. A(1, )

6 {3} A guide for teachers The eample above considered the special cases when the line interval AB is either horizontal or vertical. Pthagoras theorem is used to calculate the distance between two points when the line interval between them is neither vertical nor horizontal. The distance between the points A(1, ) and B(4, 6) is calculated below. B(4, 6) A(1, ) C(4, ) AC = 4 1 = 3 and BC = 6 = 4. B Pthagoras theorem, AB = = 5 And so AB = 5 The general case We can obtain a formula for the length of an interval. Suppose that P( 1, 1 ) and Q(, ) are two points. Q(, ) 1 P( 1, 1 ) 1 X(, 1 ) Form the right-angled triangle PQX, where X is the point (, 1 ), PX = 1 or 1 and QX = 1 or 1 depending on the positions of P and Q. B Pthagoras theorem: PQ = PX + QX = ( 1 ) + ( 1 ) Therefore PQ = QP = ( 1 ) + ( 1 ) Note that ( 1 ) is the same as ( 1 1 ) and therefore it doesn t matter whether we go from P to Q or from Q to P the result is the same.

7 The Improving Mathematics Education in Schools (TIMES) Project {4} Find the distance between the points A( 4, 3) and B(5, 7). In this case, 1 = 4, = 5, 1 = 3 and = 7. AB = ( 1 ) + ( 1 ) = (5 ( 4)) + (7 ( 3)) = = 181 Thus, AB = 181 Note that we could have chosen 1 = 5, = 4, 1 = 7 and = 3 and still obtained the same result. As long as ( 1, 1 ) refers to one point and (, ) the other point, it does not matter which one is which. EXERCISE 1 Show that the distance between the points A(a, b) and B(c, d) is the same as the distance between the points P(a, d) and Q(c, b) the points U(b, a) and V(d, c) Illustrate both of these. EXERCISE The distance between the points (1, a) and (4, 8) is 5. Find the possible values of a and use a diagram to illustrate. THE MIDPOINT OF AN INTERVAL The coordinates of the midpoint of a line interval can be found using averages as we will see. We first deal with the situation where the points are horizontall or verticall aligned. Find the coordinates of the midpoint of the line interval AB, given: a A(1, ) and B(7, ) b A(1, ) and B(1, 3)

8 {5} A guide for teachers a AB is a horizontal line interval, the midpoint is at (4, ), since 4 is halfwa between 1 and 7. A(1, ) B(7, ) Note: 4 is the average of 1 and 7, that is, 4 = b The midpoint of AB has coordinates 1, 1. Note that 1 is the average of 3 and. B(1, 3) A(1, ) When the interval is not parallel to one of the aes we take the average of the -coordinate and the -coordinate. This is proved below. 8 B(5, 8) M(, ) T A(1, ) S 1 5 Let M be the midpoint of the line AB. Triangles AMS and MBT are congruent triangles (AAS), and so AS = MT and MS = BT. Hence the -coordinate of M is the average of 1 and 5. = = 3 The coordinate of M is the average of and 8. = + 8 = 5 Thus the coordinates of the midpoint M are (3, 5). The general case We can find a formula for the midpoint of an interval. Suppose that P( 1, 1 ) and Q(, ) are two points and let M(, ) be the midpoint.

9 The Improving Mathematics Education in Schools (TIMES) Project {6} Q(, ) M(, ) T 1 P( 1, 1 ) S 1 Triangles PMS and MQT are congruent triangles (AAS), and so PS = MT and MS = QT. Hence the -coordinate of M is the average of 1 and, and -coordinate of M is the average of 1 and. Therefore = 1 + and = 1 + Midpoint of an interval The midpoint of an interval with endpoints P( 1, 1 ) and Q(, ) is + 1, + 1. Take the average of the -coordinates and the average of the -coordinates. Find the coordinates of the midpoint of the line interval joining the points (6, 8) and ( 3, ). The midpoint has coordinates, 6 + ( 3), 8 + = 3, 5 If C(3, 6) is the midpoint of line interval AB and A has coordinates ( 1, 1), find the coordinates of B. Let the coordinates of B be ( 1, 1 ). 1 + ( 1) = 3 and = = = 1 so 1 = 7 so 1 = 11. Thus B has coordinates (7, 11).

10 {7} A guide for teachers EXERCISE 3 A square has vertices O(, ), A(a, ), B(a, a) and C(, a). a Find the midpoint of the diagonals OB and CA. b Find the length of a diagonal of the square and the radius of the circle in which OABC is inscribed. c Find the equation of the circle inscribing the square. THE GRADIENT OF A LINE Gradient of an interval The gradient is a measure of the steepness of line. There are several was to measure steepness. In coordinate geometr the standard wa to define the gradient of an interval AB is rise run where rise is the change in the -values as ou move from A to B and run is the change in the -values as ou move from A to B. We will usuall the pronumeral m for gradient. B(5, 6) A(, 1) rise run Given the points A(, 1) and B(5, 6): gradient of interval AB = rise run = = 5 3 Notice that as ou move from A to B along the interval the -value increases as the -value increases. The gradient is positive. Given the points A(, 7) and B(6, 1) gradient of interval AB = rise run = = 6 4 = 3 or gradient of interval BA = rise run = = 3 A(, 7) B(6, 1)

11 The Improving Mathematics Education in Schools (TIMES) Project {8} Notice that in this case as we move from A to B the value decreases as the value increases. The gradient is negative. Similarl the gradient of BA = 3 which is the same as the gradient of AB. In general: gradient of line interval AB = rise run m = 1 1 A( 1, 1 ) B(, ) 1 (rise) 1 (run) Note that since 1 = it does not matter which point we take as the first and which point we take as the second. If the interval is vertical, the run is zero and the gradient of the interval is not defined. This is shown b interval AB. P Q B If the interval is horizontal, the rise is zero as shown b interval PQ. The gradient of the interval is zero. A Gradient of PQ is zero Gradient of AB is not defined Gradient of a line The gradient of a line is defined to be the gradient of an interval within the line. This definition depends on the fact that two intervals on a line have the same gradient. Q B P Y A X Suppose AB and PQ are two intervals on the same straight line. Draw right-angled triangles ABX and PQY with sides AX and PY parallel to the -ais and sides BX and QY parallel to the -ais. Triangle ABX is similar to triangle PQY since the corresponding angles are equal. Therefore: QY PY = BX AX That is, the intervals have the same gradient.

12 {9} A guide for teachers A line passes through the points (1, ) and (5, 1). Find its gradient. gradient = 1 1 = = (5, 1) 1 = 8 (1, ) 5 1 = 4 A line passes through the point (5, 7) and has gradient 3. Find the -coordinate of a point on the line when = 13. Gradient of the line = 6 5 = 3 18 = ( 5) 6 5. Thus, (, 13) 13 7 (5, 7) 5 9 = 5 = 14 EXERCISE 4 Find the gradient of the line passing through (a, b) and (, c) Intercepts The -intercept of a line is the point at which it crosses the -ais. The -intercept of a line is the point at which it crosses the -ais. A C B D In the diagram to the left the -intercept is at A and the -intercept at B.

13 The Improving Mathematics Education in Schools (TIMES) Project {1} The second diagram shows a line parallel to the -ais and it has a -intercept at C. The third diagram shows a line parallel to the -ais and it has an -intercept at D. EQUATION OF A STRAIGHT LINE When we plot points which satisf the equation = + 1 we find that the lie in a straight line. = + 1 (, 5) Can we find the equation of the line given suitable geometric information about the line? The following (1, 3) shows that this can be done given the gradient of the line and the -intercept. ( 1, 1) (, 1) THE LINE = 3X + Consider the line with gradient 3 and -intercept. This passes through the point A(, ). Let B(, ) be an point on this line. Gradient of interval AB = rise run = B(, ) The gradient of the line is 3. = A(, ) So, = 3 Rearranging = 3 = 3 + So the coordinates of B(, ) satisfies = 3 +. This is called the equation of the line. Conversel suppose that B(, ) satisfies the equation = 3 +, then = 3 and it passes the point (, ) so the point lies on the line with gradient 3 and -intercept. We summarise this b saing that the equation of the line is = 3 +. THE EQUATION = m + c Consider the line with gradient m and -intercept c. If passes through the point A(, c). Let B(, ) be an point on this line. Gradient of interval AB= c = c B(, ) We know the gradient of the line is m. A(, c)

14 {11} A guide for teachers Therefore c = m c = m = m + c That is, the line in the cartesian plane with gradient m and -intercept c has equation = m + c. Conversel, the points whose coordinates satisf the equation = m + c alwas lie on the line with gradient m and -intercept c. Vertical and horizontal lines Vertical lines In a vertical line all points have the same -coordinate, but the -coordinate can take an value. The equation of the vertical line through the point (6, ) is = 6. The -ais intercept is 6. All the points on this line have -coordinate 6. (6, ) In general, the equation of the vertical line through P(a, b) is = a. Because this line does not have a gradient it cannot be written in the form = m + b. Horizontal lines A horizontal line has gradient. In a horizontal line all points have the same -coordinate, but the -coordinate can take an value. The equation of the horizontal line through the point (, 5) is = 5. The equation of the horizontal line through the point (9, 5) is = 5. (, 5) In general, the equation of the horizontal line through P(a, b) is = b.

15 The Improving Mathematics Education in Schools (TIMES) Project {1} Write down the gradient and -intercept of the line with equation = 3 4. The gradient of the line is 3 and the -intercept is 4. Sometimes an equation needs to be rearranged before the gradient and -intercept can be determined. Consider the following eample. Rewrite the equation + 3 = 6 in the form = m + c and hence find the value of the gradient and -intercept. +3= 6 so that is, Thus 3= 6 = 3 = 3 + The gradient of the line is 3 and the -intercept is. Equation of a line given its gradient and a point on the line We want to find the equation of the line with gradient m and which passes through the point P( 1, 1 ). P(, ) Let P(, ) be an point with 1 on the line passing through the point A( 1, 1 ) and let m be the gradient of this line. Using gradient, m = 1 1 and 1 = m( 1 ) A( 1, 1 ) 1 (run) 1 (rise) This is the equation of the straight line with gradient m passing through the point A( 1, 1 ). Find the equation of the line that passes through the point (, 3) with gradient 4. The equation for this line is: 1 = m( 1 ) 3 = 4( ( ))

16 {13} A guide for teachers that is, 3 = 4 8 = 4 5 Note that it is usual to give the answer in the form = m + c Equation of a straight line given two points Given two points A( 1, 1 ) and B(, ) the equation of the line passing through the two points can be found. The gradient m of the line passing through A( 1, 1 ) and B(, ) = 1 1, 1 Substituting into 1 = m( 1 ) gives 1 = 1 1 ( 1), 1 GRAPHING STRAIGHT LINES Using the equation to sketch the line If ou are given an equation of a straight line and asked to draw its graph all ou need to do is find two points whose coordinates satisf the equation and plot the points. There are two commonl used methods to find two points. Using the -intercept and one other point Using the -intercept and a second point the equation can be found. Draw the graph of = + 3. The -intercept is 3 and the gradient. (1, 5) Substitute = 1, so = 5 giving the point (1, 5) lies on the line. Plot the two points and draw the line through them. (, 3) 1 This method does not work if the line is parallel to the -ais.

17 The Improving Mathematics Education in Schools (TIMES) Project {14} The gradient of a line is 6 and the -intercept is. Find the equation of the line and sketch it. Using the = m + c form for the equation of a straight line. The equation of the line is = 6 +. The point (, ) lies on the line. Substitute = 1 in = 6 + = = 4 A(1, 4) The point (1, 4) lies on the line. The graph is shown. -intercept -intercept method In this method both intercepts are found. The -intercept is found b substituting = and The -intercept is found b substituting =. This method does not work if the line is parallel to an ais or passes through the origin. Using the -intercept -intercept method sketch the graph of: a = 3 4 b = a When =, = 4 When =, 3 4 = = 4 = 3 4 = 4 3 4

18 {15} A guide for teachers b When =, = 3 = = = 3 When =, + 6 = = 6 = 3 The general form for the equation of a straight line The equations = 3, = 6 and 3 = 6 can be written as + + 3, 6 = and 3 6 = respectivel. The general form for the equation of a line is a + b + c = where a, b and c are constants and a or b. The equation of ever line can be put in general form. The general form is not unique. The equation + +1 = is the same straight line as =. EXERCISE 5 An equilateral triangle ABC has coordinates O(, ), B(a, ) and C(c, d). a Find c and d in terms of a b using the fact that OB = BC = CO. b Find the equation of the lines OB, BC and CO. PARALLEL AND PERPENDICULAR LINES Parallel lines If two lines l 1 and l are parallel then corresponding angles are equal. Conversel, if corresponding angles are equal then the lines are parallel. l 1 l Theorem Two lines are parallel if the have the same gradient and conversel, two lines with the same gradient are parallel. Proof In the diagram, two lines are drawn and the right-angled triangles PQX and ABY are added with QX = BY.

19 The Improving Mathematics Education in Schools (TIMES) Project {16} P A Q X B Y If the lines are parallel then /PQX = /ABY (corresponding angles). The two triangles are congruent b the AAS test. Therefore PX = PY and PX QX = AY BY. That is, the gradients are equal. Conversel. If the gradients are equal PX QX = AY BY. Now QX = BY and therefore PX = AY. Hence the triangles QPX and ABY are congruent b the SAS test. Hence the corresponding angles PQX and ABY are equal and the lines are parallel. Show that the line passing through the points A(6, 4) and B(7, 11) is parallel to the line passing through P(, ) and Q(, 14). Gradient of AB = = 7 Gradient of PQ = 14 = 7 The two lines have the same gradient and so are parallel. Find the equation of the line that is parallel to the line = + 6 and passing through the point A(1, 1). The gradient of the line = + 6 is. Therefore the line through the point A(1, 1) parallel to = + 6 has equation: 1 = m( 1 ) 1 = ( 1) = + 1

20 {17} A guide for teachers Perpendicular lines When we draw = 3 and draw a line perpendicular to it passing through the origin then it is clear that = a where a is a small positive number. We will show that the equation is = 1 3. We are now going to show the surprising result that if two lines are perpendicular then the product of their gradients is 1 (or if one is vertical and the other horizontal). The converse is also true. That is, If the product of the gradients of two lines is 1 then the are perpendicular. We first consider the case when both lines pass through the origin. Draw two lines passing through the origin with one of the lines having positive gradient and the other negative gradient. P Q A O B Form right-angled triangles OPQ and OAB with OQ = OB. Gradient of the line OA = AB OB Gradient of the line through OP = OQ PQ The product = OQ PQ AB OB = OQ PQ AB OQ (since OQ = OB) = AB PQ If the lines are perpendicular, /POQ = /AOB. P 9 Q A O B

21 The Improving Mathematics Education in Schools (TIMES) Project {18} Therefore OPQ OAB (AAS), so PQ = AB and the product, AB PQ, of the gradients is 1. Conversel If the product is 1, then AB = PQ, so OAB OPQ (SAS). Therefore /POQ = /AOB and so /AOP = 9. We have now proved the result for lines through the origin. If we are given two lines anwhere in the plane, we can draw lines through the origin parallel to the given two lines. The gradient of each new line is the same as the gradient of the corresponding original line. So the result holds for lines that do not necessaril pass through the origin. Show that the line through the points A(6, ) and B(, 1) is perpendicular to the line through P(8, 1) and Q(4, 8). Gradient of AB = 1 6 = Gradient of PQ = = 4 = 1 (Gradient of AB) (gradient of PQ) = 1 = 1 Hence the lines are perpendicular. Parallel and perpendicular lines If two non-vertical lines are parallel then the have the same gradient. Conversel if two non-vertical lines have the same gradient then the are parallel. If two non-vertical lines are perpendicular then the product of their gradients is 1. Conversel if the product of the gradients of two lines is 1 then the are perpendicular.

22 {19} A guide for teachers Find the equation of the line which passes through the point (1, 3) and is perpendicular to the line whose equation is = + 1. Gradient of the line = + 1 is. Gradient of a line perpendicular to this line is 1. The required equation is 3 = 1 ( 1) ( 3) = ( 1) + = 7 Thus the equation of the required line is + = 7. PROOFS WITH COORDINATE GEOMETRY Coordinate geometr can be used to prove results in Euclidean Geometr. An important aspect of doing this is placing objects on the Cartesian plane in a wa that minimises calculations. Prove that the midpoints of a parallelogram bisect each other using coordinate geometr. Let the coordinates of the vertices be O(, ), A(a, ), B(a + c, d) and C(c, d). There is no loss in generalit in placing the vertices of the parallelogram on the Cartesian plane in this wa. The midpoint M of OB = a + c, d The midpoint N of AC = a + c, d M = N and so the midpoints coincided which means that the diagonals bisect each other. Prove that the diagonals of a rhombus bisect each other at right angles using coordinate geometr. Let the coordinates of the vertices be O(, ), A(a, ), B(a + c, d) and C(c, d). Because it is a rhombus all the sides are of equal length.

23 The Improving Mathematics Education in Schools (TIMES) Project {} OA = AB = BC = CO Gradient of OB = d a + c and, Gradient of AC = d c a The product of the gradients of the diagonals = OA = a and b Pthagoras theorem, AB = c + d so a = c + d. Hence d = (c a ) Thus the product of the gradients of the diagonals = 1. d a + c d c a = d c a. EXERCISE 6 In an triangle ABC prove that AB + AC = (AD + DC ) Where D is the midpoint of BC. EXERCISE 7 Prove that set of points equidistant from two given points is a straight line. EXERCISE 8 Prove that the lines joining the midpoints of opposite sides of a quadrilateral and the lines joining the midpoints of its diagonals meet in a point and bisect each other. LINKS FORWARD Coordinate geometr leads into man other topics in school mathematics. The techniques of coordinate geometr are used in calculus, functions, statistics and man other important areas. HISTORY There were three facets of the development of coordinate geometr. The invention of a sstem of coordinates The recognition of the correspondence between geometr and algebra The graphic representation of relations and functions

24 {1} A guide for teachers The Greek mathematician Menaechmus (38 3 BC) proved theorems b using a method that was ver close to using coordinates and it has sometimes been maintained that he had introduced coordinate geometr. Apollonius of Perga (6 19 BC) dealt with problems in a manner that ma be called an coordinate geometr of one dimension; with the question of finding points on a line that were in a ratio to the others. The results and ideas of the ancient Greeks provided a motivation for the development of coordinate geometr. Coordinate geometr has traditionall been attributed to René Descartes ( ) and Pierre de Fermat ( ) who independentl provided the beginning of the subject as we know it toda. ANSWERS TO EXERCISES EXERCISE 1 AB = PQ = UV = (d b) + (c a) V(d, t) = P(a, d) B(c, d) C(b, a) B(c, d) A(a, b) Q(c, b) A(a, b) EXERCISE a = 4 or a = 1 A(1, 1) (4, 8) B(1, 4) EXERCISE 3 a a, a b diagonal = a, radius = a c a + a = a

25 The Improving Mathematics Education in Schools (TIMES) Project {} EXERCISE 4 b c a EXERCISE 5 a c = a, d = 3a b OA: =, OC: = d c, OB: = d c a ( a) EXERCISE 6 Place the triangle so D is at the origin. Then let the coordinates of B and C be ( a, ) and (a, ) respectivel. Let the coordinates of A be (d, c). AB = c + (d + a) and AC = c + (d a) So AB + AC = c + d + a AD = c + d and DC = a Hence AB + AC = (AD + DC ) EXERCISE 7 Let P(, ) be a point equidistant from A(a, b) and C(c, d) PA = PC (c a) + (d b) = d + c a b EXERCISE 8 Let the coordinates of the vertices be O(, ), A(a, c), B(m, n) and D(b, ) Show that the midpoint of all the required line segments has coordinates 1 4 (m + b + a), 1 (n + c) 4

26 The aim of the International Centre of Ecellence for Education in Mathematics (ICE-EM) is to strengthen education in the mathematical sciences at all levelsfrom school to advanced research and contemporar applications in industr and commerce. ICE-EM is the education division of the Australian Mathematical Sciences Institute, a consortium of 7 universit mathematics departments, CSIRO Mathematical and Information Sciences, the Australian Bureau of Statistics, the Australian Mathematical Societ and the Australian Mathematics Trust. The ICE-EM modules are part of The Improving Mathematics Education in Schools (TIMES) Project. The modules are organised under the strand titles of the Australian Curriculum: Number and Algebra Measurement and Geometr Statistics and Probabilit The modules are written for teachers. Each module contains a discussion of a component of the mathematics curriculum up to the end of Year 1.