COORDINATE GEOMETRY - AMSI

1 INTRODUCTION TO COORDINATE GEOMETRYIININTROThe Improving Mathematics Education in Schools (TIMES) Project NUMBER AND ALGEBRA Module 29 a guide for teachers - Years 9 10 June 2011 Introduction to COORDINATE GEOMETRY (Number and Algebra : Module 29)For teachers of Primary and Secondary Mathematics510 Cover design, Layout design and Typesetting by Claire HoThe Improving Mathematics Education in Schools (TIMES) Project 2009 2011 was funded by the Australian Government Department of Education, Employment and Workplace views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations. The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution NonCommercial NoDerivs Unported License.

2 Nc BrownMichael EvansDavid HuntJanine McIntoshBill PenderJacqui RamaggeIntroductIonitoicoordIn tCiaCoeCtrGIInIntroIntroduci CaerGmyntdmyC(NrbAl(myCiarCarg(nii:NrMIo Gbg2r9ci)t(yrIntroducIiuc Cordcu Module 29 Introdunci nCuaeGu mnyn(ua mnNbAlIntroducc{1}A guide for teachersASSUMED KNOWLEDGE Fluency with the arithmetic of the rational numbers Knowledge of ratios Congruent and similar triangles Basic algebraic notation Fluency with algebraic expressions and equations Basic plotting points in the Cartesian plane including plotting points from a table of GEOMETRY is one of the most important and exciting ideas of mathematics. In particular it is central to the mathematics students meet at school. It provides a connection between algebra and GEOMETRY through graphs of lines and curves. This enables geometric problems to be solved algebraically and provides geometric insights into invention of calculus was an extremely important development in mathematics that enabled mathematicians and physicists to model the real world in ways that was previously impossible.))))

3 It brought together nearly all of algebra and GEOMETRY using the COORDINATE plane. The invention of calculus depended on the development of COORDINATE is expected that students have met plotting points on the plane and have plotted points from tables of values of both linear and non linear number plane (Cartesian plane) is divided into four quadrants by two perpendicular axes called the x-axis (horizontal line) and the y-axis (vertical line). These axes intersect at a point called the origin. The position of any point in the plane can be represented by an ordered pair of numbers (x, y). These ordered pairs are called the coordinates of the TO COORDINATE GEOMETRY {2}The Improving Mathematics Education in Schools (TIMES) ProjectThe point with coordinates (4, 2) has been plotted on the Cartesian plane shown. The coordinates of the origin are (0, 0).Once the coordinates of two points are known the distance between the two points and midpoint of the interval joining the points can be BETWEEN TWO POINTSD istances are always positive, or zero if the points coincide.

4 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 vertically or horizontally the distance between the following pairs of A(1, 2) and B(4, 2) b A(1, 2) and B(1, 3)SOLUTIONa The distance AB = 4 1 = 3 0xyA(1, 2)B(4, 2)Note: The distance AB is obtained from the difference of the x coordinates of the two The distance AB = 3 ( 2) = 5 0xyB(1, 3)A(1, 2)Note: The distance AB is obtained from the difference of the y coordinates of the two 1 1 2 2 3 3 4 422(4, 2)x-coordinatey-coordinate3434xy{3}A guide for teachersThe example above considered the special cases when the line interval AB is either horizontal or vertical. Pythagoras theorem is used to calculate the distance between two points when the line interval between them is neither vertical nor distance between the points A(1, 2) and B(4, 6) is calculated (4, 6)C(4, 2)A(1, 2)AC = 4 1 = 3 and BC = 6 2 = Pythagoras theorem, AB2 = 32 + 42 = 25 And so AB = 5 The general caseWe can obtain a formula for the length of any interval.

5 Suppose that P(x1, y1) and Q(x2, y2) are two (x1, y1)X(x2, y1)Q(x2, y2)x2 x1y2 y1 Form the right angled triangle PQX, where X is the point (x2, y1), PX = x2 x1 or x1 x2 and QX = y2 y1 or y1 y2depending on the positions of P and Pythagoras theorem: PQ2 = PX2 + QX2 = (x2 x1)2 + (y2 y1)2 Therefore PQ = QP = (x2 x1)2 + (y2 y1)2 Note that (x2 x1)2 is the same as (x1 x1)2 and therefore it doesn t matter whether we go from P to Q or from Q to P the result is the same.{4}The Improving Mathematics Education in Schools (TIMES) ProjectEXAMPLEFind the distance between the points A( 4, 3) and B(5, 7).SOLUTIONIn this case, x1 = 4, x2 = 5, y1 = 3 and y2 = = (x2 x1)2 + (y2 y1)2 = (5 ( 4))2 + (7 ( 3))2 = 92 + 102 = 181 Thus, AB = 181 Note that we could have chosen x1 = 5, x2 = 4, y1 = 7 and y2 = 3 and still obtained the same result. As long as (x1, y1) refers to one point and (x2, y2) the other point, it does not matter which one is 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 2 The distance between the points (1, a) and (4, 8) is 5.

6 Find the possible values of a and use a diagram to MIDPOINT OF AN INTERVALThe coordinates of the midpoint of a line interval can be found using averages as we will first deal with the situation where the points are horizontally or vertically the coordinates of the midpoint of the line interval AB, given:a A(1, 2) and B(7, 2) b A(1, 2) and B(1, 3){5}A guide for teachersSOLUTIONa AB is a horizontal line interval, the 0xyA(1, 2)B(7, 2) midpoint is at (4, 2), since 4 is halfway between 1 and 7. Note: 4 is the average of 1 and 7, that is, 4 = 1 + The midpoint of AB has coordinates 1, 12. 0xyB(1, 3)A(1, 2) Note that 12 is the average of 3 and the interval is not parallel to one of the axes we take the average of the x COORDINATE and the y COORDINATE . This is proved (x, y)215y8A(1, 2)B(5, 8)Let M be the midpoint of the line AB. Triangles AMS and MBT are congruent triangles (AAS), and so AS = MT and MS = the x COORDINATE of M is the average of 1 and 5.

7 X = 5 + 12 = 3 The y COORDINATE of M is the average of 2 and 8. x = 2 + 82 = 5 Thus the coordinates of the midpoint M are (3, 5).The general caseWe can find a formula for the midpoint of any interval. Suppose that P(x1, y1) and Q(x2, y2)are two points and let M(x, y) be the midpoint.{6}The Improving Mathematics Education in Schools (TIMES) ProjectP(x1, y1)Q(x2, y2)0xySTxM(x, y)yy2y1x1x2 Triangles PMS and MQT are congruent triangles (AAS), and so PS = MT and MS = the x COORDINATE of M is the average of x1 and x2, and y COORDINATE of M is the average of y1 and y2. Therefore x = x1 + x22 and y = y1 + y22 Midpoint of an intervalThe midpoint of an interval with endpoints P(x1, y1) and Q(x2, y2) is x1 + x22 , y1 + the average of the x coordinates and the average of the y the coordinates of the midpoint of the line interval joining the points (6, 8) and ( 3, 2).SOLUTIONThe midpoint has coordinates , 6 + ( 3)2, 8 + 22 = 32, 5 EXAMPLEIf C(3, 6) is the midpoint of line interval AB and A has coordinates ( 1, 1), find the coordinates of the coordinates of B be (x1, y1).

8 X1 + ( 1)2 = 3 and y1 + 12 = 6 x1 1 = 6 y1 + 1 = 12so x1 = 7 so y1 = B has coordinates (7, 11).{7}A guide for teachersEXERCISE 3A square has vertices O(0, 0), A(a, 0), B(a, a) and C(0, a).a Find the midpoint of the diagonals OB and Find the length of a diagonal of the square and the radius of the circle in which OABC is Find the equation of the circle inscribing the GRADIENT OF A LINEG radient of an intervalThe gradient is a measure of the steepness of line. There are several ways to measure steepness. In COORDINATE GEOMETRY the standard way to define the gradient of an interval AB is riserun where rise is the change in the y values as you move from A to B and run is the change in the x values as you move from A to B. We will usually the pronumeral m for (5, 6)riserunA(2, 1)Given the points A(2, 1) and B(5, 6): gradient of interval AB = riserun = 6 15 2 = 53 Notice that as you move from A to B along the interval the y value increases as the x value increases.

9 The gradient is the points A(2, 7) and B(6, 1) gradient of interval AB = riserun = 1 76 2 = 64 = 32 0xyB(6, 1)A(2, 7) or gradient of interval BA = riserun = 7 12 6 = 32{8}The Improving Mathematics Education in Schools (TIMES) ProjectNotice that in this case as we move from A to B the y value decreases as the x value increases. The gradient is negative. Similarly the gradient of BA = 32 which is the same as the gradient of general: gradient of line interval AB = riserun 0xy(rise)(run)A(x1, y1)B(x2, y2)x2 x1y2 y1 m = y2 y1x2 x1 Note that since y2 y1x2 x1 = y1 y2x1 x2 it does not matter which point we take as the first and which point we take as the the interval is vertical, the run is zero 0xyAPQBG radient of PQ is zeroGradient of AB is not definedand the gradient of the interval is not defined. This is shown by interval the interval is horizontal, the rise is zero as shown by interval PQ. The gradient of the interval is of a lineThe gradient of a line is defined to be the gradient of any interval within the definition depends on the fact that two intervals on a line have the same AB and PQ are two intervals on the same straight line.

10 Draw right angled triangles ABX and PQY with sides AX and PY parallel to the x axis and sides BX and QY parallel to the y ABX is similar to triangle PQY since the corresponding angles are equal. Therefore: QYPY = BXAXThat is, the intervals have the same gradient.{9}A guide for teachersEXAMPLEA line passes through the points (1, 2) and (5, 10). Find its gradient = y2 y1x2 x1 0xy(5, 10)5 1 = 410 2 = 8(1, 2) = 10 25 1 = 2 EXAMPLEA line passes through the point (5, 7) and has gradient 23. Find the x COORDINATE of a point on the line when y = of the line = 6x 5. Thus, 0xy(x, 13)x 513 7(5, 7) 6x 5 = 23 18 = 2(x 5) 9 = x 5 x = 14 EXERCISE 4 Find the gradient of the line passing through (a, b) and (0, c)InterceptsThe x-intercept of a line is the point at which it crosses the x y-intercept of a line is the point at which it crosses the y the diagram to the left the y intercept is at A and the x intercept at B.{10}The Improving Mathematics Education in Schools (TIMES) ProjectThe second diagram shows a line parallel to the x axis and it has a y intercept at third diagram shows a line parallel to the y axis and it has an x intercept at OF A STRAIGHT LINEWhen we plot points which satisfy the equation 0xyy = 2x + 1(0, 1)(1, 3)(2, 5)( 1, 1)y = 2x + 1 we find that they lie in a straight we find the equation of the line given suitable geometric information about the line?