Supporting Australian Mathematics Project. A guide for teachers Years 11 and 12. Algebra and coordinate geometry: Module 2. Coordinate geometry

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1 1 Supporting Australian Mathematics Project A guide for teachers Years 11 and 1 Algebra and coordinate geometr: Module Coordinate geometr

2 Coordinate geometr A guide for teachers (Years 11 1) Principal author: Associate Professor David Hunt, Universit of NSW Peter Brown, Universit of NSW Dr Michael Evans, AMSI Associate Professor David Hunt, Universit of NSW Dr Daniel Mathews, Monash Universit Editor: Dr Jane Pitkethl, La Trobe Universit Illustrations and web design: Catherine Tan, Michael Shaw Full bibliographic details are available from Education Services Australia. Published b Education Services Australia PO Bo 177 Carlton South Vic 353 Australia Tel: (3) Fa: (3) info@esa.edu.au Website: 13 Education Services Australia Ltd, ecept where indicated otherwise. You ma cop, distribute and adapt this material free of charge for non-commercial educational purposes, provided ou retain all copright notices and acknowledgements. This publication is funded b the Australian Government Department of Education, Emploment and Workplace Relations. Supporting Australian Mathematics Project Australian Mathematical Sciences Institute Building 161 The Universit of Melbourne VIC 31 enquiries@amsi.org.au Website:

3 Assumed knowledge Motivation Content The Cartesian plane The distance between two points Midpoints and division of an interval Gradients and the angle of inclination Intercepts and equations of lines Parallel, intersecting and perpendicular lines Perpendicular distances Geometrical proofs using coordinate methods Links forward Three-dimensional coordinate geometr Quadratic equations and the conics Histor and applications Answers to eercises

4 Coordinate geometr Assumed knowledge Familiarit with the coordinate plane. Facilit with algebra see the module Algebra review. Knowledge of elementar trigonometr. Motivation Coordinate geometr is where algebra meets geometr. In secondar school mathematics, most coordinate geometr is carried out in the coordinate plane R, but threedimensional geometr can also be studied using coordinate methods. Classical Euclidean geometr is primaril about points, lines and circles. In coordinate geometr, points are ordered pairs (, ), lines are given b equations a +b +c = and circles b equations ( a) +( b) = r. Thus the simplest, most useful and most often met application of coordinate geometr is to solve geometrical problems. Parabolas, ellipses and hperbolas also regularl arise when studing geometr, and we shall discuss their equations in this module. Readers not familiar with coordinate geometr ma find the TIMES module Introduction to coordinate geometr (Years 9 1) useful as a gentle introduction to some of the ideas in this module. Content The Cartesian plane The number plane, or Cartesian plane, is divided into four quadrants b two perpendicular lines called the -ais, a horizontal line, and the -ais, a vertical line. These aes

5 A guide for teachers Years 11 and 1 {5} intersect at a point called the origin. Once a unit distance has been chosen, the position of an point in the plane can be uniquel represented b an ordered pair of numbers (, ). For the point (5,3), for eample, 5 is the -coordinate and 3 is the -coordinate, sometimes called the first and second coordinates. When developing trigonometr, the four quadrants are usuall called the first, second, third and fourth quadrants as shown in the following diagram. 3 (5,3) nd Quadrant 1 1st Quadrant rd Quadrant 1 4th Quadrant 3 There are a number of elementar questions that can be asked about a pair of points A(a,b) and B(c,d). A(a,b) B(c,d) What are the coordinates of the midpoint of the interval AB? What is the distance between the two points? What is the gradient of the interval joining the two points? What is the equation of the line joining the two points?

6 {6} Coordinate geometr The answers to all of these questions will be discussed in this module, as will answers to other questions such as: What is the equation of the line through the origin parallel to AB? What is the equation of the perpendicular bisector of AB? At what angle does the line AB meet the -ais? These questions give some idea of the scope of coordinate geometr. The distance between two points Distances in geometr are alwas positive, ecept when the points coincide. The distance from A to B is the same as the distance from B to A. In order to derive the formula for the distance between two points in the plane, we consider two points A(a,b) and B(c,d). We can construct a right-angled triangle ABC, as shown in the following diagram, where the point C has coordinates (a,d). (a,b) A b d C B (a,d) a c (c,d) Now, using Pthagoras theorem, we have AB = b d + a c = (a c) + (b d). So AB = (a c) + (b d). A similar formula applies to three-dimensional space, as we shall discuss later in this module.

7 A guide for teachers Years 11 and 1 {7} The distance formula Suppose P( 1, 1 ) and Q(, ) are two points in the number plane. Then PQ = ( 1 ) + ( 1 ) and so PQ = ( 1 ) + ( 1 ). Q(,) P(1,1) It is clear from the distance formula that: PQ = QP PQ = if and onl if P = Q. Midpoints and division of an interval Given two points A, B in the plane, it is clearl possible to create a number line on AB so as to label each point on AB with a (real) number. 1 1 B A There are (infinitel) man was in which this can be done, but it turns out not to be particularl useful for geometrical purposes.

8 {8} Coordinate geometr On the other hand, in the following diagrams, we can sa that the point P divides the interval AB in the ratio 1 : 1 and the point Q divides the interval AB in the ratio 3 :. 1 A 1 P B A 3 Q B Midpoint of an interval The midpoint of an interval AB is the point that divides AB in the ratio 1 : 1. A(1,1) P X Y B(,) Assume that the point A has coordinates ( 1, 1 ) and the point B has coordinates (, ). It is eas to see, using either congruence or similarit, that the midpoint P of AB is ( 1 +, 1 + ). This is shown in the module Introduction to coordinate geometr (Years 9 1). Internal division of an interval We now generalise the idea of a midpoint to that of a point that divides the interval AB in the ratio k : 1. Definition Suppose k > is a real number and let P be a point on a line interval AB. Then P divides AB in the ratio k : 1 if AP PB = k 1 = k. A k P 1 B Note. As k, P A and as k, P B.

9 A guide for teachers Years 11 and 1 {9} Theorem Let A( 1, 1 ) and B(, ) be two points in the plane and let P(, ) be the point that divides the interval AB in the ratio k : 1, where k >. Then = 1 + k 1 + k and = 1 + k 1 + k. Proof If 1 =, then it is clear that is given b the formula above. So we can assume that 1. Consider the points C (, 1 ) and D(, 1 ), as shown in the following diagram. k P(,) 1 B(,) A(1,1) C(,1) D(,1) The triangles AC P and ADB are similar (AAA). So AP AB = AC AD k k + 1 = 1 1 k k 1 = k k Similarl, (k + 1) = k + 1 = 1 + k 1 + k. = 1 + k 1 + k. Eercise 1 If P(, ) lies on the interval A( 1, 1 ), B(, ) such that AP : PB = a : b, with a and b positive, show that = b 1 + a b + a and = b 1 + a. b + a The formulas of eercise 1 are worth learning.

10 {1} Coordinate geometr Eample Let A be ( 3,5) and B be (5, 1). Find 1 the distance AB the midpoint P of AB 3 the point Q which divides AB in the ratio : 5. Solution 1 AB = ( 5 ( 3) ) + ( 1 5 ) = = 17, so AB = 17. P has coordinates ( 1, 5 ). 3 Q has coordinates ( , ) ( = , 5 ). 7 Eercise Give an alternative proof of the previous theorem on internal division of an interval. Use the fact that AC = AP cosθ in the diagram from the proof, where θ = PAC. Eercise 3 Suppose points A, B and C are not collinear and have coordinates ( 1, 1 ), (, ) and ( 3, 3 ). Let D be the midpoint of BC and suppose G divides the median AD in the ratio : 1. Find the coordinates of G and deduce that the medians of ABC are concurrent. Eternal division of an interval Dividing an interval AB internall in a given ratio produces a point between A and B. Eternal division produces a point outside the interval AB. 1 A 1 B 1 C D

11 A guide for teachers Years 11 and 1 {11} Suppose D, A, B and C are collinear and D A = AB = BC, as in the above diagram. Then AC CB = 1 = and AD DB = 1. We sa that C divides AB eternall in the ratio : 1 and that D divides AB eternall in the ratio 1 :. Clearl this is different from the internal division of an interval discussed in the previous subsection. In general, suppose that P(, ) is on the line AB but is eternal to the interval AB and that AP PB = k, for some k >. 1 k 1 B(,) P(,) A(1,1) Then 1 k = 1 1 = k k (1 k) = 1 k = 1 k 1 k. This suggests that we make the following definition. Definition Suppose k < is a real number with k 1 and let P be a point on a line AB. Then P divides the interval AB in the ratio k : 1 if P is eternal to the interval and AP PB = k. With this convention we have = 1 + k 1 + k and = 1 + k 1 + k for eternal division, eactl as for the internal division of an interval! These formulas are algebraicall the same as for internal division, but here k is negative.

12 {1} Coordinate geometr Eample Find the coordinates of the point P which divides the interval A( 3, 7), B( 1, 4) eternall in the ratio 4 : 3. Solution Here k = 4 3 and = 3 + ( 4 ) 3 ( 1) 1 4 = = 5, 3 = 7 + ( 4 ) 3 ( 4) = = The point P has coordinates (5,5). Now, for each point P on the line AB, we have an associated number k: for P in the interval AB, we take k = AP PB for P on the line AB eternal to the interval AB, we take k = AP PB. The following diagram shows the values of k for the marked points A 1 1 B 3 Clearl, as P in either direction, AP PB the real projective line.) 1. (We are now close to a concept called Eercise 4 Suppose M is the midpoint of AB, where A is (8,1) and B is (18,). Further suppose that P divides AB internall and Q divides AB eternall in the ratio : 3. Show that MP MQ = MB.

13 A guide for teachers Years 11 and 1 {13} Gradients and the angle of inclination Suppose l is a line in the number plane not parallel to the -ais or the -ais. B(,) l A(1,1) θ run rise θ Let θ be the angle between l and the positive -ais, where < θ < 9 or 9 < θ < 18. Suppose A( 1, 1 ) and B(, ) are two points on l. Then, b definition, the gradient of the interval AB is m = rise run = 1 1. From the diagram above, this is equal to tanθ. So tanθ is the gradient of the interval AB. Thus the gradient of an interval on the line is constant. Thus we ma sensibl define the gradient of l to be tanθ. l l θ θ positive gradient negative gradient < θ < 9 9 < θ < 18 In the case where the line is parallel to the -ais, we sa that the gradient is. In the case where the line is parallel to the -ais, we sa that θ = 9.

14 {14} Coordinate geometr = c = c zero gradient no gradient θ not defined θ = 9 Eample Find the gradients of the lines joining the following pairs of points. Also find the angle θ between each line and the positive -ais. 1 (, 3), (5, 3) (1, ), (7, 8) 3 (1, ), (7, 4) 4 (, 3), (, 7) Solution 1 The line through (,3) and (5,3) is parallel to the -ais and so the gradient is. The gradient is = 1 and θ = The gradient is = 1 and θ = The line is parallel to the -ais, so the line has no gradient and θ = 9. Eercise 5 Find the gradients of the lines joining the following pairs of points: a b c ( cp, c ) (, cq, c ) p q (ap,ap), (aq,aq) (a cosθ,b sinθ), (a cosφ,b sinφ). Intercepts and equations of lines Intercepts All lines, ecept those parallel to the -ais or the -ais, meet both coordinate aes. Suppose that a line l passes through (a,) and (,b). Then a is the -intercept and b is

15 A guide for teachers Years 11 and 1 {15} the -intercept of l. The intercepts a and b can be positive, negative or zero. All lines through the origin have a = and b =. (,b) l (a,) Equations of lines One of the aioms of Euclidean geometr is that two points determine a line. In other words, there is a unique line through an two fied points. This idea translates to coordinate geometr and, as we shall see, all points on the line through two points satisf an equation of the form a + b + c =, with a and b not both. Conversel, an linear equation a + b + c = is the equation of a (straight) line. This is called the general form of the equation of a line. Point gradient form Consider the line l which passes through the point ( 1, 1 ) and has gradient m. θ m = tanθ l (,) (1,1) Let P(, ) be an point on l, ecept for ( 1, 1 ). Then and so m = 1 1, 1 = m( 1 ). This equation is called the point gradient form of the equation of the line l. Suppose that ( 1, 1 ) = (,c). Then the equation is c = m or, equivalentl, = m +c. This is often called the gradient intercept form of the equation of the line.

16 {16} Coordinate geometr Line through two points l (,) (1,1) To find the equation of the line through two given points ( 1, 1 ) and (, ), first find the gradient m = 1 1 and then use the point gradient form 1 = m( 1 ). A special case is the line through (a,) and (,b), where a,b. (,b) l (a,) In this case, the gradient is m = b a = b a. Thus the equation of the line is = b ( a) a a + b = ab or, equivalentl, a + b = 1, which is eas to remember. This is called the intercept form of the equation of a line.

17 A guide for teachers Years 11 and 1 {17} Summar There are four different forms of the equation of a straight line which we have considered. Equation of a straight line General form a + b + c = Point gradient form 1 = m( 1 ) Gradient intercept form Intercept form = m + c a + b = 1 Eample 1 Find the equation of the line through (3,4) with gradient 5. Find the equation of the line through (3,4) and (, 3). 3 Find the equation of the line with -intercept 3 and -intercept 5. Solution 1 The equation is 4 = 5( 3) or, alternativel, = The gradient is m = 4 ( 3) 3 ( ) = 7 5. So the equation of the line is 4 = 7 ( 3) 5 or, alternativel, =. 3 The equation is 3 5 = 1 or, alternativel, =.

18 {18} Coordinate geometr Vertical and horizontal lines The equation of the vertical line through (3,4) is = 3. The equation of the horizontal line through (3, 4) is = 4. = 4 = 3 Parallel, intersecting and perpendicular lines The aioms for Euclidean geometr include: Two lines meet at a point or are parallel. l n l n Amongst those pairs of lines which meet, some are perpendicular. l n If lines l and n are not vertical, then the are parallel if and onl if the have the same gradient m = tanθ. l n θ θ

19 A guide for teachers Years 11 and 1 {19} Clearl, two horizontal lines are parallel. Also, an two vertical lines are parallel. If lines l and n are not parallel, then their point of intersection can be found b solving the equations of the two lines simultaneousl. Eample Determine whether the following pairs of lines are parallel or identical and, if the are not, find the (unique) point of intersection: 1 = + 3 = = = + 3 = = = =. Solution 1 Both lines have gradient m = 1, but the have different -intercepts. So the two lines are parallel. The lines = + 3 and = + 7 have the same gradient and are therefore parallel. 3 3 The line 5 = can also be written as = + 3. So the two lines are the same. 4 Number the equations: = + 3 (1) =. () Substituting (1) into () gives ( + 3) =, so + 6 =. Therefore = 8 and = 5, and the point of intersection is ( 8, 5). Eercise 6 a b Find the equation of the line parallel to the -ais which passes through the point where the lines = and 7 = meet. Find the gradient of the line which passes through the point (,3) and the point of intersection of the lines 3 + = and = 7. Perpendicular lines All vertical lines = a are perpendicular to all horizontal lines = b. So, in the following, we discuss lines with equations = m + b, where m.

20 {} Coordinate geometr Theorem The lines = m + b and = m 1 + c are perpendicular if and onl if mm 1 = 1. Proof It is sufficient to prove that = m is perpendicular to = m 1 if and onl if mm 1 = 1. = m1 B 1 A + = 1 1 (cosθ,sinθ) θ = m Suppose = m meets the -ais at angle θ, that is, m = tanθ. For convenience, assume that m >. Then = m meets the unit circle at (cosθ,sinθ) in the first quadrant. If = m 1 is perpendicular to = m, then AOB = θ, AB = sinθ, OA = cosθ and B is ( sinθ,cosθ). So as required. m 1 = cosθ sinθ = 1 tanθ = 1 m, Conversel, suppose mm 1 = 1 and m = tanθ. Then m 1 = 1 tanθ = cosθ sinθ = sin ( π + θ) ( π ) cos ( π = tan + θ) + θ. Hence = m is perpendicular to = m 1. This result is often stated as: The line = m 1 + c 1 is perpendicular to the line = m + c if and onl if m 1 m = 1.

21 A guide for teachers Years 11 and 1 {1} Eercise 7 Use the angle sum formulas to show that sin(a + B) = sin A cosb + cos A sinb cos(a + B) = cos A cosb sin A sinb tan(9 + θ) = 1 tanθ. Then use this identit to give an alternative proof that lines = m and = m 1 are perpendicular if and onl if mm 1 = 1. Eample Find the equation of the line l through (1,3) perpendicular to the line + 3 = 1. Find the equation of the line through (4,5) parallel to l. Solution The gradient of the line + 3 = 1 is 3, so l has gradient 3. The equation of l is 3 = 3 ( 1) 6 = =. The other line has equation 5 = 3 ( 4) or, equivalentl, 3 =. Eample For each pair of lines, determine whether the are parallel, identical or meet. If the meet, find the point of intersection and whether the are perpendicular. 1 l : 3 = + 5, n : + 3 = 7 l : 6 = 5 + 7, n : 1 = l : 3 = 1, n : 4 = l : 5 = +, n : =

22 {} Coordinate geometr Solution 1 The gradient of l is 3 and the gradient of n is 3. Since 3 3 = 1, the lines are perpendicular. We have 6 = from l, = 1 from n. Substituting, we obtain = 1 and so = 11 13, = ( 9 The point of intersection is 13, 11 ). 13 The lines l and n have the same gradient 6 5. The line l has -intercept 7, and the line 6 n has -intercept 13. So l is parallel to n, but l n. 1 3 The gradient of l is 3 and the gradient of n is 4. So l meets n, but l is not perpendic- 5 ular to n. We have 15 = 1 5 from l, 1 = 15 1 from n. Substituting, we obtain 1 = So = 13 and = 9. The point of intersection is ( 13, 9). 4 Rearranging = gives 5 = +. So l and n are the same line. Eercise 8 Consider the line l with equation a + b + c = and the point P( 1, 1 ). a Show that the line through P parallel to l is given b a + b = a 1 + b 1. b Show that the line through P perpendicular to l is given b b a = b 1 a 1.

23 A guide for teachers Years 11 and 1 {3} Perpendicular distances Given a line l and a point P not on the line, it is natural to ask the question: What is the distance d of P from l? P(1,1) d θ Q(s,1) a + b + c = θ Assume that l has equation a +b +c =, with a and b. We will also assume that the gradient m of l is positive. So the angle θ between l and the positive -ais is acute. Let P( 1, 1 ) be an point, and let Q(s, 1 ) be the point where the line l meets the line through P parallel to the -ais. Since Q lies on the line l, we have as +b 1 +c = and so s = b 1 c. a Thus PQ = 1 s = a 1 + b 1 + c. a We can draw the following triangle, as m = tanθ with θ acute. 1 + m θ 1 m From the triangle, we have sinθ = m 1 + m. Since m = a, this implies that b sin θ = m 1 + m = a a + b.

24 {4} Coordinate geometr We are assuming that θ is acute, so sinθ is positive and therefore sinθ = a a + b. From the first diagram, we now have d = PQ sinθ = a 1 + b 1 + c a + b. Theorem Let P( 1, 1 ) be a point and let l be the line with equation a + b + c =. Then the distance d from P to l is given b d = a 1 + b 1 + c a + b. Proof We have checked that the formula holds when a, b and m >. Check that the formula also holds when a =, when b = and when m <. Eample What is the distance d of the point P( 6, 7) from the line l with equation = 11? Solution Here a = 3, b = 4, c = 11, 1 = 6 and 1 = 7. So d = = Eercise 9 A circle has centre (1,) and radius 5. a b Find the perpendicular distance from the centre of the circle to the line with equation + 1 = and hence show this line is a tangent to the circle. Find the perpendicular distance from the centre of the circle to the line with equation + 1 = and hence show the line does not meet the circle. Eercise 1 Consider a line l with equation a + b + c =. Show that the epression a 1 + b 1 + c is positive for ( 1, 1 ) on one side of the line l and negative for all points on the other side.

25 A guide for teachers Years 11 and 1 {5} Geometrical proofs using coordinate methods If a point P(, ) lies on the circle with centre (a,b) and radius r, then Pthagoras theorem gives ( a) + ( b) = r. (a,b) r (,) Conversel, if P(, ) satisfies the equation above, then P lies on the circle with centre (a,b) and radius r. This is an eample of a locus problem. A locus is the set of all points satisfing a geometrical condition. For eample, a circle is the locus of all points a fied distance r from a fied point (a,b). Eample Let P(a,b) and Q(c,d) be two points in the plane. Find the equation of the line l that is the perpendicular bisector of the line segment PQ. Solution First assume that a c and b d. The gradient of PQ is b d, and so the gradient of the a c perpendicular line l is m = c a b d. ( a + c The line l goes through the midpoint, b + d ) of PQ. Thus the equation of l is b + d ( = m a + c (b d)( b d) = (c a)( a c) (b d) + (d b ) = (c a) + (a c ) (b d) + (a c) = a + b c d. ) The final equation above is a sensible smmetric form for the equation of the perpendicular bisector of PQ. This equation is also correct in the case that a = c or b = d. Alternativel, the question can be answered b finding the locus of all points X (, ) equidistant from P(a, b) and Q(c, d).

26 {6} Coordinate geometr P(a,b) X(,) Q(c,d) P X = QX P X = QX ( a) + ( b) = ( c) + ( d) a b + a + b = c d + c + d (b d) + (a c) = a + b c d. Note that the locus method involves less algebra. Eercise 11 Use the locus method illustrated in the previous eample to find the equation of the perpendicular bisector of the line segment PQ, where the coordinates of P and Q are ( 1,6) and (3,4). Eample What is the locus of all points distance d from the fied interval AB? Solution This is not meant to be a problem in coordinate geometr! The locus is two intervals parallel to AB together with two semicircles of radius d with centres A and B. A d d d d B

27 A guide for teachers Years 11 and 1 {7} Eample What is the area of the triangle with vertices A( 1, 1 ), B(, ) and C ( 3, 3 )? Assume that C is above the interval AB as shown. A(1,1) C(3,3) B(,) P R Q Solution Let P( 1,), Q(,) and R( 3,) be the projections of A, B and C onto the -ais. We epress the area of the triangle ABC in terms of three trapezia: Area ABC = Area APRC + Area CRQB Area APQB = 1 ( (1 + 3 )( 3 1 ) + ( 3 + )( 3 ) ( 1 + )( 1 ) ) = 1 ( 1 ( ) + ( ) + 3 ( ) ) = 1 ( 1 ( 3 ) + ( 1 3 ) + 3 ( 1 ) ). The formula found in the previous eample gives a negative value if C is below AB. In general, the area of the triangle with vertices A( 1, 1 ), B(, ), C ( 3, 3 ) is Area ABC = ( 1 1 ( 3 ) + ( 1 3 ) + 3 ( 1 ) ). Note. For readers familiar with determinants, the area of the triangle is 1, where = Eercise 1 Three points A(3,6), B(7,8) and C (5,) are the vertices of ABC. Find the area of this triangle using the formula given above.

28 {8} Coordinate geometr Eample Use coordinate geometr to show that the altitudes of a triangle ABC are concurrent (at the orthocentre of the triangle). Solution B choosing the -ais to be the line AB and the -ais to be the line OC, we can assume A(a,), B(b,) and C (,c) are the vertices of the triangle ABC, as shown in the following diagram. C(,c) A(a,) B(b,) The gradient of BC is c b, so the gradient of the altitude through A is b. Thus the equation of the altitude through A c is = b ( a). c To find the -intercept of the altitude, take =, which implies = ba. B swapping c a and b, we get ab c = ba. So the altitude through B has the same -intercept. The c ( altitude through C is the -ais. Thus H =, ab c ) lies on all three altitudes. Eercise 13 Find a proof that the three altitudes of a triangle are concurrent using classical Euclidean geometr. Eercise 14 Let M be the point (,3) and let l be a line through M which meets + 3 = at A and meets = at B. If M is the midpoint of AB, find the equation of l.

29 A guide for teachers Years 11 and 1 {9} Links forward Three-dimensional coordinate geometr The principal elements in solid geometr (3-dimensional geometr) are points, lines and planes. In the plane (-dimensional geometr), the incidence properties are ver straightforward: two points determine a line two lines are parallel or meet at a unique point. In 3-dimensional space, things are more complicated: two points still determine a line two lines meet or are parallel or are skew. To visualise a pair of skew lines, think of one as being behind the other. Three points which are not collinear determine a unique plane. But there are an infinite number of planes through a fied line or three collinear points. Two planes are parallel or meet in a line. Indeed, if two planes have a point in common, the must have a whole line of points in common (this is not obvious). To set up a coordinate sstem for solid geometr, choose a plane and then fi aes and as in plane geometr. z

30 {3} Coordinate geometr The ais z is the normal to the plane through the origin. Usuall we choose z to come out of the standard plane. (This is called a right-handed coordinate sstem and is ver important when vectors are used especiall in applications in phsics.) Thus, z and z. z P(,,z) (,,) (,,) (,,) Ever point in 3-space is a certain distance z above the plane (if z <, the point is below). The points in the plane are given the coordinates (,,) and, in general, P is (,, z). Midpoints and distance Let P( 1, 1, z 1 ) and Q(,, z ) be points in R 3. Then: ( 1 + The midpoint of the line segment PQ is M, 1 +, z 1 + z ). The distance between P and Q is ( 1 ) + ( 1 ) + (z 1 z ). So, for eample, in the above diagram OP = + + z. Lines in R 3 To describe the points on a line, it is easiest to use a parameter which varies as a point moves along the line. A(a,b,c) l P(,,z) B(d,e,f )

31 A guide for teachers Years 11 and 1 {31} Consider the line l through the fied points A(a,b,c) and B(d,e, f ). Let P(,, z) be an point on l. Then the vector ( a, b, z c) is a scalar multiple t of the vector (d a,e b, f c). So and thus a = t(d a) b = t(e b) z c = t(f c) = td + (1 t)a = te + (1 t)b z = t f + (1 t)c, for t R. This is the parametric form for a line in R 3. If t =, then P = A and, if t = 1, then P = B. We can obtain the Cartesian form for the line b eliminating t. It is a d a = b e b = z c f c. Parametric forms for curves in R are discussed in the module Quadratics. Planes in R 3 There are three principal planes in R 3. The plane, which contains the aes and, is given b Π z = { (,, z) R 3 z = }. The other two principal planes are = and =. z z =

32 {3} Coordinate geometr Consider the set Π = { (,, z) R = 6 }. If z =, then this restricts to a line in the plane. (,3,) (,,) (a,b,) However, if (a,b,) is an point on this line, then (a,b, z) Π for an value of z. Thus Π is a plane in R 3. Indeed Π is a plane parallel to the z-ais. In a similar wa, 3 + z = 6 is a plane parallel to the -ais, and 3 + z = 6 is a plane parallel to the -ais. Now consider the set Π = { (,, z) R z = 1 }. Given the previous eamples, it should be no surprise that this is a plane. Clearl it passes through the points (4,,), (,6,) and (,,3), so the plane has three intercepts! z (,6,) (,,3) (4,,) The general equation for a plane is Π = { (,, z) R 3 a + b + cz = d }, where at least one of a,b,c is non-zero. So, in some was, the general plane in R 3 is analogous to the line in R. To prove the above facts about planes, it is best to use vectors to represent the points in the plane.

33 A guide for teachers Years 11 and 1 {33} Quadratic equations and the conics The general linear equation a + b + c =, where a or b is non-zero, is the equation of a line in R. So it is natural to ask the question: What curves in R occur as the graph of the general quadratic equation in two variables? Consider the set S = { (, ) R a + h + b + g + f + c = }, where a,b,c, f, g,h are constants and at least one of a, h and b is non-zero. The s in the equation are somewhat msterious, but are related to completing the square. There are several approaches to studing such sets S, including conic sections and the focus directri definition, but we will leave them to the module Quadratics. In this section, we will simpl give a picture galler of tpical eamples. Eample 1 (Circle) The graph of + = r is the circle with centre the origin and radius r. r + = r r r r Eample (Empt set) The graph of + = 1 is the empt set.

34 {34} Coordinate geometr Eample 3 (Two lines) The equation = factorises as (+)( ) =, so = or =. So this eample is the union of two lines. = = Eample 4 (One line) The equation + = factorises as ( ) =, which is just one line! = Eample 5 (One point) The graph of ( f ) + ( g ) = is just the single point (f, g ). (f,g)

35 A guide for teachers Years 11 and 1 {35} Eample 6 (Rectangular hperbola) The graph of = 1 is called a rectangular hperbola. = 1 Eample 7 (Ellipse) The graph of a + = 1, for a >, b >, b is an ellipse. It is a circle if a = b. b a a b Eample 8 (Hperbola) Consider the quadratic equation a = 1, for a >, b >. b B setting =, we see there are no -intercepts. If =, then = ±a. As, ± and the curve approaches the lines = b called the asmptotes of the curve. a or = b, which are a

36 {36} Coordinate geometr b = a = b a ( a,) (a,) This is a (general) hperbola. It is a rectangular hperbola if a = b. If the rectangular hperbola = 1 from Eample 6 is rotated through 45, it becomes = 1, which is recognisable as a hperbola. Eample 9 (Parabola) The graph of = 4a, for a >, is a parabola. It has no asmptotes. =4a The set of points in R 3 satisfing + = z forms a cone (perhaps better described as a double cone), as shown in the following diagram. z A horizontal plane meets the cone in either one point or a circle. All the other eamples in our picture galler (ecept the empt set) occur when different planes intersect the cone!

37 A guide for teachers Years 11 and 1 {37} It is educative to tr and visualise them all. For this reason the ellipse, the parabola, the hperbola and the circle are often called the conic sections. Several of the Greek mathematicians followed this approach to the conics. The general quadratic in two variables and is S = { (, ) R a + h + b + g + f + c = }. If we appl a translation, a rotation or a reflection to S, then we do not change the intrinsic geometrical nature of S. Unless a = b = (in which case S is a rectangular hperbola), a rotation can be found which reduces S to either a 1 + b 1 + g 1 + f 1 + c 1 = or a 1 + g 1 + f 1 + c 1 =. In the second case, completing the square in leads to a + f + c =, which is a parabola. In the first case, completing the square in both and leads to a + b = c, which is either a hperbola, an ellipse, a circle, two lines, one line, one point or the empt set, depending on the values of a, b and c. Histor and applications There were three facets to 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. The Greek mathematician Menaechmus (38 3 BCE) proved theorems 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 BCE) dealt with problems in a manner that ma be called a 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.

38 {38} Coordinate geometr Answers to eercises Eercise 1 Let A( 1, 1 ) and B(, ) be two points. Assume that P(, ) is on the interval AB with AP : PB = a : b = a b : 1. So P divides AB in the ratio a b : 1. Thus = 1 + a b 1 + a b and, similarl, Eercise = b 1 + a. b + a = b 1 + a b + a Without loss of generalit, assume PB = 1. Then AC = AP cosθ 1 = k 1 k + 1 ( 1 )(k + 1) = k k 1 = 1 + k k + 1. From the same diagram, we have CP = AP sinθ. Proceed in a similar wa to find. Eercise 3 A(1,1) 1 B(,) 1 D G 1 C(3,3) The midpoint D of BC is ( + 3, + ) 3. The point G that divides the median AD in the ratio : 1 is ( 1 ( 1 + ( + 3 ) ), 1 ( 1 + ( + 3 ) ) ) ( =, )

39 A guide for teachers Years 11 and 1 {39} This is smmetric in 1,, 3 and in 1,, 3. So G lies on the other medians BE and CF, and the medians are concurrent. Eercise 4 Q A(8,1) P M B(18,) M is (13,15). ( ) P is, = (1,14) ( ) Q is, = ( 1, 1). 3 3 Thus Hence MB = = 5, MP = =, MQ = = 5. MP MQ = ( 5 ) 1 = 5 = MB. Eercise 5 a Gradient = c q c p cq cp = p q pq(q p) = 1 pq b c Gradient = aq ap aq ap = (q p) (q p)(q + p) = p + q b sinφ b sinθ b cos Gradient = = a cosφ a cosθ ( φ+θ a sin ( φ+θ ) ( φ θ ) sin ) sin ( θ φ ) = b a cot ( φ + θ ) Eercise 6 a b The lines = and 7 = meet where 11 + =, so = and = 3. The line through (3, ) parallel to the -ais is =. The lines 3 + = and = 7 meet where = 13 and = 8. So the gradient is = 1.

40 {4} Coordinate geometr Eercise 7 The angle sum formulas give Thus sin(θ + 9 ) = sinθ cos9 + cosθ sin9 = cosθ cos(θ + 9 ) = cosθ cos9 sinθ sin9 = sinθ. tan(9 + θ) = cosθ sinθ = 1 tanθ. Suppose = m is perpendicular to = m 1. Then m = tanθ and m 1 = tan(9 + θ). So m 1 = 1 m and therefore mm 1 = 1. Conversel, suppose mm 1 = 1, where m = tanθ. Then m 1 = 1 tanθ = tan(9 + θ). So = m is perpendicular to = m 1. Eercise 8 a b All lines parallel to a + b + c = have equation a + b + d =, for some d. If P( 1, 1 ) lies on the line, then a 1 +b 1 +d =. So the equation is a+b = a 1 +b 1. All lines perpendicular to l have equation b a +d =, for some d. If P( 1, 1 ) lies on the line, then b 1 a 1 + d =. So the equation is b a = b 1 a 1. Eercise 9 a The perpendicular distance of (1,) from the line + 1 = is = 5, which equals the radius. Therefore the line is tangent to the circle. b The perpendicular distance of (1,) from the line + 1 = is = 7 5 = 7 5 5, which is greater than the radius 5. Hence the line does not touch the circle.

41 A guide for teachers Years 11 and 1 {41} Eercise 1 (γ,δ) l (α,β) a + b + c = Note that a point ( 1, 1 ) lies on the line l if and onl if a 1 + b 1 + c =. Suppose that points (α,β) and (γ,δ) are on the same side of l. Then the line segment from (α,β) and (γ,δ) does not cross l. So a 1 + b 1 + c for each point ( 1, 1 ) on the line segment. Thus the epression a 1 + b 1 + c cannot change sign as the point ( 1, 1 ) moves along the line segment from (α,β) to (γ,δ). Hence the signs of aα + bβ + c and aγ + bδ + c are the same. Points on different sides of l must have different signs, as it is possible to choose ( 1, 1 ) so that a 1 + b 1 + c >, and choose (, ) so that a + b + c <. Eercise 11 ( + 1) + ( 6) = ( 3) + ( 4) = = = + 3 = Eercise 1 The area of the triangle is 1 ( ) (5 7) + 8(3 5) + (7 3) = = 1.

42 {4} Coordinate geometr Eercise 13 A α F H α E B D α C Let AD be the altitude perpendicular to BC, let BE be the altitude perpendicular to AC and let AD meet BE at H. Finall let C H meet AB at F. We must prove FC is perpendicular to AB. Join DE. Let α = BCF. Then ECDH is a cclic quadrilateral (supplementar opposite angles), so DE H = α (angles in the same segment). Similarl, AE B D is a cclic quadrilateral and D AB = α. From ABD, ABD = 18 9 α = 9 α. Finall, in CBF, BFC + α + 9 α = 18, so BFC = 9, as required. Eercise 14 A M(,3) B ( a + c Let A = (a,b) and B = (c,d). The midpoint M of AB is (,3) =, b + d ), so a + c = 4 (1) b + d = 6. () Since the point A(a,b) lies on the line + 3 = and the point B(c,d) lies on the line =, we also have a + b = 3 (3) 3c d = 1. (4)

43 A guide for teachers Years 11 and 1 {43} From (1) and (), we have c = 4 a and d = 6 b. Substituting these into (4) gives 3a b = 1. (5) We can now solve (3) and (5) to obtain a = 1 and b = 1. We have found that A is (a,b) = (1,1). So the gradient of the line l is m = 1 3 =. Thus 1 l has equation 1 = ( 1) or, equivalentl, = 1.

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