Mathematics Stage 5 PAS5.2.3 Coordinate geometry. Midpoint, distance and gradient

Transcription

1 Mathematics Stage 5 PAS5..3 Coordinate geometry Part 1 Midpoint, distance and gradient

2 Number: Title: PAS5..3 Coordinate Geometry This publication is copyright New South Wales Department of Education and Training (DET), however it may contain material from other sources which is not owned by DET. We would like to acknowledge the following people and organisations whose material has been used: Outcomes and indicators from Mathematics Years 7-10 Syllabus Board of Studies, NSW 00 Overview, pp. iii-v Part 1, p. 3 Part, pp. 3-4 COMMONWEALTH OF AUSTRALIA Copyright Regulations 1969 WARNING This material has been reproduced and communicated to you on behalf of the New South Wales Department of Education and Training (Centre for Learning Innovation) pursuant to Part VB of the Copyright Act 1968 (the Act). The material in this communication may be subject to copyright under the Act. Any further reproduction or communication of this material by you may be the subject of copyright protection under the Act. CLI Project Team acknowledgement: Writer: Janine Angove Editor: Dr Ric Morante Illustrator(s): Thomas Brown, Tim Hutchinson Desktop Publishing: Gayle Reddy Version date: May 13, 005 Revision date: March 30, 006 All reasonable efforts have been made to obtain copyright permissions. All claims will be settled in good faith. Published by Centre for Learning Innovation (CLI) 51 Wentworth Rd Strathfield NSW 135 Copyright of this material is reserved to the Crown in the right of the State of New South Wales. Reproduction or transmittal in whole, or in part, other than in accordance with provisions of the Copyright Act, is prohibited without the written authority of the Centre for Learning Innovation (CLI). State of New South Wales, Department of Education and Training 005.

3 Contents Part 1 Introduction Part Indicators...3 Preliminary quiz...5 Midpoint...7 Midpoint from the graph...7 Midpoint from the coordinates...9 Distance between points...13 Gradient...3 Reviewing gradient...3 Formula for gradient...6 Geometry on the number plane...33 Suggested answers Part Additional resources Part Exercises Part Part 1 Midpoint, distance and gradient 1

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5 Introduction Part 1 The coordinate system on a number plane can be used to provide a great deal of information about lines and intervals. These features can be used to determine geometrical properties and allow you to prove or disprove a variety of conjectures. Indicators By the end of Part 1, you will have been given the opportunity to work towards aspects of knowledge and skills including: understanding the meaning of terms such as conjecture, midpoint, surd, gradient and collinear establishing the formula for midpoint, gradient and distance using coordinates on the number plane using the formulas to solve problems using subscript notation. By the end of Part 1, you will have been given the opportunity to work mathematically by: using the appropriate formula to solve problems using gradient, midpoint and distance to determine properties of geometrical shapes on the number plane justifying conclusions reached about intervals and geometrical shapes on the number plane verifying that when given two points, either point can be used as ( x 1, y 1 ). Source: Adapted from outcomes of the Mathematics Years 7 10 syllabus < 710_syllabus.pdf > (accessed 04 November 003). Board of Studies NSW, 00. Part 1 Midpoint, distance and gradient 3

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7 Preliminary quiz Before you start this part, use this preliminary quiz to revise some skills you will need. Activity Preliminary quiz Try these. 1 Name the hypotenuse of this triangle. A C B Use Pythagoras theorem to find the length of the hypotenuse of the triangle below..5 cm 6 cm Part 1 Midpoint, distance and gradient 5

8 3 Plot and label the given points on the number plane. 5 y x 1 A (5, 1) B ( 3, 4) C (, 0) D ( 1, 1 ) If d = 35 find the value of d correct to one decimal place. 5 Calculate the number that is half way between 3 and can be rewritten as: 5 = 5 = 5 Rewrite 45 using the same technique. (This is often called simplifying the surd.) Check your response by going to the suggested answers section. 6 PAS5..3 Coordinate geometry

9 Midpoint If you wish to bisect an interval or check some geometrical properties of shapes, you need to be able to locate the midpoint of an interval on the number plane. Two methods will be discussed in this section: finding the midpoint using a graph finding the midpoint using the coordinates of the two endpoints. Midpoint from the graph To locate the midpoint between two other points on the number plane you move half way across and half way up (or down) from one of the points. 3 A across 3 spaces up to the interval B This example shows the midpoint between A (, 1) and B (4, 6). To locate the midpoint, you start at A and move half the distance across to B and up to the interval. The distance up is also half the distance up to B. The midpoint is 1, 3 1. Use this method to find the coordinates of the midpoints of the intervals in the following activity. Part 1 Midpoint, distance and gradient 7

10 Activity Midpoint Try these. 1 Use the graphs below to find the coordinates of the midpoint for each interval. a y x 1 b y x Plot A ( 4, 5) and B (4, 3) and determine the coordinates of the midpoint of AB. 6 y x PAS5..3 Coordinate geometry

11 Check your response by going to the suggested answers section. The drawback with this method is that you must create the graph first. This can be time consuming. The following section describes a method for finding the midpoint without needing the graph. Midpoint from the coordinates You can use the coordinates of the two endpoints to find the midpoint without graphing. To find the midpoint of an interval, you simply need to find the number half way between two numbers. To do this, you need to remember how to find an average. Average = sum of values number of values The midpoint of an interval can be described as: midpoint = (average of the x-values, average of the y-values) sum of x-values sum of y-values =, The diagram below shows how to find the midpoint of the interval joining (4, 6) and (10, 9). (4, 6) (10, 9) Midpoint = 4+10, = 7, Use this technique to complete the following task. Part 1 Midpoint, distance and gradient 9

12 Activity Midpoint Try these. 3 Find the midpoint between these pairs of points without using a graph. a (8, 1) and (4, 19) b (0, 17) and (5, 1) c ( 3, 5) and (, 7) d ( 1, 30) and (9, 1) Check your response by going to the suggested answers section. This process for finding the midpoint from the coordinates can be written as a formula. To be able to refer to different x and y-coordinates, it is useful to introduce a way of identifying each one. A common method in mathematics is to use subscripts. A subscript is a small, lowered number written after the pronumeral. x 1 subscript 10 PAS5..3 Coordinate geometry

13 Do not confuse this with a power. It is simply a way of using the same pronumeral to stand for different numbers: x 1, x, x 3, etc all represent different x- coordinates. You can use this notation to write the general endpoints of any interval as ( x 1, y 1 ) and ( x, y ). For this interval, you can create a formula for the coordinates of the midpoint. Midpoint = (average of the x-values, average of the y-values) = sum of x-values, = x 1 + x, y 1 + y sum of y-values It does not matter which endpoint you assign to be the first point. The following activity allows you to practise using the formula to find the midpoint. Activity Midpoint Try these. 4 Use the formula to find the midpoint of these pairs of points. The working for first example is partially completed. a ( 5, 18) and (9, 0) Midpoint = x 1 + x, y 1 + y =, b 3 1, 4 and ( 7, 14) Part 1 Midpoint, distance and gradient 11

14 Check your response by going to the suggested answers section. Whether you use the concept of averaging the coordinates or you prefer to remember the formula, the arithmetic remains the same. You now have three methods for finding the midpoint of an interval. But what if you know the midpoint, and need to work backwards to find one endpoint? The following activity asks you to solve just such a problem. Activity Midpoint Try this. 5 (Harder) One endpoint of an interval AB is A (5, ). If the midpoint of AB is (1, 3), what are the coordinates of the other endpoint, B? You may use any method to solve this problem, including drawing a graph. Graph paper has been provided in the additional resources section of this part. Check your response by going to the suggested answers section. Use your understanding of midpoint to complete the following exercise. Go to the exercises section and complete Exercise 1.1 Midpoint. 1 PAS5..3 Coordinate geometry

15 Distance between points Lengths on the number plane are not described in terms of centimetres or millimetres. The term units is used to describe distance on the number plane. To find the distance between two points, you are already familiar with drawing a right-angled triangle and using Pythagoras theorem to calculate the length of the interval. The diagram and working below demonstrate this method. 4 y 3 A 4 units x 1 B units Using Pythagoras theorem AB = 4 + = = 0 AB = units to one decimal place Review this method by completing the following activity. Activity Distance between points Try these. 1 Use the graphs below to find the length of each interval. 4 3 y C A x 1 D 3 B Part 1 Midpoint, distance and gradient 13

16 a Interval AB (correct to one decimal place) b Interval CD (exact length) Draw some axes and plot the points ( 5, 3) and ( 1, 0) on the number plane below. Use your graph to find the distance between these two points. This method works well, but is time consuming because you have to graph the points. Can you find the answer faster if you work straight from the coordinates of the endpoints? 14 PAS5..3 Coordinate geometry

17 If you re-examine the earlier example, you can see that for these two points, the length of the right-angled triangle is the difference between the two x-coordinates and the height is the difference between the y- coordinates y A 4 units B units y A (1, ) B (5, 4) Height = 4 Length = x x But will the differences between coordinates give you the lengths you need if the interval goes across the axes? Explore this question by completing the following activity. Activity Distance between points Try these. 3 Draw the right-angled triangle between A ( 4, 3) and B (, 5). Write the length and height of the triangle on the diagram. 6 y x Part 1 Midpoint, distance and gradient 15

18 Now work out the values you get when you work with the coordinates. Beware of negative signs and subtractions. Difference between the x-coordinates = = Difference between the y-coordinates = = Comment on any similarities or differences bewteen the numbers you obtained from the graph and those you obtained just using the coordinates. Check your response by going to the suggested answers section. Distances must be positive numbers. When you work with the coordinates alone, the answers are sometimes negative. But using a little common sense, if the calculation gives you an answer of 3 then the length is really 3 units. So you can use the coordinates of the endpoints to find the length and height of the right-angled triangle between the two points. You can use these two lengths, and Pythagoras theorem, to find the distance between the two points. And you don t have to graph anything! The activity below lets you use this technique to find some distances on the number plane without needing a number plane at all. 16 PAS5..3 Coordinate geometry

19 Activity Distance between points Try these. 4 Complete the working below to find the distance between the two points (9, 4) and (17, ). Difference between the x-coordinates = = So length of triangle = Difference between the y-coordinates = = So height of triangle = Using Pythagoras theorem: distance = The distance between the two points is units. 5 Find the length of CD if C has coordinates (0, 7) and D has coordinates ( 1, ). (Choose your own setting out for your working.) Part 1 Midpoint, distance and gradient 17

20 Check your response by going to the suggested answers section. You can save time by not having to graph the points. But the working above is still quite lengthy. You can reduce the working further by using algebra to write a formula. Follow the explanation below to see how the formula can be created. Let the two endpoints be ( x 1, y 1 ) and ( x, y ). Let the distance = d units Difference between the x-coordinates = x 1 x Difference between the y-coordinates = y 1 y Using Pythagoras theorem: d = length + height = ( x 1 x ) + ( y 1 y ) Stop for a minute and look at the formula so far. Remember that x 1 x sometimes gives a negative answer and so it doesn t really give the true length. This won t matter now because you will square the number. For example, ( 3) gives the same answer as 3. You now know how to calculate d. To find d, you must take the square root of both sides. 18 PAS5..3 Coordinate geometry

21 This formula looks very complex. To use it, it is best to remember what it means: d = (distance across) + (distance up). The example below shows how to set out your working when using this formula. Follow through the steps in this example. Do your own working in the margin if you wish. Find the distance between (3, 7) and (5, ). Solution d = (x 1 x ) + (y 1 y ) = ( 3 5) + ( 7 ) = ( ) + 9 = = 85 So the distance is 85 units which is approximately 9. units. The formula reduces the amount of work dramatically, but it needs to be used very carefully. Notice that in this example, the exact answer is a surd. In practical situations you need to round the answer off to something sensible. Practise using this formula by completing this activity. Part 1 Midpoint, distance and gradient 19

22 Activity Distance between points Try these. 6 Complete the working to find the distance between ( 10, 9) and (, 7). ( 10, 9) (, 7) d = ( ) + ( ) = ( ) + ( ) = + = = 7 Use the formula to find the distance between these pairs of points. Answer correct to one decimal place where necessary. a (0, 5) and (6, 1) b ( 3, 4) and (10, 0) 0 PAS5..3 Coordinate geometry

23 c (1 1, 18) and ( 3, 54) Check your response by going to the suggested answers section. This distance formula is frequently used when designing computer activities. In games, for instance, the distance between two objects is often an important factor. The position of each object on the screen is represented by adjustable coordinates. Below is a small sample of Java script that would be used to position a player at the origin, move then 0 spaces to the right. The next section of the code would determine the distance between the player and, say, a tennis ball in a tennis game. Part 1 Midpoint, distance and gradient 1

24 This last line uses the distance formula. Here, the formula uses multiplication instead of the power of because the computer will perform a multiplication faster, and as you know speed in a game is vital. You now have several strategies you can use to find lengths of intervals on a number plane. Use these to complete the following exercise. Go to the exercises section and complete Exercise 1. Distance between points. PAS5..3 Coordinate geometry

25 Gradient The gradient of a line is its slope. It is a very important feature of a line because it tells you how fast things are changing. Look at the graphs below to find the meaning of gradient in two different situations. Sue s line is steeper than Felix s so she is travelling at a faster rate. Distance Sue s journey Felix s journey Time Georgie s line is going down so she is losing money, while Ivan s line is going up so he is gaining money. Dollars Ivan s savings Georgie s savings Time Reviewing gradient Some facts you already know about gradients are reviewed below. Lines that go up from left to right are said to have a positive gradient, and lines that go down have a negative gradient. y 4 Positive gradient x 1 Negative gradient 3 Part 1 Midpoint, distance and gradient 3

26 Lines with the same gradient are parallel y x 1 3 All these lines have a gradient of 3. The gradient can be described using a number. This number not only indicates whether the line is steep or gradual, but also whether it goes up or down from left to right. In the past you have calculated this number using the formula: Gradient = + or ( ) rise run 4 PAS5..3 Coordinate geometry

27 The rise is the distance you move up the graph from one point to another. The run is the distance you move across from one point to the other. y 4 3 B rise = A run = x 1 Notice that this diagram is the same one you would use to the find the distance between the two points A and B. In this section, the sides of the right-angled triangle are used to find the gradient. Gradient =+ 4 = 1 This line has a positive slope because it goes up from left to right. But which points on the line do you use? Can you just select any two points on the line to find the gradient? Complete the following activity to answer these questions. Activity Gradient Try these. 1 A y B 3 C x D Part 1 Midpoint, distance and gradient 5

28 a Use the points A and B to find the gradient of the line shown in the graph above. Remember to check its sign. b Use the points C and D to find the gradient of the line. Remember to check its sign. c Comment on the answers you got in parts a and b. d Select a third pair of points on the line and check that you still get the same result. The two points are Gradient = Check your response by going to the suggested answers section. Formula for gradient You have discovered that to find the gradient of a straight line you can select any two points on the line. But do you really need to draw the graph? Can you find the gradient simply using the coordinates? Complete the steps in the following activity to develop a formula for calculating the gradient of a line passing through any two points. 6 PAS5..3 Coordinate geometry

29 Activity Gradient Try this. The diagram below shows no scale on the axes. Two points have been plotted on the line: A ( x 1, y 1 ) and B ( x, y ). y A ( x 1, y 1 ) B ( x, y ) x Use the coordinates of the points to write an algebraic expression for: the rise = the run = Combine these expressions to write a formula for the gradient: Gradient = rise run = Check your response by going to the suggested answers section. And so you can find the gradient of a line using two points on it, without every having to draw a graph. The formula is: Part 1 Midpoint, distance and gradient 7

30 But what about the sign of the gradient? Don t you need the graph to decide whether the gradient should be positive or negative? The example below shows that the formula gives you the sign and the number without needing to graph, provided you use it carefully,. Follow through the steps in this example. Do your own working in the margin if you wish. Find the gradient of the line that passes through the points (, 7) and (6, 1). Solution The two solutions below show that you get the same gradient no matter which point you decide will be ( x 1, y 1 ). The solutions also show that the sign of the gradient is found by the formula. Method 1 Using (, 7) as the first point ( x 1, y 1 ) Gradient = y 1 y x 1 x = (, 7) (6, 1) = 6 4 = 3 8 PAS5..3 Coordinate geometry

31 Method Using (6, 1) as the first point ( x 1, y 1 ) Gradient = y 1 y x 1 x = (6, 1) (, 7) = 6 4 = 3 Notice that the result was the same no matter which point you chose to use as ( x 1, y 1 ). Also notice that the sign (+ or ) is automatically determined by the formula. So the line joining (6, 1) and (, 7) would slope down because it has a negative gradient. One final point about this example: gradients are left as improper fractions because the numerator (top) tells you the rise and the denominator (bottom) tells you the run. This is one of the rare times when an improper fraction is a better answer than a mixed numeral. Activity Gradient Try these. 3 Use the formula to calculate the gradient of the line that passes through each pair of points. a (5, ) and (0, 7) Part 1 Midpoint, distance and gradient 9

32 b (, 3) and ( 1, ) c ( 6, 10) and (, 4) 4 a Use the formula to calculate the gradient of the line that passes through (3, 5) and (3, ). b Plot the points (3, 5) and (3, ) and comment on why the gradient is unusual. 30 PAS5..3 Coordinate geometry

33 5 I used the points (3, 8) and (4, 5) and got a gradient of 3. But the correct answer is 3. Please check my working and tell me what I did wrong. Gradient = = 3 1 = 3 6 Show that the lines AB and CD are parallel given the coordinates below. A (5, ), B (8, 4), C ( 10, 4) and D ( 7, ). Part 1 Midpoint, distance and gradient 31

34 7 (Harder) The gradient of a line is 5. If one point on the line is (3,1), write the coordinates of another point on the line. (Graph paper can be found in the additional resources section if needed.) Check your response by going to the suggested answers section. Gradient is a very important feature of lines and intervals. It tells you how things are changing. You now have several methods for calculating the gradient of lines. The formula is often the fastest method, but you need to be very careful when using it. Use these skills to complete the following exercise. Go to the exercises section and complete Exercise 1.3 Gradient. 3 PAS5..3 Coordinate geometry

35 Geometry on the number plane By drawing shapes on a number plane, you can use coordinates to prove or disprove geometrical facts. Midpoints, gradients and lengths can all be used to show properties of shapes. In this section, you will be asked to complete various investigations and then interpret the results you find. Activity Geometry on the number plane Try these. 1 Triangle ABC has coordinates A (1, 8), B (9, 14) and C (7, 0). Plot these points and draw the triangle on the number plane below. y x 1 Part 1 Midpoint, distance and gradient 33

36 a b c Use the distance formula to find the length of each side. From these lengths, is ABC scalene, isosceles or equilateral? Use Pythagoras theorem to determine if ABC is a right-angled triangle or not. Show your working to support your conclusion. d Use the information you already have to calculate the area of ABC correct to decimal places. 34 PAS5..3 Coordinate geometry

37 Check your response by going to the suggested answers section. One advantage of creating a shape on a coordinate system like a number plane is that you can use the coordinates to explore properties. The number plane provides a separate measuring system that can be used to find lengths and slopes. Explore another shape by working through the following activity. Activity Geometry on the number plane Try these. The graph below shows a circle with diameter AB. 6 y B (6, 3) x 1 A (, 1) 3 4 a Use the diameter AB to calculate the coordinates of the centre of the circle. Call the centre C. b Calculate the radius of the circle. Part 1 Midpoint, distance and gradient 35

38 c (Harder) Does the point D (6., 0) lie inside, outside or on the circle? Show working to support your answer. (Hint: use the centre and the radius.) d (Harder) Plot the points E (4, 5) and F (, 3). Prove that ABEF is not a trapezium. Check your response by going to the suggested answers section. 36 PAS5..3 Coordinate geometry

39 The next investigation again asks you to draw some complex conclusions from some quite simple calculations. Activity Geometry on the number plane Try these. 3 The coordinates of the points A, B and C are A (1, 1 ), B ( 4, 13) and C(6, 1). a Find the gradient of the line through AB. b Find the gradient of the line through BC. c Look at the results above and comment on what this tells you about the three points. Check your response by going to the suggested answers section. Part 1 Midpoint, distance and gradient 37

40 This final investigation below asks you to explore midpoints of sides of triangles. Activity Geometry on the number plane Try these. 4 The triangle on the graph below has coordinates A ( 8, 5), B ( 4, 7) and C (4, 1). y 8 B C x A 5 6 a Find the coordinates of D which is the midpoint of AB and E which is the midpoint of AC. 38 PAS5..3 Coordinate geometry

41 b Calculate the length of BC and the length of DE. What do you notice? c Find the gradient of BC and the gradient of DE. What do you notice? d You have found that when you join the midpoints of sides AB and AC, you get an interval that is parallel to the third side and half its length. Part 1 Midpoint, distance and gradient 39

42 Also, since D and E are the midpoints, then AD is half AB and AE is half AC. All three pairs of sides are in equal ratio, so the triangles ABC and ADE are similar. Repeat these processes to show that this is true when using another two sides of the triangle. Check your response by going to the suggested answers section. Coordinates provide information that can be used to find lengths, midpoints and slopes. These features can be used to prove properties of figures. Continue to explore geometry on the number plane by completing the following exercise. Go to the exercises section and complete Exercise 1.4 Geometry on the number plane. 40 PAS5..3 Coordinate geometry

43 Suggested answers Part 1 Check your responses to the preliminary quiz and activities against these suggested answers. Your answers should be similar. If your answers are very different or if you do not understand an answer, contact your teacher. Activity Preliminary quiz 1 The hypotenuse is the longest side of a right-angled triangle. So for this triangle, the hypotenuse is BC (or CB). hypotenuse = (.5) = 4.5 hypotenuse = 4.5 = 6.5 cm x 1 y C A B 3 4 D 4 d = 5.9 or 5.9 (use your calculator to find the square root ). 5 One way to find the middle value between two numbers is to find their average by adding them together and divide by. Middle = = 90 = 45 Part 1 Midpoint, distance and gradient 41

44 6 45 = 9 5 = 9 5 = 3 5 Activity Midpoint 1 The midpoint is shown on the diagrams. a y (, 4 1 ) x 1 b 3 1 y x 1 1 (, ) The midpoint is shown on the diagram. A y (0, 1) x 1 3 B PAS5..3 Coordinate geometry

45 3 Your setting out may be different to this. a Average x-value = Average y-value = 1+19 = 0 = 1 = 6 = 10 Therefore midpoint is (6, 10). b Average x-value = = 1 Average y-value = = Therefore midpoint is ( 1, 9) c Midpoint = 3 +, = 1, 1 d Average x-value = Average y-value = = (half of 9 1 ) = 4 = 1 Therefore the midpoint is 4 3 4, a Midpoint = x + x 1 = = 4, 18 =, 9 ( ), y 1 + y 5 + 9, Part 1 Midpoint, distance and gradient 43

46 3 1 b Midpoint = + 7, 3 1 =, 18 = 1 3 4, Two methods are shown below. You may have chosen to use yet another method. Method one Using a graph Draw some axes on a piece of grid paper (provided in additional resources) to create the graph below. Your graph might look slightly different to this one. B 5 up y across Midpoint 1 5 up x across A To move from A to the midpoint you move 4 across and 5 up. So you must move these distances again to find the other endpoint. Therefore the coordinates of B are ( 3, 8). 44 PAS5..3 Coordinate geometry

47 Method Using the formula and equation solving This method uses each part of the formula to create an equation for the x-coordinate and another for the y-coordinate. These equations are then solved to find the coordinates of B. You may have used a similar method, but your working may be different to that shown below. x-coordinate of midpoint = x 1 + x Therefore: 1 = 5 + x = 5 + x (after doubling both sides) 3 = x (after subtracting 5 from both sides) So the x-coordinate of B is 3. The working below shows how to find the y-coordinate. y-coordinate of midpoint = y 1 + y 3 = + y 6 = + y 8 = y Therefore the coordinates of B are ( 3, 8). Activity Distance between points 1 a Run is from 5 to 3 = 8 units. Rise is from 3 to 1 = 4 units Distance = = 80 Distance = 80 = 8.9 units to one decimal place b This is a vertical interval so the length is just the rise = 5 units. Part 1 Midpoint, distance and gradient 45

48 Your working may be different to that shown below, but your answer should be the same. y Distance = units x 4 units 1 = = 5 Distance = 5 = 5 units 3 There are two possible right-angled triangles: one above the interval and one below. The one above is shown here. 6 y 5 6 units x units If you used ( 4, 3) as your first point and (, 5) as your second point your answer would be: Difference between the x-coordinates = 4 = 6 Difference between the y-coordinates = 3 5 = 8 If you used (, 5) as your first point and ( 4, 3) as your second point your answer would be: 46 PAS5..3 Coordinate geometry

49 Difference between the x-coordinates = 4 = 6 Difference between the y-coordinates = 5 8 = 8 Your comments should mention that the numerals are the same (that is 6 and 8) but when working with the coordinates you can get negative answers. 4 Two solutions are shown below. One uses (9, 4) as the first point, the other uses (17, ) as the first point. Look at both to see if it changes the answers. Using (9, 4) as the first point Difference between the x-coordinates = 9 17 = 8 So length of triangle = 8 Difference between the y-coordinates = 4 = 6 So height of triangle = 6 Using (17, ) as the first point Difference between the x-coordinates = 17 9 = 8 So length of triangle = 8 Difference between the y-coordinates = 4 = 6 So height of triangle = 6 Using Pythagoras theorem Distance = = 100 Distance = 100 = 10 The distance between the two points is 10 units. Part 1 Midpoint, distance and gradient 47

50 5 Your setting out might be different to this. Distance = = 169 Distance = 169 = 13 units 6 Beware: a common mistake is to mix up the x and y-coordinate. d = ( 10 ) + ( 9 7) = ( 1) + ( 16) = 400 = 0 The distance is 0 units. 7 Use the formula to find the distance between these pairs of points. Answer correct to one decimal place where necessary. a d = ( 0 6) + ( 5 1) = = 5 7. The distance is approximately 7. units. b d = ( 3 10) + ( 4 0) = ( 13) + ( 16) = The distance is approximately 0.6 units. c d = = 4 1 = ( 18 54) + ( 36) The distance is approximately 36.3 units. 48 PAS5..3 Coordinate geometry

51 Activity Gradient 1 a Rise =, run = 1 therefore gradient =. b Rise = 8, run = 4 therefore gradient = 8 4 = c d The two pairs of points gave the same gradient. You could select any pair of points on the line and the gradient would be the same. If you cannot work with the algebraic coordinates, put in numbers and work out what you would do to find the rise and run. Then go back to working with the pronumerals. The rise = distance from y up to y 1 = y 1 y The run = x 1 x Gradient = rise run = y 1 y x 1 x 3 a Gradient = = 5 5 = 1 b Gradient = 3 1 = 1 3 Part 1 Midpoint, distance and gradient 49

52 c 10 4 Gradient = 6 = 14 4 = 7 4 a Gradient = = 7 0 = error You cannot divide by zero so you cannot find a value for the gradient. b When you draw in some axes and plot the points, you find they lie on a vertical line. Vertical lines are said to have no gradient or an undefined gradient. 5 Lauren mixed the coordinates. She used the y-coordinate from (3, 8) as her first y-value, but then used the x-coordinates from (4, 5) as the first x-value. The correct working is: Gradient = = 3 1 or Gradient = = 3 1 = 3 = 3 6 Parallel lines have the same gradient. Gradient of AB = = 6 3 Gradient of CD = = 6 3 = = Since the gradients are the same, AB CD. (Remember, means is parallel to.) 50 PAS5..3 Coordinate geometry

53 7 There are an infinite number of solutions to this because there are an infinite number of points on the line. The best way to check your own answer is to use it in the formula to make sure you get a gradient of 5. Two methods of finding some answers are shown below. Method 1 Using a graph Plot the point (3, 1). Then draw in a right-angled triangle with a rise of 5 and a run of. Remember to draw the triangle so the line would have a positive slope. 6 y (5, 6) 1 y (3, 1) rise = x 1 rise = 5 (3, 1) 1 run = x (1, 4) run = From these graphs you can see two answers are (5, 6) and (1, 4). You could also use a rise of 10 and a run of 4 because 10 4 = 5 giving you the points (7, 11) and ( 1, 9). Or you could use a rise of 1 and a run of 1. These equivalent fractions would lead to all the other points on the line. Method Using the formula Let the missing point be ( x 1, y 1 ). Gradient = y 1 y x 1 x, therefore: 5 = y 1 1 x 1 3 Looking at the numerators, if y 1 1 = 5 then y 1 = 6. Part 1 Midpoint, distance and gradient 51

54 Looking at the denominators: if x 1 3 = then x 1 = 5. So one point is (5, 6). But any fraction equivalent to 5 will also give the correct answer. For example, you can use any of these equations to find a point = y 1 1 x = y 1 1 x = y 1 1 x = y 1 1 x 1 3 Activity Geometry on the number plane y B 9 8 A C x a AB = ( 1 9) + ( 8 14) = = 10 units BC = ( 9 7) + ( 14 0) = = 00 units (also written as 10 units) AC = ( 1 7) + ( 8 0) = = 10 units 5 PAS5..3 Coordinate geometry

55 b Isosceles (two sides equal, AC = AB) c BC = ( 00 ) = 00 AB + AC = = 00 Since BC = AB + AC then it is a right-angled triangle. d Since AB and AC are perpendicular (at right angles) then you can use these as the base and height of the triangle. A = 1 bh = = 50 Area of triangle ABC is 50 square units. a The centre will be the midpoint of the diameter. Midpoint = + 6, 1+ 3 = 4, = (, 1) C is (, 1). (Check on the graph that this looks reasonable.) b The radius is the distance from the centre to one end of the diameter. (You could also calculate the length of the diameter and halve it.) Radius = ( 6 ) + ( 3 1) = = 0 units (This can also be written as 5 units.) Part 1 Midpoint, distance and gradient 53

56 c Find the distance from the centre to the point D (6., 0). If it is exactly equal to the radius, then the point is on the circle. If the distance is less than the radius then the point is insides the circle, or it is outside if the distance is more than the radius. CD = ( 6.) + ( 1 0) = ( 4.) + 1 = units The radius is 0 units so CD is too short. That means D is inside the circle. d A trapezium is a quadrilateral that has at least one pair of opposite sides parallel. So you need to show that none of the sides are parallel. The graph seems to show this, but in a proof you must show that they are definitely not parallel by looking at their gradients. Gradient AB = = 4 8 = 1 Gradient EF = = 6 = 1 3 Gradient BE = = = 1 Gradient AF = 3 1 (So AF is vertical.) = 4 0 = undefined None of the lines have the same gradient, so none are parallel and therefore ABEF is not a trapezium. 54 PAS5..3 Coordinate geometry

57 3 a Gradient AB = = 5 = 5 10 (after doubling the top and bottom) = 5 b 1 13 Gradient BC = 6 4 = 5 10 = 5 c The gradients are the same, and both share the point B. So A, B and C all lie in the same line. (There is only one line going through B with a slope of 5. Every other line through B will have a different slope.) 4 a D = = 1, = 6, 1 ( ), b BC = ( 4 4) + ( 7 1) = ( 8) + 6 = = 10 units E = 8 + 4, = 4, 4 =, DE = ( ) ( 6 ) + ( 1 ) = ( 4) + 3 = 5 = 5 units BC is twice the length of DE. Part 1 Midpoint, distance and gradient 55

58 c Gradient BC = = 6 8 = 3 4 Gradient DE = 1 6 = 3 4 = 3 4 BC is parallel to DE because the gradients are the same. d You need to find the midpoint of BC. You may have called it F. You should find the DF is parallel to AC and half its length. You should also find that DE is parallel to AB and half its length. By joining the midpoints of the sides of any triangle you form intervals that are parallel to the third side and half its length. This means that the small triangle formed is similar to the original large triangle. 56 PAS5..3 Coordinate geometry

59 Additional resources Part 1 Part 1 Midpoint, distance and gradient 57

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61 Part 1 Midpoint, distance and gradient 59

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63 Part 1 Midpoint, distance and gradient 61

64 6 PAS5..3 Coordinate geometry

65 Exercises Part 1 Exercises 1.1 to 1.4 Name Teacher Exercise 1.1 Midpoint 1 Use the graph to find the coordinates of the midpoint of interval EF y F x E 1 Find the midpoint between these pairs of points without using a graph. a (3, 0) and (15, 0) b (8, 9) and (3, 1) c ( 3 1, 0) and (4 1, 7) Part 1 Midpoint, distance and gradient 63

66 3 (Harder) The midpoint of CD is ( 6, 3). If D is the point (, 7), find the coordinates of C. 64 PAS5..3 Coordinate geometry

67 Exercise 1. Distances between points 1 Use the graphs below to find the length of each interval. Answer correct to one decimal place where necessary. y 4 A 3 B C x 1 D 3 a Interval AB b Interval CD Use any method you prefer to find the distance between these pairs of points. Show enough working to explain your method. If you wish to use graph paper, you can find some in the additional resources section in this part. Attach any graphs to this exercise. a ( 18, 7) and (6, 17) Part 1 Midpoint, distance and gradient 65

68 b (4, ) and (1, ) c (6 1, 0) and (4, 1) 3 Use the formula to find the distance from ( 8, 15) to ( 4, ). Show your working. Answer correct to two decimal places. 66 PAS5..3 Coordinate geometry

69 Exercise 1.3 Gradient 1 Calculate the gradient of the line that passes through each pair of points. a (8, 0) and (1, 6) b (0, 5) and ( 3, ) c ( 15, ) and ( 5, 6) a Show that the line passing through (9, 5) and (4, 5) has a gradient of zero. b Plot (7, ) and (1, ) on the grid below and use your graph to explain what a gradient of zero tells you about the line. Part 1 Midpoint, distance and gradient 67

70 3 One way to calculate the gradient of the line through (5, 17) and ( 7, 13) is to use (5, 17) as the first point ( x 1, y 1 ). The working is shown below. Gradient = y y 1 x 1 x = 5 7 = 4 3 = 1 8 Show that you get the same result when you select ( 7, 13) as the first point ( x 1, y 1 ). 4 Show that the lines AB and CD are parallel given the following coordinates: A (18, 7), B (6, 3), C ( 4, 1) and D ( 1, 11). 68 PAS5..3 Coordinate geometry

71 Exercise 1.4 Geometry on the number plane Use your own paper to record your answers to this exercise. Staple your solutions to this page. Graph paper is provided in the additional resources section of this booklet if you wish to use it. 1 The points A (4, 5), B (5, ), C (0, 3) and D ( 1, 0) form a quadrilateral. Your task is to prove ABCD is a parallelogram using three different methods. Explain your reasoning in each case. a b c Use the lengths of the sides to show ABCD is a parallelogram. Use the gradients of the sides to show ABCD is a parallelogram. Use the midpoints of the diagonals to show ABCD is a parallelogram. (Harder) The coordinates of three points are A (4, 10), B (1, 1) and C (, 8). Calculate the three lengths AB, BC and AC. Use these lengths to decide whether A, B and C are all collinear (in the same straight line). Give your reasons. Part 1 Midpoint, distance and gradient 69