Distance. Dollars. Reviewing gradient

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1 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 2 Negative gradient 3 Part 1 Midpoint, distance and gradient 1

2 Lines with the same gradient are parallel y x 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 2 PAS5.2.3 Coordinate geometry

3 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 B rise = 2 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 =+ 2 4 = 1 2 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 2 C x D Part 1 Midpoint, distance and gradient 3

4 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. 4 PAS5.2.3 Coordinate geometry

5 Activity Gradient Try this. 2 The diagram below shows no scale on the axes. Two points have been plotted on the line: A (, y 1 ) and B ( x 2, y 2 ). y A (, y 1 ) B ( x 2, y 2 ) 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 5

6 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 (2, 7) and (6, 1). Solution The two solutions below show that you get the same gradient no matter which point you decide will be (, y 1 ). The solutions also show that the sign of the gradient is found by the formula. Method 1 Using (2, 7) as the first point (, y 1 ) Gradient = y 1 y 2 x 2 = (2, 7) (6, 1) = 6 4 = PAS5.2.3 Coordinate geometry

7 Method 2 Using (6, 1) as the first point (, y 1 ) Gradient = y 1 y 2 x 2 = (6, 1) (2, 7) = 6 4 = 3 2 Notice that the result was the same no matter which point you chose to use as (, y 1 ). Also notice that the sign (+ or ) is automatically determined by the formula. So the line joining (6, 1) and (2, 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, 2) and (0, 7) Part 1 Midpoint, distance and gradient 7

8 b (2, 3) and ( 1, 2) c ( 6, 10) and ( 2, 4) 4 a Use the formula to calculate the gradient of the line that passes through (3, 5) and (3, 2). b Plot the points (3, 5) and (3, 2) and comment on why the gradient is unusual. 8 AS5.2.3 Coordinate geometry

9 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, 2), B (8, 4), C ( 10, 4) and D ( 7, 2). Part 1 Midpoint, distance and gradient 9

10 7 (Harder) The gradient of a line is 5. If one point on the line is 2 (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. 10 PAS5.2.3 Coordinate geometry

11 Activity Gradient 1 a Rise = 2, run = 1 therefore gradient = 2. b Rise = 8, run = 4 therefore gradient = 8 4 = 2 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. 2 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 2 up to y 1 = y 1 y 2 The run = x 2 Gradient = rise run = y 1 y 2 x 2 3 a Gradient = = 5 5 = 1 b Gradient = = 1 3 Part 1 Midpoint, distance and gradient 11

12 c 10 4 Gradient = 6 2 = 14 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 = 2 = 2 Since the gradients are the same, AB CD. (Remember, means is parallel to.) 12 PAS5.2.3 Coordinate geometry

13 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 2. 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 2. Remember to draw the triangle so the line would have a positive slope. 6 y (5, 6) 2 1 y (3, 1) rise = x 1 rise = (3, 1) 1 run = x (1, 4) run = 2 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 2 giving you the points (7, 11) and ( 1, 9). Or you could use a rise of and a run of 1. These equivalent fractions would lead to all the other points on the line. Method 2 Using the formula Let the missing point be (, y 1 ). Gradient = y 1 y 2 x 2, therefore: 5 2 = y Looking at the numerators, if y 1 1 = 5 then y 1 = 6. Part 1 Midpoint, distance and gradient 13

14 Looking at the denominators: if 3 = 2 then = 5. So one point is (5, 6). But any fraction equivalent to 5 2 will also give the correct answer. For example, you can use any of these equations to find a point = y = y = y = y PAS5.2.3 Coordinate geometry