Topic 4.3.1

1. Slope of a Line Topic 4.3.1

2. Topic 4.3.1 Slope of a Line California Standard: 7.0:Students verify that a point lies on a line, given an equation of the line.Students are able to derive linear equationsby using the point-slope formula. What it means for you: You’ll find the slope of a line given any two points on the line. • Key words: • slope • steepness • horizontal • vertical • rise over run

3. Topic 4.3.1 Slope of a Line By now you’ve had plenty of practice in plotting lines. Any line can be described by its slope — which is what this Topic is about.

4. Dx is pronounced “delta x” and just means “change in x.” Topic 4.3.1 Slope of a Line The Slope of a Line is Its Steepness The slope (or gradient) of a line is a measure of its steepness. The slope of a straight line is the ratio of the vertical change to the horizontal change between any two points lying on the line. The vertical change is usually written Dy, and it’s often called the rise. Dy = rise In the same way, the horizontal change is usually written Dx, and it’s often called the run. Dx = run

5. rise Dy Dy , provided Dx¹ 0 = = Slope = run Dx Dx vertical change horizontal change y2 – y1 , provided x2 – x1¹ 0 = m = x2 – x1 There is an important difference between positiveand negativeslopes — a positiveslope means the line goes “uphill” , whereas a line with a negativeslope goes “downhill” . Topic 4.3.1 Slope of a Line If you know the coordinates of any two points on a line, you can find the slope. The slope, m, of a line passing through points P1 (x1, y1) and P2 (x2, y2) is given by this formula:

6. 4 – 1 3 y2 – y1 m = = = 7 – 2 5 x2 – x1 3 So the slope is . 5 Topic 4.3.1 Slope of a Line Example 1 Find the slope of the line that passes through the points (2, 1) and (7, 4) and draw the graph. Solution You can use the formula to find the slope of the line. Solution continues… Solution follows…

7. 3 5 Notice how the line has a positiveslope, meaning it goes “uphill” from left to right. In fact, since the slope is ,the line goes 3 units upfor every 5 units across. Topic 4.3.1 Slope of a Line Example 1 Find the slope of the line that passes through the points (2, 1) and (7, 4) and draw the graph. Solution (continued) You know that the line passes through (2, 1) and (7, 4), so just join those two points up to draw the graph.

8. 3.m = 1.m = 2.m = 4.m = …….. …….. …….. …….. 1 3 2 4 3 4 – – 3 2 Topic 4.3.1 Slope of a Line Guided Practice In Exercises 1–4, find the slope of the line on the graph below. Solution follows…

9. 3 m = – 2 Topic 4.3.1 Slope of a Line Guided Practice 5. Find the slope of the line that passes through the points (1, 5) and (3, 2), and draw the graph on a copy of the coordinate plane opposite. 5. 6. 6. Find the slope of the line that passes through the points (3, 1) and (2, 4), and draw the graph on a copy of the coordinate plane opposite. m = –3 Solution follows…

10. Example 2 Be extra careful if any of the coordinates are negative. 4 + 2 6 4 – (–2) –3 –3 3 – 6 m = = = Topic 4.3.1 Slope of a Line Find the slope of the line that passes through the points (3, 4) and (6, –2). Solution = –2 So the slope is –2. Solution follows…

11. Here the slope is –2, which means that the line goes 2 units downfor every 1 unit across. 1 2 Topic 4.3.1 Slope of a Line This time the line has a negativeslope, meaning it goes “downhill” from left to right.

12. 3 + 6 3 3 9 1 a) m = = = = = = 3 – (–6) 5 – 2 2 2 + 4 6 2 2 3 – 1 2 – (–4) b) m = Topic 4.3.1 Slope of a Line Example 3 Find the slope of the lines through: a) (2, 5) and (–4, 2) b) (1, –6) and (3, 3) Solution Solution follows…

13. y2 – y1 y2 – y1 y2 – y1 y2 – y1 y2 – y1 x2 – x1 x2 – x1 x2 – x1 x2 – x1 x2 – x1 1 –(–5) –4 – (–3) –7 – (–7) 6 2 – 2 –6 – 0 1 – (–1) –3 – 5 –6 3 –(–1) m = = = 0 5 – (–1) –3 – 4 m = = = = –1 m = = = 0 1 2 6 m = = = – 7 m = = = – Topic 4.3.1 Slope of a Line Guided Practice Find the slope m of the line through each pair of points below. 7. (–1, 2) and (3, 2) 8. (0, –5) and (–6, 1) 9. (5, –7) and (–3, –7) 10. (4, –1) and (–3, 5) 11. (–1, –3) and (1, –4) Solution follows…

14. y2 – y1 y2 – y1 y2 – y1 y2 – y1 y2 – y1 x2 – x1 x2 – x1 x2 – x1 x2 – x1 x2 – x1 –1 – 0 7 – 2 –14 – 0 –32 –18 m = = = = m = = = = 15 –12 – 7 –19 –15 –1 5 19 1 –11 – 5 m = = = = – –14 –16 –50 –1 16 1 –17 – (–2) m = = = = 1 1 –3 – (–2) 10 m = = = = 0 – (–1) 1 1 – 0 14 Topic 4.3.1 Slope of a Line Guided Practice Find the slope m of the line through each pair of points below. 12. (5, 7) and (–11, –12) 13. (–2, –2) and (–3, –17) 14. (18, 2) and (–32, 7) 15. (0, –1) and (1, 0) 16. (0, 0) and (–14, –1) Solution follows…

15. 2k + 1 y2 – y1 2k – (–1) m = , which means that m = = x2 – x1 6 – 4 2 2k + 1 But the slope is 3, so = 3 2 5 Þk = 2 Topic 4.3.1 Slope of a Line Example 4 If the slope of the line that passes through the points (4, –1) and (6, 2k) is 3, find the value of k. Solution Even though one pair of coordinates contains a variable, k, you can still use the slopeformula in exactly the same way as before. Þ 2k + 1 = 6 Þ 2k = 5 Solution follows…

16. y2 – y1 y2 – y1 y2 – y1 y2 – y1 y2 – y1 x2 – x1 x2 – x1 x2 – x1 x2 – x1 x2 – x1 –c – (–2c) 7b – 4b –5k –2k m = = = –5 – 2 –3 – 1 7 – 1 9d – 7d 3b – b 2q – 3q 5d – 3d m = = = = – –7k –4 2d 2 c m = = = = k –7 2b 2d 3 b 6 m = = = – q m = = = = 1 Topic 4.3.1 Slope of a Line Guided Practice Find the slope m of the lines through the points below. 17. (7, –2c) and (10, –c) 18. (b, 1) and (3b, –3) 19. (2, 2k) and (–5, –5k) 20. (3q, 1) and (2q, 7) 21. (3d, 7d) and (5d, 9d) Solution follows…

17. y2 – y1 y2 – y1 y2 – y1 y2 – y1 y2 – y1 x2 – x1 x2 – x1 x2 – x1 x2 – x1 x2 – x1 15v – 12v 14t –(–3s) –7 –14d 14t + 3s 14d + 7 12c – 9c 18s – 2t 18s – 2t d – 10 10 – d m = = = = – 7k – 5k 3v 3c 2a – 4a m = = = = p – q m = = 2k k v q – p –2a a c m = = = m = = = Topic 4.3.1 Slope of a Line Guided Practice Find the slope m of the lines through the points below. 22. (4a, 5k) and (2a, 7k) 23. (9c, 12v) and (12c, 15v) 24. (p, q) and (q, p) 25. (10, 14d) and (d, –7) 26. (2t, –3s) and (18s, 14t) Solution follows…

18. y2 – y1 y2 – y1 y2 – y1 –3t –4 16 –4 2 3 2 2 13 –7 –7 –4 – 3 x2 – x1 x2 – x1 x2 – x1 3b 3b 7 7 3 4 9 5 2 3k + 2 3k + 2 3k –(–2) –10 –(–6) –8t –(–5t) 7b – 4b 7 – 4 2 2 3 m = = = = Þ (–7)(5) = 2(3k + 2) Þ –35 = 6k + 4 Þ 6k = –39 Þk = – m = = = = Þ –16 = 9b Þb = – 5 7 4 m= = = = –t –t = Þt = – Topic 4.3.1 Slope of a Line Guided Practice In Exercises 27–29 you’re given two points on a line and the line’s slope, m. Find the value of the unknown constant in each Exercise. 27. (–2, 3) and (3k, –4), m = 28. (4, –5t) and (7, –8t), m = 29. (4b, –6) and (7b, –10), m = Solution follows…

19. y2 – y1 y2 – y1 y2 – y1 10 58 58 2 1 1 1 2 1 5 1 x2 – x1 x2 – x1 x2 – x1 12 39 20 39 3k 3k 2 2 4 3 5 41 – (–17) –1 – (–3) 40 – 1 k – 7k 4 – (–6) m = = = = – 4 = Þ 12k = –1Þk = – m= = = = – v= Þ –4v= 5 Þv= – 2 12 – (–8) 5 2 –6k m = = = – t = Þt = – = – = – 174 58 × 78 4524 174 78 39 × 174 6786 78 Topic 4.3.1 Slope of a Line Guided Practice In Exercises 30–32 you’re given two points on a line and the line’s slope, m. Find the value of the unknown constant in each Exercise. 30. (–8, –6) and (12, 4), m = – v 31. (7k, –3) and (k, –1), m = 4 32. (1, –17) and (40, 41), m = – t Solution follows…

20. (i) m = …….. 1 1 (ii) m = …….. 9 2 (iii) m = …….. (iv) m = …….. (v) m = …….. Topic 4.3.1 Slope of a Line Independent Practice 1. Find the slope m of the lines shown below. 2 –2 0 Solution follows…

21. m = m = m = m = 3 1 4 1 2 4 3 2 Topic 4.3.1 Slope of a Line Independent Practice In Exercises 2–5, find the slope of the line that passes through the given points, and draw the graph on a copy of the coordinate plane below. 2. (–2, 1) and (0, 2) 3. (4, 4) and (1, 0) 4. (–5, 2) and (–1, 3) 5. (3, –3) and (7, 3) 5. 2. 3. 4. Solution follows…

22. m = – m = – t 4 s 5 3 m = 2d Topic 4.3.1 Slope of a Line Independent Practice In Exercises 6–10, find the slope of the line through each of the points. 6. (–3, 5) and (2, 1) 7. (0, 4) and (–4, 0) 8. (2, 3) and (4, 3) 9. (6d, 2) and (4d, –1) 10. (2s, 2t) and (s, 3t) m = 1 m = 0 Solution follows…

23. 1 1 5 1 1 2 3 2 4 3 1 32 25 12 d = t = – d = – Topic 4.3.1 Slope of a Line Independent Practice In Exercises 11–15, you’re given two points on a line and the line’s slope, m. Find the value of the unknown constant in each Exercise. 11. (3t, 7) and (5t, 9), m = – 12. (3k, 1) and (2k, 7), m = 13. (0, 14d) and (10, –6d), m = –1 14. (2t, –3) and (–3t, 5), m = 15. (0, 8d) and (–1, 4d), m = – t = –2 k = –18 Solution follows…

24. Topic 4.3.1 Slope of a Line Round Up Slope is a measure of how steep a line is — it’s how many units up or down you go for each unit across. If you go up or down a lot of units for each unit across, the line will be steep and the slope will be large (either large and positive if it goes up from left to right, or large and negative if it goes down from left to right).