Slope of a Line Topic 4.3.1

Slope of a Line 4.3.1 Topic 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 equations by 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

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.

Slope of a Line 4.3.1 Topic 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 Dx is pronounced “delta x” and just means “change in x.”

Slope of a Line 4.3.1 Topic Slope = , provided Dx ¹ 0 vertical change horizontal change = rise run Dy Dx 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: = m = , provided x2 – x1 ¹ 0 Dy Dx y2 – y1 x2 – x1 There is an important difference between positive and negative slopes — a positive slope means the line goes “uphill” , whereas a line with a negative slope goes “downhill” .

Slope of a Line 4.3.1 Topic Solution 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. m y2 – y1 x2 – x1 = = 4 – 1 7 – 2 = 3 5 So the slope is . 3 5 Solution continues… Solution follows…

Slope of a Line 4.3.1 Topic Solution (continued) 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. Notice how the line has a positive slope, meaning it goes “uphill” from left to right. In fact, since the slope is , the line goes 3 units up for every 5 units across. 3 5

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

Slope of a Line 4.3.1 Topic 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. 6. 5. 3 2 m = – 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…

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

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

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 a) m = 5 – 2 2 – (–4) = 3 2 + 4 = 3 6 = 1 2 b) m = 3 – (–6) 3 – 1 = 3 + 6 2 = 9 2 Solution follows…

Slope of a Line 4.3.1 Topic 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) m = = = 0 y2 – y1 x2 – x1 2 – 2 3 –(–1) m = = = = –1 y2 – y1 x2 – x1 1 –(–5) –6 – 0 6 –6 m = = = 0 y2 – y1 x2 – x1 –7 – (–7) –3 – 5 m = = = – 6 7 y2 – y1 x2 – x1 5 – (–1) –3 – 4 m = = = – y2 – y1 x2 – x1 –4 – (–3) 1 – (–1) 1 2 Solution follows…

Slope of a Line 4.3.1 Topic 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) m = = = = y2 – y1 x2 – x1 –12 – 7 –11 – 5 –19 –16 16 19 m = = = = 15 y2 – y1 x2 – x1 –17 – (–2) –3 – (–2) –15 –1 m = = = = – y2 – y1 x2 – x1 7 – 2 –32 –18 5 –50 1 10 m = = = = 1 y2 – y1 x2 – x1 0 – (–1) 1 – 0 1 m = = = = y2 – y1 x2 – x1 –1 – 0 –14 – 0 –1 –14 1 14 Solution follows…

Slope of a Line 4.3.1 Topic Þ 2k + 1 = 6 Þ 2k = 5 Þ k = 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 slope formula in exactly the same way as before. m y2 – y1 x2 – x1 = , which means that m = 2k – (–1) 6 – 4 2k + 1 2 But the slope is 3, so = 3 2k + 1 2 Þ 2k + 1 = 6 Þ 2k = 5 Þ k = 5 2 Solution follows…

Slope of a Line 4.3.1 Topic 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) m = = = –c – (–2c) 7b – 4b c 3 y2 – y1 x2 – x1 m = = = = – –3 – 1 3b – b –4 2b 2 b y2 – y1 x2 – x1 m = = = = k –5k –2k –5 – 2 –7k –7 y2 – y1 x2 – x1 m = = = – 7 – 1 2q – 3q 6 q y2 – y1 x2 – x1 2d m = = = = 1 9d – 7d 5d – 3d y2 – y1 x2 – x1 Solution follows…

Slope of a Line 4.3.1 Topic 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) m = = = = – y2 – y1 x2 – x1 7k – 5k 2a – 4a 2k –2a k a m = = = = v c 3v 3c 15v – 12v 12c – 9c y2 – y1 x2 – x1 m = = p – q q – p y2 – y1 x2 – x1 m = = = –7 –14d d – 10 14d + 7 10 – d y2 – y1 x2 – x1 m = = = 14t –(–3s) 18s – 2t 14t + 3s y2 – y1 x2 – x1 Solution follows…

Slope of a Line 4.3.1 Topic 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 = m = = = = Þ (–7)(5) = 2(3k + 2) Þ –35 = 6k + 4 Þ 6k = –39 Þ k = – 2 5 y2 – y1 x2 – x1 –7 3k + 2 –4 – 3 3k –(–2) 13 2 5 m = = = = –t –t = Þ t = – y2 – y1 x2 – x1 –8t –(–5t) 7 – 4 2 7 –3t 3 2 7 3 4 m = = = = Þ –16 = 9b Þ b = – –4 3b 3 4 16 9 y2 – y1 x2 – x1 –10 –(–6) 7b – 4b Solution follows…

Slope of a Line 4.3.1 Topic 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 m = = = = – v = Þ –4v = 5 Þ v = – y2 – y1 x2 – x1 1 2 5 4 10 20 4 – (–6) 12 – (–8) 2 5 m = = = = – 4 = Þ 12k = –1 Þ k = – y2 – y1 x2 – x1 –1 – (–3) k – 7k 1 3k 12 2 –6k m = = = – t = Þ t = – = – = – y2 – y1 x2 – x1 174 78 58 39 41 – (–17) 40 – 1 58 × 78 39 × 174 4524 6786 2 3 174 78 Solution follows…

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

Slope of a Line 4.3.1 Topic 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) 1 2 m = 2. 5. 4. 3. 4 3 m = 1 4 m = 3 2 m = Solution follows…

Slope of a Line 4.3.1 Topic 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) 4 5 m = – m = 1 3 2d m = m = 0 t s m = – Solution follows…

Slope of a Line 4.3.1 Topic 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 = – 1 2 t = –2 1 3 k = –18 d = 1 2 5 4 t = – 32 25 1 3 d = – 1 12 Solution follows…

Slope of a Line 4.3.1 Topic 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).