Lesson 5 MI/Vocab

1. Write the equation of a line in point-slope form. • Write linear equations in different forms. • point-slope form Lesson 5 MI/Vocab

2. Lesson 5 KC 1

3. Write the point-slope form of an equation for a line that passes through (–2, 0) with slope Answer: Write an Equation Given Slope and a Point Point-slopeform (x1, y1) = (–2, 0) Simplify. Lesson 5 Ex1

4. Write the point-slope form of an equation for a line that passes through (4, –3) with slope –2. • A • B • C • D A.y – 4 = –2 (x + 3) B.y + 3 = –2 (x – 4) C.y – 3 = –2 (x – 4) D.y + 4 = –2 (x – 3) Lesson 5 CYP1

5. Write an Equation of a Horizontal Line Write the point-slope form of an equation for a horizontal line that passes through (0, 5). y – y1 = m(x – x1) Point-slopeform y – 5 = 0 (x – 0) (x1, y1) = (0, 5) y – 5 = 0 Simplify. Answer: The equation is y – 5 = 0. Lesson 5 Ex2

6. Write the point-slope form of an equation for a horizontal line that passes through (–3, –4). • A • B • C • D A.y + 4 = 0 B.y + 3 = 0 C.y – 4 = 0 D.x + 3 = 0 Lesson 5 CYP2

7. Lesson 5 CS 1

8. Write an Equation in Standard Form In standard form, the variables are on the left side of the equation. A, B, and C are all integers. Original equation Multiply each side by 4 to eliminate the fraction. Distributive Property Lesson 5 Ex3

9. Write an Equation in Standard Form 4y – 3x = 3x – 20 – 3x Subtract 3x from each side. –3x + 4y = –20 Simplify. Answer: The standard form of the equation is –3x + 4y = –20 or 3x – 4y = 20. Lesson 5 Ex3

10. Write y – 3 = 2(x + 4) in standard form. • A • B • C • D A. –2x + y = 5 B. –2x + y = 7 C. –2x + y = 11 D. 2x + y = 11 Lesson 5 CYP3

11. Write an Equation in Slope-Intercept Form Original equation Distributive Property Add 5 to each side. Lesson 5 Ex4

12. Answer: The slope-intercept form of the equation is Write an Equation in Slope-Intercept Form Simplify. Lesson 5 Ex4

13. A. B.y = –3x + 6 C.y = –3x + 3 D.y = 2x + 3 Write 3x + 2y = 6 in slope-intercept form. • A • B • C • D Lesson 5 CYP4

14. A. GEOMETRYThe figure shows trapezoid ABCD with bases AB and CD. Write an Equation in Point-Slope Form Write the point-slope form of the lines containing the bases of the trapezoid. Lesson 5 Ex5

15. Step 1 First find the slopes of AB and CD. AB: CD: Write an Equation in Point-Slope Form Slope formula (x1, y1) = (–2, 3) (x2, y2) = (4, 3) Slope formula (x1, y1) = (1, –2) (x2, y2) = (6, –2) Lesson 5 Ex5

16. AB: AB: Method 1 Use (–2, 3). Method 2 Use (4, 3). Write an Equation in Point-Slope Form Step 2 You can use either point for (x1, y1) in the point-slope form. y – y1 = m(x – x1) y – y1 = m(x – x1) y – 3 = 0(x + 2) y – 3 = 0(x – 4) y – 3 = 0 y – 3 = 0 Lesson 5 Ex5

17. CD: Method 1 Use (1, –2). CD: Method 2 Use (6, –2). Answer: The point-slope form of the equation containing AB is y – 3 = 0. The point-slope form of the equation containing CD is y + 2 = 0. Write an Equation in Point-Slope Form y – y1 = m(x – x1) y – y1 = m(x – x1) y + 2 = 0(x – 1) y + 2 = 0(x – 6) y + 2 = 0 y + 2 = 0 Lesson 5 Ex5

18. AB: y – 3 = 0 CD: y + 2 = 0 Write an Equation in Point-Slope Form B.Write each equation in standard form. Original equation y – 3 + 3 = 0 + 3 Add 3 to each side. Answer:y = 3 Simplify. Original equation y + 2 – 2 = 0 – 2 Subtract 2 from each side Answer:y = –2 Simplify. Lesson 5 Ex5

19. A. The figure shows right triangle ABC. Write the point-slope form of the line containing the hypotenuse AB. • A • B • C • D A.y – 6 = 1(x – 4) B.y – 1 = 1(x – 3) C.y + 4 = 1(x + 6) D.y – 4 = 1(x – 6) Lesson 5 CYP5

20. B. The figure shows right triangle ABC. Write the equation in standard form of the line containing the hypotenuse. • A • B • C • D A. –x + y = 10 B. –x + y = 3 C. –x + y = –2 D.x – y = 2 Lesson 5 CYP5

21. End of Lesson 5

22. Five-Minute Check (over Lesson 4-5) Main Ideas and Vocabulary Targeted TEKS Key Concept: Scatter Plots Example 1: Analyze Scatter Plots Example 2: Find a Line of Fit Example 3: Linear Interpolation Lesson 6 Menu

23. Interpret points on a scatter plot. • Use lines of fit to make and evaluate predictions. • scatter plot • positive correlation • negative correlation • line of fit • best-fit line • linear interpolation Lesson 6 MI/Vocab

24. Lesson 6 Key Concept 1

25. Analyze Scatter Plots TECHNOLOGY Determine whether the graph shows a positive correlation, a negative correlation, or no correlation. If there is a positive or negative correlation, describe it. The graph shows the average students per computer in Maria’s school. Answer: The graph shows a negative correlation. With each year, more computers are in Maria’s school, making the students-per-computer rate smaller. Lesson 6 Ex1

26. Determine whether the graph shows a positive correlation, a negative correlation, or no correlation. If there is a positive or negative correlation, describe it.The graph shows the number of mail-order prescriptions. • A • B • C • D A. Positive correlation; with each year, the number of mail-order prescriptions has increased. B. Negative correlaton; with each year, the number of mail-order prescriptions has decreased. C. No correlation D. Cannot be determined Lesson 6 CYP1

27. Find a Line of Fit POPULATIONThe table shows the world population growing at a rapid rate. Interactive Lab:Analyzing Linear Equations Lesson 6 Ex2

28. Find a Line of Fit A. Draw a scatter plot and determine what relationship exists, if any, in the data. Let the independent variable x be the year and let the dependent variable y be the population (in millions). The scatter plot seems to indicate that as the year increases, the population increases. There is a positive correlation between the two variables. Lesson 6 Ex2

29. Find a Line of Fit B. Draw a line of fit for the scatter plot. No one line will pass through all of the data points. Draw a line that passes close to the points. A line is shown in the scatter plot. Lesson 6 Ex2

30. Find a Line of Fit C. Write the slope-intercept form of an equation for equation for the line of fit. The line of fit shown passes through the data points (1850, 1000) and (2004, 6400). Step 1 Find the slope. Slope formula Let (x1, y1) = (1850, 1000) and (x2, y2) = (2004, 6400) Simplify. Lesson 6 Ex2

31. Find a Line of Fit Step 2 Use m= 35.1 and either the point-slope form or the slope-intercept form to write the equation. You can use either data point. We chose (1850, 1000). Point-slope form y – y1 = m(x – x1) y – 1000  35.1 (x – 1850) y – 1000  35.1x – 64,935) y 35.1x – 63,935 Lesson 6 Ex2

32. Find a Line of Fit Slope-intercept form y = mx + b 1000 = 35.1 (1850) + b 1000  64,935 + b –63,935 b y 35.1x – 63,935 Answer: The equation of the line is y  35.1 – 63,935. Lesson 6 Ex2

33. The table shows the number of bachelor’s degrees received since 1988. • A • B • C • D Lesson 6 CYP2

34. A. Draw a scatter plot and determine what relationship exists, if any, in the data. • A • B • C • D A. There is a positive correlation between the two variables. B. There is a negative correlation between the two variables. C. There is no correlation between the two variables. D. Cannot be determined Lesson 6 CYP2

35. B. Draw a line of best fit for the scatter plot. • A • B • C • D A.B. C.D. Lesson 6 CYP2

36. C. Write the slope-intercept form of an equation for the line of fit. • A • B • C • D A.y = 8x + 1137 B.y = –8x + 1104 C.y = 6x + 47 D.y = 8x + 1104 Lesson 6 CYP2

37. Linear Interpolation Use the prediction equation y 35.1x – 63,935, where x is the year and y is the population (in millions), to predict the world population in 2010. y 35.1x – 63,935 Original equation y 35.1 (2010) – 63,935 Replace x with 2010. y 6616 Simplify. Answer: 6,616,000,000 Lesson 6 Ex3

38. Use the equation y = 8x + 1104, where x is the years since 1998 and y is the number of bachelor’s degrees (in thousands), to predict the number of bachelor’s degrees that will be received in 2015. • A • B • C • D A. 1,204,000 B. 1,104,000 C. 1,104,008 D. 1,264,000 Lesson 6 CYP3