1. Chapter 4 Point-slope form

2. Write equations for lines using point-slope form.
I CAN:
Write equations for lines using point-slope form.

3. Point-slope form y – y1 = m (x – x1) Point: (3,5) Slope: -2
y – 5 = -2 (x – 3)
y – y1 = m (x – x1)

4. y – y1 = m (x – x1) m = slope (x1, y1) = a given point
Point-slope form
y – y1 = m (x – x1)
m = slope
(x1, y1) = a given point
This is easy to use if you know a
point and the slope

5. Point-slope form y – y1 = m (x – x1) Step 1: Write :
Step 2: Substitute the values for:
x1, y1 and m.
Step 3: Distribute.
Step 4: Solve for: y
y – y1 = m (x – x1)

6. Example 1 y – y1 = m (x – x1) y – 4 = -3 (x – 2) y – 4 = -3x + 6
Write an equation for the line with a slope of –3 passing through (2, 4).
Step 1:
Step 2:
Step 3:
Step 4:
y – y1 = m (x – x1)
y – 4 = -3 (x – 2)
y – 4 = -3x + 6
y = -3x + 10

7. Example 2 y – y1 = m (x – x1) y – 1 = ¾(x – (-8)) y – 1 = ¾(x + 8)
Write an equation for the line with a slope of ¾ passing through (–8, 1)
Step 1:
Step 2:
Step 3:
Step 4:
y – y1 = m (x – x1)
y – 1 = ¾(x – (-8))
y – 1 = ¾(x + 8)
y – 1 = ¾x + 6
y = ¾x + 7

8. y = -1x – 5 y = -4 y – y1 = m (x – x1) y – y1 = m (x – x1)
Your turn
1. (1, -6) m = -1
2. (-3, -4) m = 0
y – y1 = m (x – x1)
y – y1 = m (x – x1)
y – -6 = -1 (x – 1)
y – -4 = 0 (x – -3)
y + 6 = -1x + 1
y + 4 = 0
y = -1x – 5
y = -4

9. Example 3 y – 3 = 2 (x – 4) y – 3 = 2x - 8 y = 2x - 5
Simplify equations in point-slope form:
Step 2:
Step 3:
Step 4:
y – 3 = 2 (x – 4)
y – 3 = 2x - 8
y = 2x - 5

10. Example 4 y + 9 = -5 (x – 1) y + 9 = -5x + 1 y = -5x - 8
Simplify equations in point-slope form:
Step 2:
Step 3:
Step 4:
y + 9 = -5 (x – 1)
y + 9 = -5x + 1
y = -5x - 8

11. Your turn y + 3 = -2 (x – 5) y – 9 = -2 (x + 1) y – 1 = 7 (x + 2)
Simplify:
y + 3 = -2 (x – 5)
y – 9 = -2 (x + 1)
y – 1 = 7 (x + 2)
y = -2x + 7
y = -2x + 7
y = 7x + 15

12. 4.2 SLOPE & ONE POINT

13. 4.2 Learning Target
I can:
write equations for lines if given the slope and one point on the line.

14. DO YOU REMEMBER??? What is standard form?
What is direct variation?
What is standard form?
What is slope-intercept form?
What 2 things do you need for both graphing a line and writing an equation in slope-intercept form?
DO YOU REMEMBER???

15. Step 1: Write _______________________.
SLOPE AND ONE POINT
Step 1: Write _______________________.
Step 2: Fill in the slope (____) and the coordinates of the given point. (____ , ____)
Step 3: Solve to find _______.
y – y1 = m (x – x1) m x1 y1 y

16. y = 2x + 1 EXAMPLE #1: y – y1 = m (x – x1) y – 7 = 2 (x – 3)
Write the equation for the line with a slope of 2 that passes through (3, 7).
y – y1 = m (x – x1)
y – 7 = 2 (x – 3)
y – 7 = 2x – 6
y = 2x + 1

17. y = ½x EXAMPLE #2: y – y1 = m (x – x1) y – -3 = ½(x – -6)
Write the equation for the line with a slope of ½ that passes through (–6, –3).
y – y1 = m (x – x1)
y – -3 = ½(x – -6)
y + 3 = ½(x + 6)
y + 3 = ½x + 3
y = ½x

18. y = -4x - 1 YOUR TURN #2: y – y1 = m (x – x1)
Write the equation for the line with a slope of –4 that passes through (1, –5).
y – y1 = m (x – x1)
y = -4x - 1

19. y = 7 EXAMPLE #3: y – y1 = m (x – x1) y – 7 = 0 (x – 3) y – 7 = 0
Write the equation for the line with a slope of 0 that passes through (3, 7). Then graph it.
y – y1 = m (x – x1)
y – 7 = 0 (x – 3)
y – 7 = 0
y = 7

20. y = ¼x - 1 YOUR TURN #3: y – y1 = m (x – x1)
Write the equation for the line with a slope of ¼ that passes through (–8, –3).
y – y1 = m (x – x1)
y = ¼x - 1

22. 4.3 TWO POINTS

23. Write an equation for a line given two points
4.3 Learning Target
I CAN:
Write an equation for a line given two points

24. TWO POINTS Step 1: Calculate the _____________ of the line. SLOPE
Step 2: Use point-slope form to write an equation for the line.
SLOPE

25. y = 2x + 1 EXAMPLE #1: y – y1 = m (x – x1) y – 3 = 2 (x – 1)
Write the equation for the line that passes through
(1, 3) and (–3, –5).
Step 1: Calculate the slope: 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = −5−3 −3−1 = −8 −4 =2
Step 2: Find the y-intercept using the slope and one point.
y – y1 = m (x – x1)
y – 3 = 2 (x – 1)
y – 3 = 2x – 2
y = 2x + 1

26. y = -3/4x + 2 EXAMPLE #2 y – y1 = m (x – x1) y – -1 = -3/4 (x – 4)
Write the equation for the line that passes through
(4, –1) and (–4, 5).
Step 1: Calculate the slope: 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 5−−1 −4−4 = 6 −8 =− 3 4
Step 2: Find the y-intercept using the slope and one point.
y – y1 = m (x – x1)
− 3 4 ∗− 4 1 = 12 4 =3
y – -1 = -3/4 (x – 4)
y +1 = -3/4x + 3
y = -3/4x + 2

27. y = 6/5x - 4 YOUR TURN #2 y – y1 = m (x – x1)
Write the equation for the line that passes through
(0, –4) and (5, 2).
Step 1: Calculate the slope: 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1
Step 2: Find the y-intercept using the slope and one point.
y – y1 = m (x – x1)
y = 6/5x - 4

28. EXAMPLE #3
Write the equation for the line that has an x-intercept of 4 and a y-intercept of –7.
Let’s write these as coordinate pairs: (4,0) & (0, -7)
Calculate the slope: 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = −7−0 0−4 = −7 −4 = 7 4
Let’s think…..
We know the slope: 7 4
We know the y-intercept: -7
PERFECT! That’s all we need 
y = 7/4x - 7

29. YOUR TURN #3
Write the equation for the line that passes through
(–2, 4) and (0, 6). Graph it.
Step 1: Calculate the slope: 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1
Step 2:
y – y1 = m (x – x1)

30. Notes: Lesson 5-6 A
Learning Target:
I can write an equation for a line through a given point that is parallel to a given line.
Lines that don’t touch because they have the same slope.

31. vOCAB
Parallel lines

32. Let’s take a look:
Let’s graph y = ½ x – 1 and include the point (4, -2) on the graph.

33. Example 1:
Find the equation of the line that is parallel to the given line y = ½ x -1 but that goes through the point (4, -2)
What is the equation of this line?

34. Example 2: Find the equation of the line parallel to
-3x + y = 4 that goes through the point (-2, 2).
-3x + y = 4
+3x = +3x
y = 3x + 4
m = ll m = 3
y – y1 = m ( x – x1)
y – 2 = 3 ( x – (-2))
y = 3 (x +2)
y = 3x + 8

35. Your Turn y = 2x - 1 m = 2 ll m = 2 y – y1 = m ( x – x1)
Find the equation of the line parallel to
y = 2x – 1 that goes through the point (1, -3).
y = 2x - 1
m = ll m = 2
y – y1 = m ( x – x1)
y – (-3) = 2 ( x – 1)
y = 2x - 2
y = 2x - 5

36. Your Turn 2 y = (4/3)x – 7/3 m = 4/3 ll m = 4/3 y – y1 = m ( x – x1)
Find the equation of the line parallel to
4x - 3y = 7 that goes through the point (6, -5).
y = (4/3)x – 7/3
m = 4/3 ll m = 4/3
y – y1 = m ( x – x1)
y – (-5) = 4/3 ( x – 6)
y = (4/3)x - 8
y = (4/3)x - 13

37. 4.5 TWO POINTS

38. Learning Target: Notes: Lesson 4.5
I can write an equation for a line that is perpendicular to a given line through a given point.
900
900
900
Lines that meet at 900 angles
900
900
900
900
900

39. VocaB
Perpendicular

40. Both lines have same slope
Parallel lines
Both lines have same slope
Ex. Given: slope = m = 1/2
parallel slope = ll m = 1/2
Perpendicular lines
Lines have opposite reciprocal slopes
Ex. Given: slope = m = 1/2
perpendicular slope = m = -2/1 T

41. Given m: ½ 7 4/5 10 ll m: ½ 7 4/5 10 m: -2 -1/7 5/4 1/10 T
For each given slope, provide the slope that is parallel and perpendicular
Given m: 7
4/5 10
ll m: 7
4/5 10 m: -2
-1/7
5/4
1/10 T

42. Lets look
let’s graph y = 2/3 x – 3 and include the point (2, 4) on the graph.

43. Example 1:
Find the equation of the line that is perpendicular to the given line y = 2/3x – 3 passing through the point (2, 4)
What is the equation of the blue line?
opposite reciprocal!
We know…..
Slope :
-3/2
Point :
(2, 4)

44. Given: y = 2/3x – 3 T y – y1 = m ( x – x1) y – 4 -3/2 (x – 2) =
(2, 4)
y – 4
-3/2
(x – 2) =
y – 4 = -3/2 x +3
y = -3/2x + 7

45. Example 2:
Find the equation of the line that is perpendicular to the given line
x + 2y = 10 passing through the point
(1, -3)
2y = -1x + 10
y = - ½ x + 5

46. Your turn! Try these and check with your table partner
Find the equation for the line that is:
Parallel to y = -4x + 7, passing through (8, 0)
y = -4x + 32
Perpendicular to y = -4x + 7, passing through (8, 0)
y = ¼ x – 2
Perpendicular to 2x + 4y = 12 , passing through (-1, 3)
y = 2x + 5

47. LINES OF BEST FIT

48. An effective way to see a relationship in data is to display the information as a __________________.
It shows how two variables relate to each other by showing how closely the data points _______ to a line.
scatter plot
fit
The following table presents information on tornado occurrences.
Make a scatter plot for the table.
Year
1950
1955
1960
1965
1970
1975
1980
1985
1990
1995
# of Tornadoes
201
593
616
897
654
919
866
684
1133
1234

49. Do you notice a trend? Year 1950 1955 1960 1965 1970 1975 1980 1985
1990
1995
# of Tornadoes
201
593
616
897
654
919
866
684
1133
1234
1200
1000
800
600
Do you notice a trend?
400
200
1950
1955
1960
1965
1970
1975
1980
1985
1990
1995

50. Scatter plots provide a convenient way to determine whether a ___________ exists between two variables.
correlation
A __________ correlation occurs when both variables increase.
positive
negative
A ___________ correlation occurs when one variable increases and the other variable decreases.
If the data points are randomly scattered there is _______ or no correlation.
little

51. Negative
Correlation
Positive
Correlation No
Correlation

52. 2 1 3 Which one? (Not in notes): 0 to 18 years old. 1. age and height
The scatter plots of data relate characteristics of children from
0 to 18 years old.
Match each scatter plot with the appropriate variables studied.
1. age and height
2. age and eye color
3. age and time needed to run a certain distance
no correlation between age and eye color
as your age increases your height also increases 2 1
as your age increases the time will decrease 3

53. Sometimes points on a scatter plot are represented by
a trend line or a _______________________.
You can study the line to see how the data behaves. You may have a basis predict what the data might be for values not given.
line of best fit
Example 2: Find the line of best fit for the scatter plot you made on the first page. To fit the line to the points, choose your line so that it best matches the overall trend. The line does not have to pass through any of the points.

54. Use the line of best fit to predict how many tornadoes may be reported in the United States in 2015 if the trend continues.
1200
1000
800
600
If the trend continues we predict that there will be 1200 tornadoes reported in 2015.
400
200
1950
1955
1960
1965
1970
1975
1980
1985
1990
1995
2000
2005
2010
2015

55. If the data points are close to the line of best fit, it is said to have a ___________correlation.
strong
strong positive
weak positive
strong negative
weak negative

56. Height and Arm Span Activity