1. More About Linear Equations Lesson 2-4 Part 2
Algebra 2
More About Linear Equations
Lesson 2-4 Part 2

2. Goals Goal Rubric To write and graph the equation of a line.
To write equations of parallel and perpendicular lines.
Level 1 – Know the goals.
Level 2 – Fully understand the goals.
Level 3 – Use the goals to solve simple problems.
Level 4 – Use the goals to solve more advanced problems.
Level 5 – Adapts and applies the goals to different and more complex problems.

3. Essential Question Big Idea: Equivalence
How are the slopes of two lines related?

4. Vocabulary
Parallel lines
Perpendicular lines

5. Review - Axial Intercepts
A y-intercept is the point where a graph intersects the y-axis. The x-coordinate of this point is always 0.
An x-intercept is the point where a graph intersects the x-axis. The y-coordinate of this point is always 0.

6. Example: Finding Intercepts From a Graph
Find the x- and y-intercepts.
The graph intersects the y-axis at (0, 1).
The y-intercept is (0, 1).
The graph intersects the x-axis at (–2, 0).
The x-intercept is (–2, 0).

7. Example: Intercepts From Equation
Find the x- and y-intercepts.
5x – 2y = 10
To find the x-intercept, replace y with 0 and solve for x.
To find the y-intercept, replace x with 0 and solve for y.
5x – 2y = 10
5x – 2(0) = 10
5x – 2y = 10
5(0) – 2y = 10
5x – 0 = 10
5x = 10
0 – 2y = 10
–2y = 10
x = 2
The x-intercept is (2, 0).
y = –5
The y-intercept is (0, –5).

8. Your Turn: Find the x- and y-intercepts.
The graph intersects the y-axis at (0, 3).
The y-intercept is (0, 3).
The graph intersects the x-axis at (–2, 0).
The x-intercept is (–2, 0).

9. Your Turn: Find the x- and y-intercepts. –3x + 5y = 30 –3x + 5y = 30
To find the y-intercept, replace x with 0 and solve for y.
To find the x-intercept, replace y with 0 and solve for x.
–3x + 5y = 30
–3(0) + 5y = 30
–3x + 5y = 30
–3x + 5(0) = 30
0 + 5y = 30
5y = 30
–3x – 0 = 30
–3x = 30
x = –10
The x-intercept is (–10, 0).
y = 6
The y-intercept is (0, 6).

10. Your Turn: Find the x- and y-intercepts. 4x + 2y = 16
To find the y-intercept, replace x with 0 and solve for y.
To find the x-intercept, replace y with 0 and solve for x.
4x + 2y = 16
4(0) + 2y = 16
4x + 2y = 16
4x + 2(0) = 16
0 + 2y = 16
2y = 16
4x + 0 = 16
4x = 16
x = 4
The x-intercept is (4, 0).
y = 8
The y-intercept is (0, 8).

11. Graphing Using Standard Form
For any two points, there is exactly one line that contains them both. This means you need only two ordered pairs to graph a line.
It is often simplest to find the ordered pairs that contain the x and y intercepts.
The x and y intercepts can easily be found from standard form. Therefore, to graph a linear equation in standard form, use the x and y intercepts.

12. Example: Graphing Standard Form
Use intercepts to graph the line given by the equation.
3x – 7y = 21
Step 1 Find the intercepts.
y-intercept:
x-intercept:
3x – 7y = 21
3x – 7(0) = 21
3x – 7y = 21
3(0) – 7y = 21
3x = 21
–7y = 21
x = 7
y = –3

13. Example: Continued
x-intercept: x = (7, 0) and y-intercept: y = (0, -3)
Step 2 Graph the line. x
Plot (7, 0) and (0, –3).
Connect with a straight line.

14. Example: Graphing Standard Form
Use intercepts to graph the line given by the equation.
y = –x + 4
Step 1 Write the equation in standard form.
y = –x + 4
+x +x
Add x to both sides.
x + y = 4

15. Example: Continued y-intercept: x-intercept: x + y = 4
Step 2 Find the intercepts.
y-intercept:
x-intercept:
x + y = 4
0 + y = 4
x + y = 4
x + 0 = 4
y = 4
x = 4

16. Example: Continued x-intercept: x = (4, 0) and y-intercept: y = (0, 4)
Step 3 Graph the line.
Plot (4, 0) and (0, 4).
Connect with a straight line.

17. Your Turn: x-intercept: y-intercept:
Use intercepts to graph the line given by the equation.
–3x + 4y = –12
Step 1 Find the intercepts.
x-intercept:
y-intercept:
–3x + 4y = –12
–3x + 4(0) = –12
–3x = –12
–3x + 4y = –12
–3(0) + 4y = –12
4y = –12
y = –3
x = 4

18. Your Turn: Continued
x-intercept: x = (4, 0) and y-intercept: y = (0, -3)
Step 2 Graph the line.
Plot (4, 0) and (0, –3).
Connect with a straight line.

19. Your Turn: Use intercepts to graph the line given by the equation.
Step 1 Write the equation in standard form.
Multiply both sides by 3, the LCD of the fractions, to clear the fraction.
3y = x – 6
Write the equation in standard form.
–x + 3y = –6

20. Your Turn: Continued y-intercept: x-intercept: –x + 3y = –6
Step 2 Find the intercepts.
y-intercept:
x-intercept:
–x + 3y = –6
–x + 3y = –6
–(0) + 3y = –6
–x + 3(0) = –6
3y = –6
–x = –6
x = 6
y = –2

21. Your Turn: Continued
x-intercept: x = (6, 0) and y-intercept: y = (0, -2)
Step 3 Graph the line.
Plot (6, 0) and (0, –2).
Connect with a straight line.

22. Equations of Horizontal and Vertical Lines

23. Equations of Horizontal and Vertical Lines
Equation of a Horizontal Line
A horizontal line is given by an equation of the form
y = b
where b is the y-intercept. Note: m = 0.
Equation of a Vertical Line
A vertical line is given by an equation of the form
x = a
where a is the x-intercept. Note: m is undefined.

24. Example: Horizontal LineY X
Let’s find the equation for the line passing through the points (0,2) and (5,2)
y = mx + b ( Slope Intercept Form ).
Where m is:
Y-axis
X-axis =
(2 – 2) DY
m =
= 0 DX
(5 – 0)
DY = 0 DX
(0,2)
(5,2)

25. Example: Horizontal LineY X
Because the value of m is 0,
y = 0x + 2 becomes
Y-axis
X-axis
y = 2 (A Constant Function)
(0,2)
(5,2)

26. Your Turn: Y X
Find the equation for the lines passing through the following points.
1.) (3,2) & ( 8,2)
2.) (-5,4) & ( 10,4)
3.) (1,-2) & ( 7,-2)
4.) (4,3) & ( -2,3)
y = 2
y = 4
y = -2
y = 3

27. Example: Vertical LineY X
Let’s look at a line with no y-intercept b, an x-intercept a, passing through (3,0) and (3,7).
Y-axis
X-axis
(3,0)
(3,7)

28. Example: Vertical LineY X
The equation for the vertical line is:
x = 3 ( 3 is the X-Intercept of the line).
Because m is:
Y-axis
X-axis
(3,0)
(3,7) DY =
(7 – 0) = 7
m =
= Undefined DX
(3 – 3)

29. Your Turn: Y X
Find the equation for the lines passing through the following points.
1.) (3, 5) & (3, -2)
2.) (-5, 1) & (-5, -1)
3.) (1, -6) & (1, 8)
4.) (4, 3) & (4, -4)
x = 3
x = -5
x = 1
x = 4

30. Equations of Vertical and Horizontal lines
An equation of the vertical line through the point (a, b) is x = a.
An equation of the horizontal line through the point (a, b) is y = b.

31. Let x = selling price and y = rent.
Example: Application
The table shows the rents and selling prices of properties from a game.
Selling Price ($)
Rent ($) 75 9 90 12
160 26
250 44
Express the rent as a function of the selling price.
Let x = selling price and y = rent.
Find the slope by choosing two points. Let (x1, y1) be (75, 9) and (x2, y2) be (90, 12).

32. Example: Continued
To find the equation for the rent function, use point-slope form.
y – y1 = m(x – x1)
Use the data in the first row of the table.
Simplify.

33. Example: Continued
Graph the relationship between the selling price and the rent. How much is the rent for a property with a selling price of $230?
To find the rent for a property, use the graph or substitute its selling price of $230 into the function.
Substitute.
y = 46 – 6
y = 40
The rent for the property is $40.

34. Your Turn:
Express the cost as a linear function of the number of items.
Let x = items and y = cost.
Find the slope by choosing two points. Let (x1, y1) be (4, 14) and (x2, y2) be (7, 21.50).
Items
Cost ($) 4
14.00 7
21.50 18

35. Continued
To find the equation for the number of items, use point-slope form.
y – y1 = m(x – x1)
Use the data in the first row of the table.
y – 14 = 2.5(x – 4)
y = 2.5x + 4
Simplify.

36. Continued
Graph the relationship between the number of items and the cost. Find the cost of 18 items.
To find the cost, use the graph or substitute the number of items into the function.
y = 2.5(18) + 4
Substitute.
y =
y = 49
The cost for 18 items is $49.

37. Parallel and Perpendicular Lines
By comparing slopes, you can determine if the lines are parallel or perpendicular. You can also write equations of lines that meet certain criteria.

38. Geometric Definition
Parallel Lines - are lines in the same plane that never intersect.
Line WX is parallel to line YZ.
WX || YZ.

39. Parallel Lines
These two lines are parallel. Parallel lines are lines in the same plane that have no points in common. In other words, they do not intersect.
As seen on the graph, parallel lines have the same slope.
For y = 3x + 100, m = 3 and for y = 3x + 50, m = 3.

40. Algebraic Definition

41. Geometric Definition
Perpendicular Lines – lines that intersect to form 90o angles, or right angles.
Line RS is perpendicular to line TU.
RS | TU. __

42. Algebraic Definition
Perpendicular lines have slopes that are opposite reciprocals.

43. Opposite Reciprocals Helpful Hint
If you know the slope of a line, the slope of a perpendicular line will be the "opposite reciprocal.”

44. Example: Writing Equations of Parallel Lines
Write an equation in slope-intercept form for the line that passes through (4, 10) and is parallel to the line described by y = 3x + 8.
Step 1 Find the slope of the line.
y = 3x + 8
The slope is 3.
The parallel line also has a slope of 3.
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Use the point-slope form.
Substitute 3 for m, 4 for x1, and 10 for y1.
y – 10 = 3(x – 4)

45. Example: Continued Step 3 Write the equation in slope-intercept form.
y – 10 = 3(x – 4)
y – 10 = 3x – 12)
Distribute 3 on the right side.
y = 3x – 2
Add 10 to both sides.

46. Your Turn:
Write an equation in slope-intercept form for the line that passes through (5, 7) and is parallel to the line described by y = x – 6.
Step 1 Find the slope of the line.
The slope is .
y = x –6
The parallel line also has a slope of
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Use the point-slope form.

47. Your Turn: Continued
Step 3 Write the equation in slope-intercept form.
Distribute on the right side.
Add 7 to both sides.

48. Example: Writing Equations of Perpendicular Lines
Write an equation in slope-intercept form for the line that passes through (2, –1) and is perpendicular to the line described by y = 2x – 5.
Step 1 Find the slope of the line.
y = 2x – 5
The slope is 2.
The perpendicular line has a slope of because
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Use the point-slope form.
Substitute for m, –1 for y1, and 2 for x1.

49. Example: Continued Step 3 Write the equation in slope-intercept form.
Distribute on the right side.
Subtract 1 from both sides.

50. Your Turn:
Write an equation in slope-intercept form for the line that passes through (–5, 3) and is perpendicular to the line described by y = 5x.
Step 1 Find the slope of the line.
y = 5x
The slope is 5.
The perpendicular line has a slope of because .
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Use the point-slope form.

51. Your Turn: Continued
Write an equation in slope-intercept form for the line that passes through (–5, 3) and is perpendicular to the line described by y = 5x.
Step 3 Write in slope-intercept form.
Distribute on the right side.
Add 3 to both sides.

52. a. parallel to the line 2x + 5y = 4
Your Turn:
Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition.
a. parallel to the line 2x + 5y = 4
b. perpendicular to the line 2x + 5y = 4
Solution (3, 5) is on the line so we need to find the slope to use the point-slope form. Write the equation in the slope-intercept form (solve for y).

53. a. parallel to the line 2x + 5y = 4
Continued
Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition.
a. parallel to the line 2x + 5y = 4
Solution
Subtract 2x.
Divide by 5.

54. a. parallel to the line 2x + 5y = 4
Continued:
Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition.
a. parallel to the line 2x + 5y = 4
Solution
The slope is – 2/5. Since the lines are parallel, – 2/5 is also the slope of the line whose equation is to be found.

55. a. parallel to the line 2x + 5y = 4
Continued:
Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition.
a. parallel to the line 2x + 5y = 4
Solution
Point-slope form
m = – 2/5, x1 = 3, y1 = 5
Distributive property

56. a. parallel to the line 2x + 5y = 4
Continued:
Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition.
a. parallel to the line 2x + 5y = 4
Solution
Distributive property
Add 5 (25/5).

57. b. perpendicular to the line 2x + 5y = 4
Continued:
Find the equation in slope-intercept form of the line that passes through the point (3, 5) and satisfies the given condition.
b. perpendicular to the line 2x + 5y = 4
Solution We know the slope of the line, so the slope of any line perpendicular to it is 5/2.

58. b. perpendicular to the line 2x + 5y = 4
Continued:
b. perpendicular to the line 2x + 5y = 4
Solution
Distributive property
Add 5 (10/2).

59. Essential Question Big Idea: Equivalence
How are the slopes of two lines related?
The slopes of two lines in the same plane indicate how the lines are related. Two lines are parallel if they have the same slope (and different y-intercepts). Two lines are perpendicular if their slopes are negative reciprocals.

60. Assignment
Section 2-4 Part 2, Pg95 – 97; #1 – 5 all, 6 – 44 even.