1. coordinates, lines and increment
P.2
coordinates, lines and increment

2. Rectangular Coordinate System
x-axis
y-axis
origin
Quadrant I
Quadrant II
Quadrant III
Quadrant IV
The horizontal line is called the x-axis.
The vertical line is called the y-axis.
The point of intersection is the origin.

3. The four regions in the x-y plane are known as quadrants, labeled as follows:
Quadrant I
x > 0, y > 0
Quadrant IV
x > 0, y < 0
Quadrant III
x < 0, y < 0
Quadrant II
x < 0, y > 0

4. Example : Plotting Points
Plot the point (3,2). Start at the origin and move 3 units to the right. From that point,move 2 units up. Now plot your point

5. Plotting Points
2 units
4 units
Each point in the xy-plane corresponds to a unique rdered pair (a, b).
Plot the point (2, 4). Move 2 units right
Move 4 units up

6. Increment and distance
When a particle moves from one points in the plane to another, the net changes in its coordinates are called increments.

7. Definition
An increment in a variable is a net change in the that variable. If x changes from x1 to x2, the increment in x is

8. ( ) Theorem: Distance Formula The distance between two points P 1 ( )x y 1 = , ( ) P x y 2 = ,
and
, denoted by ( ) d P 1 2 , is

9. Example: Find the distance between the points (3,8) and (-1,2)

10. Midpoint Formula
The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is

11. Midpoint Formula
Find the midpoint M of the segment with endpoints (10, 5) and (6, 4).

12. Finding Ordered Pairs that are Solutions of Equations
For the following equation find three ordered pairs that are solutions Of the equation y = 5x+2
Let y = 3
3 = 5x + 2
5 = 5x
1 = x (1,3)
Let x = 0
y = 5(0) + 2
y = 2
(0, 2)
Let x = 1
y = 5(1) + 2
y = 7
(1, 7)

13. Definitions: X-Intercept
The x-intercept is a point on any graph where the graph touches the x-axis.
The y-coordinate of the x-intercept is always zero.
The x-intercept is denoted by the point (x ,0), where x is any real number.
The x-intercept is also known as the zero or root of an equation.

14. Definitions: Y-Intercept
The y-intercept is a point on any graph where the graph touches the y-axis.
The x-coordinate of the y-intercept is always zero.
The y-intercept is denoted by the point (0, y), where y is any real number.

15. Procedure for Finding Intercepts
1. To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in the equation and solve for x.
2. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in the equation and solve for y.

16. Example: Find Intercepts
Example: Find the x- and y-intercepts of the graph of y = x2 + 4x – 5.
To find the x-intercepts, let y = 0 and solve for x.
0 = x2 + 4x – 5 Substitute 0 for y.
0 = (x – 1)(x + 5) Factor.
x – 1 = 0 x + 5 = 0
x = x = – Solve for x.
So, the x-intercepts are (1, 0) and (–5, 0).
Example: Find Intercepts

17. Example: Find Intercepts
Example: Find the x- and y-intercepts of the graph of y = x2 + 4x – 5.
To find the y-intercept, let x=0 and solve for y.
y = (0) – 5 = –5
So, the y-intercept is (0, –5).
Example: Find Intercepts

18. Graph of the Functions

19. The graph of an equation in two variables x and y consists of the set of points in the xy-plane whose coordinates (x,y) satisfy the equation.

20. Graphing
The graph of an equation is found by plotting points that are solutions of the equation.
The intercepts of the graph are good points to find first.
x-intercept is an x-value where the graph intersects the x-axis. y = 0
y-intercept is a y-value where the graph intersects the y-axis. x = 0

21. Graphing an Equation by Point Plotting
Step 1 Find the intercepts.
Step 2 Find as many additional ordered pairs as needed.
Step 3 Plot the ordered pairs from Steps 1 and 2.
Step 4 Connect the points from Step 3 in a smooth line or curve.

22. Example: Sketch the graph of y = –2x + 3.
Step 1: Find the intercepts.
Step 2: Find as many additional ordered pairs as needed. x y
(x, y) –2 7
(–2, 7) –1 5
(–1, 5) 3
(0, 3) 1
(1, 1)
3/2
(3/2, 0)
Example continued

23. Example: Sketch the graph of y = –2x + 3.
3. Plot the points in the coordinate plane. y 8 x y
(x, y) –2 7
(–2, 7) –1 5
(–1, 5) 3
(0, 3) 1
(1, 1)
3/2
(3/2, 0) 4 x 4 4 8 –4
Example continued

24. Example: Sketch the graph of y = –2x + 3.
4. Connect the points with a straight line. y 8 4 x 4 4 8 –4
Example continued

25. Example 2: Graphing Intercepts
Graph 4y + 5x = 20.
Substitute zero for x: 4y = 20 or y = 5.
Hence, the y-intercept is (0,5).
Substitute zero for the y: 5x = 20 or x = 4.
Hence, the x-intercept is (4,0).

27. Example
Graph the equation
y = 5x + 2 3 1
-2/5 2 y x

28. Example: Sketch Graph (Quadratic Function)
Example: Sketch the graph of y = x 2. y x y
(x, y) –2 4
(–2, 4) –1 1
(–1, 1)
(0, 0)
(1, 1) 2
(2, 4) 3 9
(3, 9) 16
(4, 16) 8 6 2 x –2 2 4
Example: Sketch Graph (Quadratic Function)

29. Example: Sketch Graph (Absolute Value Function)
Example: Sketch the graph of y = | x | . y x y
(x, y) –2 2
(–2, 2) –1 1
(–1, 1)
(0, 0)
(1, 2)
(2, 2) 4 2 x –2 2
Example: Sketch Graph (Absolute Value Function)

30. Example: Is (3,5) on the graph of ?
Substitute x = 3 and y = 5 into the equation:
True!
Therefore, (3,5) is on the graph of the equation.

31. Definition:
The standard form of an equation of a circle with radius r and center (h, k) is

32. Definition: A circle is a set of points in the xy-plane that are a fixed distance r from a fixed point (h, k). The fixed distance r is called the radius, and the fixed point (h, k) is called the center of the circle. y
(x, y) r
(h, k) x

33. Standard form of an equation of a circle
where the center of the circle is at the origin (0,0) and with a radius of r.

34. Example
Graph x2 + y2 = 16  3 3 4 y x

35. Unit Circle equation
where the center of the circle is at the origin (0,0) and with a radius of 1, is called the unit circle.
Since the radius = 1, use the center (0,0) as a reference point and then move 1 point to the left, right, up and down.

37. Continued.

38. Continued.

39. Equation of Straight lines

40. Straight lines Definition:

42. Possibilities for a Line’s Slope

43. Possibilities for a Line’s Slope

44. Possibilities for a Line’s Slope

45. Possibilities for a Line’s Slope

47. Example: Find the Slope

48. Solution

49. Practice Exercise

50. Answer

51. Equation of a Horizontal Line
Y-intercept
is 4

52. Equation of a Vertical Line
X-intercept is -5

53. Example: Draw the graph of the equation x = 2.y
x = 2 x

54. Example : Graphing a Horizontal Line
Y-intercept is 5.

55. Practice Exercises

56. Answers to Practice Exercises

57. The Equation of the line
1- With two points P1 (x1 , y1 ) and
P2 (x2 , y2 )

58. Example

59. Solution

60. The Equation of the line
2- Point-slope Form of the Equation of a Line

61. Example : Writing the Point-Slope Equation of a Line
Write the point-slope form of the equati on
of the line passing through (1,3) with a
slope
of 4. Then solve the equation for . y

63. Practice Exercises 1. Write the point-slope form of the equ ation
of the line passing through (4,-1) with a
slope
of 8. Then solve the equation for . y
2. Write the point-slope form of the equ
ation
of the line passing through the points (
2,0)
and (0,2). Then solve the equation for . y -

64. Answers to Practice Exercises

65. 3. The Slope-Intercept Form of the Equation of a Line
Y-intercept is b
Slope is m

66. Example : Graphing by Using the Slope and y-Intercept

67. First use the -intercept 2, to
plot the point (0,2). Starting
at (0,2), move 3 units up and
1 unit to the right. This gives
us the second point of the line.
Use a straightedge to draw a
line through the tw y
o points.

68. Text Example
Graph the line whose equation is y = 2/3 x + 2.
Solution:
y = 2/3 x + 2
The slope is 2/3.
The y-intercept is 2.

69. Text Example cont. Graph the line whose equation is y = 2/3x + 2.
We plot the second point on the line by starting at
(0, 2), the first point. Then move 2 units up (the rise) and 3 units to the right (the run). This gives us a second point at (3, 4). -5 -4 -3 -2 -1 1 2 3 4 5

70. Practice Exercises

71. Answers to Practice Exercises

72. General Form of the Equation of a Line

73. Equations of Lines

74. Example : Finding the Slope and the y-Intercept

76. Practice Exercises

77. Answers to Practice Exercises

78. Answers to Practice Exercises

79. Definitions: Parallel Lines
Two lines are said to be parallel if they do not have any points in common.
Two distinct non-vertical lines are parallel if and only if they have the same slope and have different y-intercepts.

80. Definitions: Perpendicular Lines
Two lines are said to be perpendicular if they intersect at a right angle.
Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.

81. Parallel and Perpendicular Lines
Example the following lines are perpendicular

82. Text Example
Write an equation of the line passing through (-3, 2) and parallel to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.

83. Text Example cont.
Solution . Notice that the line passes through the point (-3, 2). Using the point-slope form of the line’s equation, we have x1 = -3 and y1 = 2. -5 -4 -3 -2 -1 1 2 3 4 5
(-3, 2)
Rise = 2
Run = 1
y = 2x + 1
y – y1 = m(x – x1)
y1 = 2
x1 = -3

84. Text Example cont.
Solution Parallel lines have the same slope. Because the slope of the given line is 2, m = 2 for the new equation. -5 -4 -3 -2 -1 1 2 3 4 5
(-3, 2)
Rise = 2
Run = 1
y = 2x + 1
y – y1 = m(x – x1)
y1 = 2
m = 2
x1 = -3

85. Text Example cont.
Solution The point-slope form of the line’s equation is
y – 2 = 2[x – (-3)]
y – 2 = 2(x + 3)
Solving for y, we obtain the slope-intercept form of the equation.
y – 2 = 2x + 6
y = 2x + 8

86. Text Example
Write an equation of the line passing through (-3, 2) and perpendicular to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.

87. Text Example cont.
Solution Using the point-slope form of the line’s equation, we have x1 = -3 and y1 = 2.
y – y1 = m(x – x1)
y1 = 2
x1 = -3

88. Text Example cont.
Solution perpendicular lines have the product of their slopes is -1.
Because the slope of the given line is 2, m = -1/2 for the new equation.
y – y1 = m(x – x1)
y1 = 2
m = -1/2
x1 = -3

89. Text Example cont.
Solution The point-slope form of the line’s equation is
y – 2 = -1/2[x – (-3)]
y – 2 = -1/2(x + 3)
Solving for y, we obtain the slope-intercept form of the equation.
y – 2 = -1/2x -3/2
y = -1/2x + 1/2

90. Find the equation of the line parallel to y = -3x + 5 passing through (1,5).
Since parallel lines have the same slope, the slope of the parallel line is m = -3.

91. Example: Find the equation of the line perpendicular to y = -3x + 5 passing through (1,5).
Slope of perpendicular line: