1. GRE: Graphical Representations
COORDINATE GEOMETRY

2. The Coordinate Plane y
Coordinate planes are formed by two axes: the x-axis and the y-axis
The point where the two axes meet is called the origin.
The coordinates of the origin are (0, 0).
Coordinate planes are divided into 4 quadrants. x

3. Plotting Points in the Coordinate Plane
Points in a coordinate plane are identified according to their coordinates.
Coordinates are written as a pair: (x-coordinate, y-coordinate)
The x-coordinate indicates how far horizontally (left or right) the point is from the origin.
The y-coordinate indicates how far vertically (up or down) the point is from the origin
Example – the coordinates of point A in the coordinate plane is (3, 1) since it is 3 spaces to the right of the origin and 1 space above the origin A

4. You Try
Plot the following points in the coordinate plane on the left: A(-2, 4) B(0, -1) C(3, 0) D(5, -2) E(-6, -5)

5. You Try
State the coordinates of each point in the coordinate plane to the left. A B C D E

6. Finding the Slope Between Two Points
What is slope?
- Slope is the steepness of a line
- Slope is rate of change or the rate at which “y” changes as “x” increases by 1
- Slope is the rise (vertical change) of a line divided by its run (horizontal change)
- “m” is the symbol used to represent slope
Slope Formula: where (x1, y1) and (x2, y2) are points on the line

7. More Important Points About Slope
If a line goes down going from left to right, then the slope of the line is negative.
If a line goes up going from left to right, then the slope of the line is positive.
The steeper the line, the greater the slope.
The slope of horizontal lines is 0.
The slope of vertical lines is undefined.

8. Using the Slope Formula
Example 1 – Find the slope of the line that passes through (8, 2) and (-3, -4). Solution: Example 2 – Find the slope of the line that passes through (-7, 4) and (-5, -10). Which line in the examples is steeper? How do you know?

9. You Try
Find the slope of the line that passes through (5, -3) and (2, 3).
Find the slope of the line that passes through (-11, -9) and (11, -11).
Which of the lines is steeper?

10. Finding the Slope of a Line
When finding the slope of a line,
Find two points through which the line passes.
Plug those points into the slope formula.
Example 1 –
The line passes through
(-6, 5) and (0, -2)
2. m = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = −2 −5 0 −(−6) =
−𝟕 𝟔

11. You Try
1. Find the slope of the line in the coordinate plane on the right.

12. Finding Slope of Line from Tablex y -3 10 -2 5 -1 -5 1
-10
Example 1 – Find the slope of the line created from the data in the table on the left. Solution – Step 1 – pick any two points from the table. (-3, 10) and (-2, 5) Step 2 – plug those two points into the slope formula m = 𝒚 𝟐 − 𝒚 𝟏 𝒙 𝟏 − 𝒙 𝟐 = 𝟓 −𝟏𝟎 −𝟐 −(−𝟑) = −𝟓 𝟏 =−𝟓

13. You Try x y -2 -3 -9 2
-15 4
-21 6
-27
Find the slope of the line created from the data in the table on the left. Solution -

14. Writing Equations of Lines
When writing an equation of a line, all we need are the slope of the line and the coordinates of a point through which the line passes.
We can then plug the slope and point into the following formula: y – y1 = m(x – x1)
where m is the slope and (x1, y1) are the coordinates of the point
Lastly, we can manipulate the equation to get “y” by itself on the left side of the equation.

15. Examples of Writing an Equation of a Line
Example 1 – Write an equation of the line that passes through (3, 3) and has a slope of -1/3. Solution – Step 1 – Plug the slope and point into the formula y – y1 = m(x – x1) y – 3 = − 1 3 (x – 3) Step 2 – Manipulate the equation to get “y” by itself y – 3 = − 1 3 x y = − 𝟏 𝟑 x + 4

16. Examples of Writing an Equation of a Line
Example 1 – Write an equation of the line that passes through (-3, -5) and has a slope of 10/3. Solution – Step 1 – Plug the slope and point into the formula y – y1 = m(x – x1) y + 5 = 10 3 (x + 3) Step 2 – Manipulate the equation to get “y” by itself y + 5 = 10 3 x y = 𝟏𝟎 𝟑 x + 5

17. You Try
Write an equation of the line that passes through (5, 3) and has a slope of -1/5. Solution -

18. Writing an Equation of a Line from Tablex y -3 10 -2 5 -1 -5 1
-10
Example 1 - Write an equation of the line that is created from the data in the table on the left. Solution – Step 1 – find the slope of the line by first picking any two points from the table: (5, -2), (10, -3) m = 𝒚 𝟐 − 𝒚 𝟏 𝒙 𝟐 − 𝒙 𝟏 = 𝟓 −𝟏𝟎 −𝟐 −(−𝟑) = −𝟓 𝟏 =−𝟓 Step 2 – use the slope and any of the points in the table to plug into the formula y – y1 = m(x – x1) y – 10 = -5(x + 3) Step 3 – manipulate equation to get “y” by itself y – 10 = -5x – y = -5x - 5

19. You Try x y -2 -3 -9 2
-15 4
-21 6
-27
Write an equation of the line that is created from the data in the table on the left. Solution -

20. Write an Equation of a Line from Graph
Example 1 – Write an equation of the line to the right. Solution – Step 1 – find the slope of the line by first finding two points through which the line passes: (0, 1), (5, -5) m = 𝒚 𝟐 − 𝒚 𝟏 𝒙 𝟐 − 𝒙 𝟏 = −𝟓 −𝟏 𝟓 −𝟎 = −𝟔 𝟓 Step 2 – plug the slope of the line and a point the line passes through into the formula y – y1 = m(x – x1) y – 1 = −𝟔 𝟓 (x – 0) Step 3 – manipulate equation to get “y” by itself y – 1 = −𝟔 𝟓 x y = −𝟔 𝟓 x + 1

21. Example of Writing An Equation of A Linear Graph
Example 1: Write an equation of the line.
Solution:
Step 1 – find two points that the line passes though
(-3, -2) and (0, -4)
Step 2 – find slope of line using the two points
m = −4−(−2) 0−(−3) = −2 3
Step 3 – plug either point and slope into point-slope form
y + 2 = − 2 3 (x + 3)
Step 4 – put in slope-intercept form
y + 2 = − 2 3 x – 2
y = − 𝟐 𝟑 x – 4

22. We Try
Write an equation of the line.

23. You Try
Write an equation of the line.

24. You Try
Write an equation of the line to the right. Solution -

25. Forms of Linear Equations
Lines in coordinate planes can be represented by equations
There are 3 basic forms of linear equations:
1. slope-intercept form: y = mx + b
2. point-slope form: y – y1 = m(x – x1)
3. standard form: ax + by = c
Each form has its own unique characteristics

26. Slope-Intercept Form y = mx + b
This form explicitly gives the slope (m) of the line and the y-intercept (b)
The y-intercept is the point where the line crosses the y-axis (0, b)
x and y are general variables and are not replaced by numbers
However, m and b are replaced by numbers

27. Examples of Slope-Intercept Form
Example 1: y = -4x + 5 m = -4 and b = 5, so the slope of the line is -4 and the point where the line crosses the y-axis is (0, 5) Example 2: y = 7x m = 7 and b = 0, so the slope of the line is 7 and the point where the line crosses the y-axis is (0, 0)

28. We Try
What is the slope and y-intercept of the line whose equation is y = 9x + 4?
2. What is the slope and y-intercept of the line whose equation is y = -6x?

29. You Try
y – 10 = -6(x + 4)
What is the slope? Through what point does the line pass?
y + 3 = -(x – 2)

30. Form of Linear Equation Key Info In Equation
Slope-Intercept
y = mx + b
Slope (m), y-intercept (0, b)
Point-Slope
y – y1 = m(x – x1)
Slope (m), point through which line passes (x1, y1)
Standard Form
ax + by = c
Slope (m) = -a/b

31. Special Linear Equations
y = constant
Horizontal line
slope = 0
Example: y = 4
A horizontal line that goes through (0, 4)
x = constant
Vertical line
slope is undefined
Example: x = 4
A vertical line that goes through (4, 0)

32. Graphing Linear Equations
In order to graph linear equations, we must either know (1) two points that the line passes through or (2) the slope of the line and a point that the line passes through.
For example, if we want to graph the equation y = 3x + 4, then we know the slope is 3 and the y-intercept is (0, 4). So, we’ll plot the point (0, 4) and find another point using the slope.

33. More Notes on Graphing Linear Equations
If the slope is positive, then go up the amount in the numerator and right the amount in the denominator, or go down the amount in the numerator and left the amount in the denominator.
If the slope is negative, then go down the amount in the numerator and right the amount in the denominator, or go up the amount in the numerator and go left the amount in the denominator.

34. y = 3x + 4
y-intercept: (0, 4)
m = 3/1

35. You Try Graph the following equations: y = -3x + 5 y = 1 2 x – 4

36. Graphing Real-World Linear Equations
Example 1 – Jonathan began the year with $400 in his savings account. Every month he deposited $200 into his account. Graph the relationship between the number of months elapsed and the amount of money in Jonathan’s savings account.
Solution – First we must make a table that reflects this relationship.
Time (months)
Money ($)
400 1
600 2
800 3
1000 4
1200 5
1400

37. Graphing Real-World Linear Equations
Label your axes – “Time” goes on the x-axis and “Money” goes on the y-axis.
Make sure that the values on both axes encompass those in the table and that you scale your axes appropriately.
Since all you need are two points to graph a line, pick two coordinate pairs from the table to plot. Then draw your line.

38. Graphing Real-World Linear Equations
Example 2 – Kyla is 900 miles away from home. She is driving home at an average speed of 60 miles per hour. Graph the relationship between the time that Kyla is driving and the distance she is away from home.
Solution – First we must make a table that reflects this relationship.
Time (hours)
Distance (miles)
900 1
840 2
780 3
720 4
660 5
600

39. Graphing Real-World Linear Equations
Label your axes – “Time” goes on the x-axis and “Distance” goes on the y-axis.
Make sure that the values on both axes encompass those in the table and that you scale your axes appropriately.
Since all you need are two points to graph a line, pick two coordinate pairs from the table to plot. Then draw your line.

40. You Try
Aaliyah was given $500 for her birthday. Every week since, she has spent $25. Graph the relationship between the number of weeks that have elapsed since her birthday and the amount of money that she has left.
Solution –
Time (weeks)
Money ($) 1 2 3 4 5 6 7 8

41. You Try

42. Solving Systems of Linear Equations
What does it mean to “solve” a system of linear equations?
- To solve a system of linear equations is to find the point where the lines intersect.
How do we solve systems of linear equations?
- We solve systems of linear equations either graphically or algebraically.

43. Solving Systems of Linear Equations Graphically
When we solve systems of linear equations graphically, we are looking for the point at which the two lines intersect.
If the two lines do not intersect (parallel lines), then there is no solution.
If the two lines are identical, then there are infinite solutions.

44. Solving Systems of Linear Equations Graphically
Step 1 – Graph the equations by finding a point that each line passes through and the slope of each line. Step 2 – Identify the point at which both lines intersect.

45. Example of Solving Systems of Linear Equations Graphically
Find the point at which the lines having the equations
y = -3x + 2 and
y = x – 6 intersect.
Solution: (2, -4)

46. Examples of Solving Systems of Linear Equations Graphically
Example 1 – Solve the system of linear equations on the left. Solution – 1. Graph the two equations. 2. Find the point at which the two lines intersect. 3. Write your solution as a coordinate pair. (-7, -4)

47. Examples of Solving Systems of Linear Equations Graphically
Example 2 – Solve the system of linear equations on the left. Solution – 1. Graph the two equations. 2. Find the point at which the two lines intersect. 3. Write your solution as a coordinate pair. (-2, -7)

48. You Try
Solve the system of linear equations on the left. Solution –

49. You Try
Find the point at which the lines having equations y = -5x + 1 and y = -3x – 1 intersect.

50. Solving Systems of Real-World Linear Equations
Example 1 – Skating Rink A charges $200 up front and $5 per guest for skating parties. Skating Rink B has no up-front charge but charges $7 per guest for skating parties. Graphically show where the costs would be the same for the same amount of guests for both skating rinks.
Solution – First construct a table for both rinks.
# of guests
Skating Rink A Cost ($)
Skating Rink B Cost ($)
200 25
325
175 50
450
350 75
575
525
100
700
125
825
875

51. Solving Systems of Real-World Linear Equations
Solution –
Graph the lines for the two skating rinks in the coordinate plane on the left.
If you already noticed, the table also will indicate your solution as well.
The two skating will both charge $700 for 100 guests.

52. Parallel Lines
Do not intersect
Have the same slope

53. Examples of Parallel Lines
Example 1: equation of line a is y = 2x + 4 equation of line b is y = 2x – 3 Lines a and b are parallel because they have the same slope: 2 Example 2: equation of line c is 3x – y = 4 equation of line d is y – 3 = 3(x + 6) Lines c and d are parallel because they have the same slope: 3

54. You Try
Determine whether or not the following pairs of lines are parallel:
6x + 9y = 18
y = 2/3 x – 4
y – 8 = -2/5 (x + 6)
-4x – 10 y = 20

55. Writing Equations of Lines When Given Point and Parallel Line
What are the two key pieces of information that we need to write an equation of a line? 1. the slope of the line 2. a point through which the line passes What do we know about parallel lines? 1. they never intersect 2. they have the same slope

56. Example of Writing Equation of Line When Given Point and Parallel Line
Example 1: Write an equation of the line that passes through (-8, 3) and is parallel to the line having the equation y = -4x + 2 Solution: Step 1 – Determine the slope of the parallel line m = -4 Step 2 – Since parallel lines have the same slope, the slope that we will use for our equation will be the same. So, plug the slope and the point that the line passes through into point-slope form. y – 3 = -4(x + 8) Step 3 – Put equation into slope-intercept form y – 3 = -4x – y = -4x – 29

57. Another Example of Writing Equation of Line When Given Point and Parallel Line
Example 2: Write an equation of the line that passes through (-5, -2) and is parallel to the line whose equation is -3x – y = 15 Solution: -3x – y = 15 +3x +3x - y = x /-1 /-1 /-1 y = -15 – 3x m = -3 y + 2 = -3(x + 5) y + 2 = -3x – y = -3x - 17

58. You Try
Write an equation of the line that passes through (-10, 2) and is parallel to the line having the equation y = -3x – 2
Write an equation of the line that passes through (12, 3) and is parallel to the line having the equation 6x – 2y = -12

59. Perpendicular Lines Intersect at a right angle
Slopes are opposite reciprocals

60. Examples of Perpendicular Lines
Example 1: What is the slope of line b if it is perpendicular to line a whose equation is y = -3x + 6? ma = -3  mb = -(1/-3) = 1/3 Example 2: What is the slope of line d if it is perpendicular to line c whose equation is 10x – 5y = 15? mc = -10/-5 = 2  md = -(1/2) = -1/2

61. You Try
If lines t and u are perpendicular and the equation of line u is 4x – 8y = 12, then what is the slope of line t?
2. If lines e and f are perpendicular and the equation of line f is y – 1 = -3/7(x + 4), then what is the slope of line e?

62. Think About It…
If a line is horizontal, then what would be the slope of the line that is perpendicular to it?
If a line is vertical, then what would be the slope of the line that is perpendicular to it?

63. Writing Equations of Lines When Given Point and Perpendicular Line
What do we know about perpendicular lines? 1. They intersect at a right angle 2. Their slopes are opposite reciprocals

64. Examples
Write an equation of the line that passes through (-3, 7) and is perpendicular to the line having the equation y = − 1 3 x – 4.
Write an equation of the line that passes through (-2, 8) and is perpendicular to the line having the equation y = -4x – 1.

65. Rectangles Characteristics of rectangles - opposite sides are parallel
- adjacent sides are perpendicular (4 right angles)
- opposite sides are congruent
- Area = length x width
- Perimeter = 2L + 2W

66. Proving Rectangles in the Coordinate Plane
When proving rectangles in the coordinate plane, 1. find the slope of each side of the rectangle 2. determine if the slopes of the opposite sides of the rectangle are equal 3. determine if the slopes of the adjacent sides of the rectangle are opposite reciprocals of each other

67. Example of Proving Rectangles in the Coordinate Plane
Example 1 – Prove that figure ABCD to the right is a rectangle. Solution – 𝑚 𝐴𝐵 = 3−3 3−(−3) = 0 6 = 0 𝑚 𝐷𝐶 = −4 −(−4) 3 −(−3) = 0 6 =0 𝑚 𝐴𝐷 = 3 −(−4) −3−(−3) = 7 0 =𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑚 𝐵𝐶 = 3 −(−4) 3−3 = 7 0 =𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 Since the slopes of 𝐴𝐵 and 𝐷𝐶 are equal, then these two sides are parallel. Since the slopes of 𝐴𝐷 and 𝐵𝐶 are equal, then these two sides are parallel. Since the slopes of 𝐴𝐵 and 𝐵𝐶 are opposite reciprocals of each other, then these two sides are perpendicular. Therefore, since opposite sides are parallel and adjacent sides are perpendicular, then figure ABCD is a rectangle. A B C D

68. Another Example of Proving Rectangles in the Coordinate Plane
Example 2 – Prove that figure EFGH to the right is a rectangle. Solution – 𝑚 𝐸𝐹 = 4 −3 −2 −1 =− 1 3 𝑚 𝐻𝐺 = −2 −(−3) −4−(−1) =− 1 3 𝑚 𝐸𝐻 = 4 −(−2) −2 −(−4) = 6 2 =3 𝑚 𝐹𝐺 = 3 −(−3) 1 −(−1) = 6 2 =3 Since the slopes of 𝐸𝐹 and 𝐻𝐺 are equal, then those two sides are parallel. Since the slopes of 𝐸𝐻 and 𝐹𝐺 are equal, then those two sides are parallel. Since the slopes of 𝐸𝐹 and 𝐸𝐻 are opposite reciprocals, then those two sides are perpendicular. Therefore, figure EFGH is a rectangle since opposite sides are parallel and adjacent sides are perpendicular. E F H G

69. You Try
Prove that figure IJKL to the right is a rectangle. J I K L

70. The Midpoint Formula
To find the midpoint or the halfway point between two points, we use the Midpoint Formula:
Example 1 – find the midpoint between (-2, 5) and (4, -1)
Solution: x1 = -2, x2 = 4, y1 = 5, and y2 = -1
So, (1, 2) is the midpoint between (-2, 5) and (4, -1)

71. Visual Representation of Midpoint Formula
Notice that the midpoint is always on the line segment that connects the two endpoints.

72. Another Example
Example 2 – Find the midpoint between (0, 6) and (-4, -2). Solution: (-2, 2) So, (-2, 2) is the midpoint between (0, 6) and (-4, -2)

73. You Try Find the midpoint between (-7, 2) and (5, -8).

74. Extensions of the Midpoint Formula
How do we find the other endpoint of a line segment when given one endpoint and the midpoint?
- (x2, y2) = (2xm – x1, 2ym – y1) where (x2, y2) is the missing endpoint.
Example 1 – Find the other endpoint of a line segment if one endpoint is (-2, 7) and the midpoint is (3, -1).
Solution: (x2, y2) = (2·3 - -2, 2·-1 – 7) = (6 + 2, -2 – 7) = (8, -9)

75. More Examples
Example 2 – Find the other endpoint of a line segment if one endpoint is (-5, -6) and the midpoint is (2, 3). Solution: (x2, y2) = (2 · , 2 · ) = (4 - -5, ) = (9, 12) Example 3 – Find the other endpoint of a line segment if one endpoint is (8, 4) and the midpoint is (-1, 10) Solution: (x2, y2) = (2 · -1 – 8, 2 · 10 – 4) = (-2 – 8, 20 – 4) = (-10, 16)

76. You Try
1. Find the other endpoint of a line segment if one endpoint is (-11, -3) and the midpoint is (-7, 7). 2. Find the other endpoint of a line segment if one endpoint is (15, -12) and the midpoint is (-3, -5).

77. Finding the Distance Between Two Points
Draw a line segment connecting the two points.
Count the number of units going horizontally from one endpoint to the other. (8)
Count the number of units going vertically from endpoint to the other. (8)
Square both of those numbers and add them together. (8² + 8² = 128)
Find the square root of that sum. Simplify the radical if possible. (√128 = 8√2)
So, the distance from (-5, -3) to (3, 5) is 8√2 units.

78. The Distance Formula What is the distance formula?
- We can use the distance formula to find the distance between two points or to determine how long a line segment is.
- Distance Formula:
- (x1, y1) and (x2, y2) are the endpoints
Example 1 – Find the distance between (-5, -3) and (3, 5).
Solution:

79. More Examples
Example 2 – Find the distance between (3, 7) and (-2, -6). Solution: Example 3 – Find the distance between (-1, -5) and (2, 0).

80. You Try Find the distance between (-4, 5) and (0, -1).