GRE: Graphical Representations

COORDINATE GEOMETRY

The Coordinate Plane

• 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.

y

x

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

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)

You Try

State the coordinates of each point in the coordinate plane to the left.

ABCD E

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 line12

12

xx

yym

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.

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).Solution:

Which line in the examples is steeper? How do you know?

11

6

11

6

83

24

72

14

75

410

You Try

1. Find the slope of the line that passes through (5, -3) and (2, 3).

2. Find the slope of the line that passes through (-11, -9) and (11, -11).

Which of the lines is steeper?

Finding the Slope of a Line

When finding the slope of a line,

1. Find two points through which the line passes.

2. Plug those points into the slope formula.

Example 1 – 3. The line passes through

(-6, 5) and (0, -2)

2. m =

You Try

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

Finding Slope of Line from Table

x y

-3 10

-2 5

-1 0

0 -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 formulam =

You Tryx y

-2 -3

0 -9

2 -15

4 -21

6 -27

Find the slope of the line created from the data in the table on the left.

Solution -

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.

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 = (x – 3)Step 2 – Manipulate the equation to get “y” by itself

y – 3 = x + 1 + 3 + 3y = x + 4

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 = (x + 3)Step 2 – Manipulate the equation to get “y” by itself

y + 5 = x + 10 - 5 - 5y = x + 5

You Try

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

Solution -

Writing an Equation of a Line from Table

x y

-3 10

-2 5

-1 0

0 -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: (-2, 5), (-3, 10)m =

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 itselfy – 10 = -5x – 15 + 10 + 10y = -5x - 5

You Tryx y

-2 -3

0 -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 -

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 =

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 = (x – 0)

Step 3 – manipulate equation to get “y” by itselfy – 1 = (x – 0)y – 1 = x + 0 + 1 + 1y = x + 1

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 = = Step 3 – plug either point and slope into point-slope form

y + 2 = (x + 3)Step 4 – put in slope-intercept form

y + 2 = x – 2 - 2 - 2y = x – 4

We Try

Write an equation of the line.

You Try

Write an equation of the line.

You Try

Write an equation of the line to the right.

Solution -

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 + b2. point-slope form: y – y1 = m(x – x1)

3. standard form: ax + by = c• Each form has its own unique

characteristics

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

Examples of Slope-Intercept Form

Example 1: y = -4x + 5m = -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 = 7xm = 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)

We Try

1. 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?

You Try

1. y – 10 = -6(x + 4)What is the slope? Through what point does the line pass?

2. y + 3 = -(x – 2) What is the slope? Through what point does the line pass?

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 Formax + by = c

Slope (m) = -a/b

Special Linear Equations

y = constant• Horizontal line• slope = 0Example: y = 4

A horizontal line that goes through (0, 4)

x = constant• Vertical line• slope is

undefinedExample: x = 4

A vertical line that goes through (4, 0)

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.

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.

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

You Try

Graph the following equations:

1. y = -3x + 52. y = x – 43. 3x – y = 24. y – 4 = (x – 6) 5. -6x – 2y = -10

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 ($)

0 400

1 600

2 800

3 1000

4 1200

5 1400

Graphing Real-World Linear Equations

1. Label your axes – “Time” goes on the x-axis and “Money” goes on the y-axis.

2. Make sure that the values on both axes encompass those in the table and that you scale your axes appropriately.

3. Since all you need are two points to graph a line, pick two coordinate pairs from the table to plot. Then draw your line.

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)

0 900

1 840

2 780

3 720

4 660

5 600

Graphing Real-World Linear Equations

1. Label your axes – “Time” goes on the x-axis and “Distance” goes on the y-axis.

2. Make sure that the values on both axes encompass those in the table and that you scale your axes appropriately.

3. Since all you need are two points to graph a line, pick two coordinate pairs from the table to plot. Then draw your line.

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 ($)

0

1

2

3

4

5

6

7

8

You Try

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.

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.

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.

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)

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)

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)

You Try

Solve the system of linear equations on the left.

Solution –

You Try

Find the point at which the lines having equations

y = -5x + 1 and y = -3x – 1 intersect.

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

guestsSkating Rink A

Cost ($)

Skating Rink B

Cost ($)

0 200 0

25 325 175

50 450 350

75 575 525

100 700 700

125 825 875

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.

Parallel Lines

• Do not intersect

• Have the same slope

Examples of Parallel Lines

Example 1: equation of line a is y = 2x + 4equation 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 = 4equation of line d is y – 3 = 3(x + 6)

Lines c and d are parallel because they have the same slope: 3

You Try

Determine whether or not the following pairs of lines are parallel:

1. 6x + 9y = 18y = 2/3 x – 4

2. y – 8 = -2/5 (x + 6)-4x – 10 y = 20

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 line2. a point through which the line

passesWhat do we know about parallel lines?

1. they never intersect2. they have the same slope

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 + 2Solution:Step 1 – Determine the slope of the parallel line

m = -4Step 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 – 32 +3 + 3y = -4x – 29

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 = 15Solution:

-3x – y = 15 +3x +3x

- y = 15 + 3x /-1 /-1 /-1

y = -15 – 3x

m = -3 y + 2 = -3(x + 5)y + 2 = -3x – 15 - 2 - 2y = -3x - 17

You Try

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

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

Perpendicular Lines

• Intersect at a right angle

• Slopes are opposite reciprocals

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

You Try

1. 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?

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?

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

Examples

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

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

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

Proving Rectangles in the Coordinate Plane

When proving rectangles in the coordinate plane,1. find the slope of each side of the rectangle2. determine if the slopes of the opposite sides of the rectangle are equal3. determine if the slopes of the adjacent sides of the rectangle are opposite reciprocals of each other

Example of Proving Rectangles in the Coordinate Plane

Example 1 – Prove that figure ABCD to the right is a rectangle.

Solution – = 0

Since the slopes of and are equal, then these two sides are parallel.

Since the slopes of and are equal, then these two sides are parallel.

Since the slopes of and 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

CD

Another Example of Proving Rectangles in the Coordinate PlaneExample 2 – Prove that figure EFGH to

the right is a rectangle.

Solution –

Since the slopes of and are equal, then those two sides are parallel.

Since the slopes of and are equal, then those two sides are parallel.

Since the slopes of and 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

G

H

You Try

Prove that figure IJKL to the right is a rectangle.

I

J

K

L

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)

2,

22121 yyxx

2,12

4,2

2

2

15,

2

42

Visual Representation of Midpoint Formula

• Notice that the midpoint is always on the line segment that connects the two endpoints.

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)

2

4,

2

4

2

26,

2

40

You Try

1. Find the midpoint between (-7, 2) and (5, -8).

2. Find the midpoint between (6, 9) and (-8, 0).

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)

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 - -5, 2 · 3 - -6) =

(4 - -5, 6 - -6) = (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)

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).

Finding the Distance Between Two Points

1. Draw a line segment connecting the two points.

2. Count the number of units going horizontally from one endpoint to the other. (8)

3. Count the number of units going vertically from endpoint to the other. (8)

4. Square both of those numbers and add them together. (8² + 8² = 128)

5. Find the square root of that sum. Simplify the radical if possible. (√128 = 8√2)

6. So, the distance from (-5, -3) to (3, 5) is 8√2 units.

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:

2212

21 yyxx

1286464)8(85335 2222

28128

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).Solution:

19416925135)67(23 2222

34259530521 2222

You Try

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

2. Find the distance between (-2, 0) and (3, 5).