Download GRE: Graphical Representations

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.
Plotting Points in the Coordinate
Plane
• Points in a coordinate plane are
located according to their
coordinates.
• Coordinates are written as a pair:
(x-coordinate, y-coordinate)
• The x-coordinate indicates how far
left or right the point is from the
origin.
• The y-coordinate indicates how far
up or down the point is from the
origin
• Example – the coordinates of the
point on the coordinate plane is
(3, 1)
The Midpoint Formula
• To find the midpoint or the halfway point between
two points, we use the Midpoint Formula:  x 2 x , y 2 y
• Example 1 – find the midpoint between (-2, 5) and
(4, -1)
Solution: x1 = -2, x2 = 4, y1 = 5, and y2 = -1
1
2
1
2
  2  4 5  1   2 4 
,

   ,   1,2
2  2 2
 2
So, (1, 2) is the midpoint between (-2, 5) and (4, -1)



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).
 0  4 6  2    4 4 
Solution:  2 , 2    2 , 2   (-2, 2)
So, (-2, 2) is the midpoint between (0, 6) and
(-4, -2)
You Try
1. Find the midpoint between (-7, 2) and (5, -8).
2. Find the midpoint between (6, 9) and (-8, 0).
Homework
Do all of the problems on GRE Practice Problems 1.
For 3rd, 7th , and 8th periods #6 is extra credit.
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).
Homework
Do all of the problems on GRE Practice Problems 2.
Number 7 is extra credit for all classes.
Finding the Distance Between Two
Points
1.
2.
3.
4.
5.
6.
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.
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: x  x    y  y 
- (x1, y1) and (x2, y2) are the endpoints
• Example 1 – Find the distance between (-5, -3) and
(3, 5).
Solution:  5  3   3  5   8  (8)  64  64  128
2
1
2
128  8 2
2
2
1
2
2
2
2
More Examples
Example 2 – Find the distance between (3, 7) and
(-2, -6).
Solution: 3  2  (7  6)  5 13  25 169  194
2
2
2
2
Example 3 – Find the distance between (-1, -5) and
(2, 0).
Solution: 1  2   5  0   3   5  9  25  34
2
2
2
2
You Try
1. Find the distance between (-4, 5) and (0, -1).
2. Find the distance between (-2, 0) and (3, 5).
Homework
Do all of the problems on GRE Practice Problems 3.
For 3rd , 7th, and 8th periods #6 is extra credit.
Finding the Slope Between Two
Points
• What is slope?
- Slope is the steepness of a line
- Slope is the rate at which “y” changes as “x”
increases by 1
- Slope is the rise of a line divided by its run
- “m” is the symbol used to represent slope
y y
• Slope Formula: m  x  x where (x1, y1) and (x2,
y2) are points on the line
2
1
2
1
More 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).
42 6 6
Solution:  3  8   11  11
Example 2 – Find the slope of the line that passes
through (-7, 4) and (-5, -10).
Solution:  10  4   14  7
 5  7
2
Which line in the examples is steeper? How do you
know?
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?
Homework
Do all of the problems on GRE Practice Problems 4.
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 –
1. The line passes through (6, 5) and (0, -2)
2. -2-5/0 - -6 = -8/6 = -4/3
You Try
1. Find the slope of the line
in the coordinate plane on
the right.
Homework
Do all of the problems on GRE Practice Problems 5.
For 3rd, 7th, and 8th periods #5 is extra credit.
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
2. point-slope form
3. standard form
• 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 yaxis (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 + 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)
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?
Point-Slope Form
y – y1 = m(x – x1)
• This form explicitly gives
the slope (m) of the line
and a point through which
the line passes (x1, y1).
• x and y are general
variables and are not
replaced by numbers
• However, m, x1, and y1 are
replaced by numbers
Examples of Point-Slope Form
Example 1: y – 3 = 5(x – 1)
m = 5, x1 = 1, and y1 = 3
So, the slope of the line is 5 and it passes through the
point (1, 3)
Example 2: y + 8 = -2(x + 4)
m = -2, x1 = -4, and y1 = -8
So, the slope is -2 and the line passes through the
point (-4, -8)
*Note: the coordinates of the point are always the
opposite of the numbers in the equation.
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?
Standard Form
ax + by = c
• This form does not explicitly
give anything
• x and y are general variables
and are not replaced by
numbers
• However, a, b, and c are
replaced by numbers where a
and b are coefficients and c is
a constant
• m = -a/b  the slope of the
line in standard form is the
opposite of a divided by b
Examples of Standard Form
Example 1: 8x – 4y = 12
a = 8, b = -4, and c = 12
m = -8/-4 = 2
So, the slope of the line is 2.
Example 2: 7x + y = 11
a = 7, b = 1, and c = 11
m = -7/1 = -7
So, the slope of the line is -7.
You Try
1. Find the slope of the line whose equation is
-4x – 2y = 16
2. Find the slope of the line whose equation is
x – 6y = 12
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
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)
You Try Some More…
Identify the form and find the key info from the
following equations:
1. y – 2 = -(x + 5)
2. 4x – 3y = 15
3. y = -7x – 2
4. -x + 4y = -10
5. y + 7 = 3(x – 1)
Homework
Complete all of the problems on GRE Practice
Problems 6.
Parallel Lines
• Do not intersect
• Have the same slope
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
You Try
Determine whether or not the following pairs of
lines are parallel:
1. 6x + 9y = 18
y = 2/3 x – 4
2. y – 8 = -2/5 (x + 6)
-4x – 10 y = 20
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?
Homework
Complete all of the problems on GRE Practice
Problems 7.
Converting From Point-Slope to
Slope-Intercept
Point-Slope Form: y – y1 = m(x – x1)
Slope-Intercept Form: y = mx + b
What do these two equations have in common?
What is different about these two equations?
We must get “y” by itself on the left side of
the equal sign.
Example of Converting From PointSlope to Slope-Intercept
Since point-slope form and slope-intercept form both
have “x”, “y”, and “m”, we just need to rearrange the
variables using some algebraic properties.
Example 1: Convert y – 3 = 4(x + 5) to slope-intercept
form.
Solution:
Step 1
y – 3 = 4x + 20 Distributive Property
Step 2
y – 3 = 4x + 20
+3
+ 3 Inverse Operations
y
= 4x + 23
More Examples of Converting From
Point-Slope to Slope-Intercept
Example 2: Convert y + 6 = -2(x – 8) to slopeintercept form.
Solution:
Step 1
y + 6 = -2x + 16
Dist Prop
Step 2
y + 6 = -2x + 16
-6
- 6
Inverse Op
y
= -2x + 10
More Examples of Converting From
Point-Slope to Slope-Intercept
Example 3: Convert y + 8 = -3/4(x – 12) to
slope-intercept form.
Solution:
Step 1
y + 8 = -3/4 x + 9 Dist. Prop
Step 2
y + 8 = -3/4 x + 9
-8
- 8 Inverse Op.
y
= -3/4 x + 1
You Try
1. Convert y + 7 = -4(x – 3) to slopeintercept form.
2. Convert y – 1 = 2/5(x + 10) to
slope-intercept form.
3. Convert y + 2 = -3/7(x – 7) to slopeintercept form.
Homework
Do all of the problems on GRE Practice
Problems 9.
Converting From Standard Form to
Slope-Intercept Form
Standard Form: ax + by = c
Slope-Intercept Form: y = mx + b
To convert from standard form to slopeintercept form, we must get “y” by itself on
the left side of the equal sign by using inverse
operations.
Example of Converting from Standard
Form to Slope-Intercept
Example 1: Convert 8x – 4y = 16 to slope-intercept
form.
Solution:
Step 1
8x – 4y = 16
-8x
- 8x
-4y = 16 – 8x
Step 2
-4y/-4 = 16/-4 – 8x/-4
y = -4 + 2x
Step 3
y = 2x – 4
More Examples of Converting from
Standard Form to Slope-Intercept
Example 2: Convert -12x – 3y = -24 to slopeintercept form.
Solution:
Step 1
-12x – 3y = -24
+12x
+ 12x
-3y = -24 + 12x
Step 2
-3y/-3 = -24/-3 + 12x/-3
y = 8 + -4x
Step 3
y = -4x + 8
Recap on Converting from Standard
Form to Slope-Intercept
Step 1 – add/subtract the “x” term from both
sides of equation
Step 2 – divide all terms by the coefficient of y
Step 3 – rearrange the terms on the right side of
the equation so that the “x” term is first
You Try
1. Convert 10x + 2y = 20 to slope-intercept
form.
2. Convert -9x – 3y = 18 to slope-intercept
form.
3. Convert -2x – y = 6 to slope-intercept
form.
4. Convert x + y = 3 to slope-intercept form.
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 yintercept 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.
• If the slope is negative, then go down the
amount in the numerator and right the
amount in the denominator.
y = 3x + 4
y-intercept:
(0, 4)
m = 3/1
You Try
Graph the following equations:
1. y = -3x + 7
2. y = ½ x – 8
3. 3x – y = 12
4. y – 4 = -1/6(x – 6)
5. -6x – 2y = -10
Homework
Do all of the problems on GRE Practice
Problems 11. Study for quiz.
Writing Equations of Lines When
Given Two Points
• There are two key pieces of information
that we need when writing an equation of
a line:
1. The slope of the line
2. A point through which the line
passes
Example of Writing An Equation of a
Line When Given Two Points
Example 1: Write an equation of the line that passes through points (4, 0)
and (6, -4).
Solution:
Step 1 – Find the slope of the line
𝑦2 −𝑦1
−4 − 0
−4
m=
=
= = -2
𝑥2 −𝑥1
6−4
2
Step 2 – Plug slope and either point into point-slope form
y – y1 = m(x – x1)
y + 4 = -2(x – 6)
Step 3 – Put in slope-intercept form
y + 4 = -2x + 12
-4
- 4
y
= -2x + 8
Another Example of Writing An
Equation of a Line When Given Two
Points
Example 2: Write an equation of the line that passes
through points (-1, -5) and (2, 4).
Solution:
𝑦2 −𝑦1
𝑥2 −𝑥1
4−(−5)
2−(−1)
m=
=
y – y1 = m(x – x1)
y + 5 = 3(x + 1)
y + 5 = 3x + 3
-5
-5
y
= 3x - 2
9
3
= =3
You Try
1. Write an equation of the line that passes
through points (4, -1) and (9, 14).
2.
Write an equation of the line that passes
through points (10, -3) and (9, -8).
Homework
Complete all of the problems on GRE Practice
Problems 12.
Writing Equations of Linear Graphs
Again, the two key pieces of information
that we need in order to write an equation
of a line are (1) the slope of the line and (2)
a point through which the line passes.
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
2
y + 2 = − 3(x + 3)
Step 4 – put in slope-intercept form
2
y + 2 = − 3x – 2
-2
-2
y
𝟐
= − 𝟑x – 4
Another Example of Writing An
Equation of A Linear Graph
1.
2.
3.
The two points that
the line passes
through are (2, -2)
and
(-2, -4).
−4−(−2)
=
−2−2
−2
1
=
−4
2
1
y + 4 = (x + 2)
2
1
y+4= x+1
2
m=
-4
y
-4
𝟏
𝟐
= x–3
Another Example
Write an
equation of the
line.
Another Example
Write an
equation of the
line.