Coordinate Plane - Ninth Grade Math
1. Coordinate Plane Review
Coordinate Plane (Cartesian Plane): A two-dimensional plane formed by two perpendicular number lines: the x-axis (horizontal) and y-axis (vertical)
Origin: The point where the x-axis and y-axis intersect, represented as $(0, 0)$
Ordered Pair: A pair of numbers $(x, y)$ that represents a point's location on the plane
Coordinates: The numbers in an ordered pair
x-coordinate (Abscissa): The first number in an ordered pair (horizontal position)
y-coordinate (Ordinate): The second number in an ordered pair (vertical position)
Origin: The point where the x-axis and y-axis intersect, represented as $(0, 0)$
Ordered Pair: A pair of numbers $(x, y)$ that represents a point's location on the plane
Coordinates: The numbers in an ordered pair
x-coordinate (Abscissa): The first number in an ordered pair (horizontal position)
y-coordinate (Ordinate): The second number in an ordered pair (vertical position)
The Axes
X-Axis (Horizontal Axis):
• Runs horizontally (left to right)
• Positive values are to the RIGHT of the origin
• Negative values are to the LEFT of the origin
• Points on the x-axis have y-coordinate = 0
• Example: $(5, 0)$, $(-3, 0)$
Y-Axis (Vertical Axis):
• Runs vertically (up and down)
• Positive values are ABOVE the origin
• Negative values are BELOW the origin
• Points on the y-axis have x-coordinate = 0
• Example: $(0, 4)$, $(0, -6)$
• Runs horizontally (left to right)
• Positive values are to the RIGHT of the origin
• Negative values are to the LEFT of the origin
• Points on the x-axis have y-coordinate = 0
• Example: $(5, 0)$, $(-3, 0)$
Y-Axis (Vertical Axis):
• Runs vertically (up and down)
• Positive values are ABOVE the origin
• Negative values are BELOW the origin
• Points on the y-axis have x-coordinate = 0
• Example: $(0, 4)$, $(0, -6)$
Plotting Points
How to Plot a Point $(x, y)$:
Step 1: Start at the origin $(0, 0)$
Step 2: Move right if $x$ is positive, left if $x$ is negative
Step 3: From there, move up if $y$ is positive, down if $y$ is negative
Step 4: Mark the point
Memory Tip: "Go along the corridor (x) before going up the stairs (y)"
Step 1: Start at the origin $(0, 0)$
Step 2: Move right if $x$ is positive, left if $x$ is negative
Step 3: From there, move up if $y$ is positive, down if $y$ is negative
Step 4: Mark the point
Memory Tip: "Go along the corridor (x) before going up the stairs (y)"
Example 1: Plot the point $(3, 2)$
• Start at origin $(0, 0)$
• Move 3 units to the RIGHT (positive x)
• Move 2 units UP (positive y)
• Mark the point
• Start at origin $(0, 0)$
• Move 3 units to the RIGHT (positive x)
• Move 2 units UP (positive y)
• Mark the point
Example 2: Plot the point $(-4, -3)$
• Start at origin $(0, 0)$
• Move 4 units to the LEFT (negative x)
• Move 3 units DOWN (negative y)
• Mark the point
• Start at origin $(0, 0)$
• Move 4 units to the LEFT (negative x)
• Move 3 units DOWN (negative y)
• Mark the point
The Four Quadrants
Quadrant: One of four regions created when the x-axis and y-axis divide the coordinate plane
Numbering: Quadrants are numbered I, II, III, and IV in counterclockwise order, starting from the upper right
Numbering: Quadrants are numbered I, II, III, and IV in counterclockwise order, starting from the upper right
| Quadrant | Location | Sign Convention | Examples |
|---|---|---|---|
| Quadrant I | Upper Right | $(+, +)$ → Both positive | $(3, 5)$, $(7, 2)$ |
| Quadrant II | Upper Left | $(-, +)$ → x negative, y positive | $(-4, 6)$, $(-2, 3)$ |
| Quadrant III | Lower Left | $(-, -)$ → Both negative | $(-5, -3)$, $(-1, -7)$ |
| Quadrant IV | Lower Right | $(+, -)$ → x positive, y negative | $(6, -2)$, $(4, -8)$ |
Important Notes:
• Points on the axes do NOT belong to any quadrant
• The origin $(0, 0)$ is not in any quadrant
• Points with $x = 0$ lie on the y-axis
• Points with $y = 0$ lie on the x-axis
• Quadrants are numbered counterclockwise: I, II, III, IV
• Points on the axes do NOT belong to any quadrant
• The origin $(0, 0)$ is not in any quadrant
• Points with $x = 0$ lie on the y-axis
• Points with $y = 0$ lie on the x-axis
• Quadrants are numbered counterclockwise: I, II, III, IV
Example 3: Identify the quadrant for each point:
a) $(5, 7)$ → Quadrant I (both positive)
b) $(-3, 8)$ → Quadrant II (x negative, y positive)
c) $(-6, -4)$ → Quadrant III (both negative)
d) $(2, -9)$ → Quadrant IV (x positive, y negative)
e) $(0, 5)$ → On the y-axis (not in any quadrant)
f) $(-7, 0)$ → On the x-axis (not in any quadrant)
a) $(5, 7)$ → Quadrant I (both positive)
b) $(-3, 8)$ → Quadrant II (x negative, y positive)
c) $(-6, -4)$ → Quadrant III (both negative)
d) $(2, -9)$ → Quadrant IV (x positive, y negative)
e) $(0, 5)$ → On the y-axis (not in any quadrant)
f) $(-7, 0)$ → On the x-axis (not in any quadrant)
Quick Memory Tool for Quadrants:
Start at Quadrant I (upper right) and go counterclockwise:
• Quadrant I: All Students (All positive) - $(+, +)$
• Quadrant II: Take Calculus (x negative, y positive) - $(-, +)$
• Quadrant III: And Physics (All negative) - $(-, -)$
• Quadrant IV: Too! (x positive, y negative) - $(+, -)$
Start at Quadrant I (upper right) and go counterclockwise:
• Quadrant I: All Students (All positive) - $(+, +)$
• Quadrant II: Take Calculus (x negative, y positive) - $(-, +)$
• Quadrant III: And Physics (All negative) - $(-, -)$
• Quadrant IV: Too! (x positive, y negative) - $(+, -)$
2. Midpoint Formula: Find the Midpoint
Midpoint: The point that is exactly halfway between two endpoints on a line segment
Properties:
• Divides the line segment into two equal parts
• Is equidistant from both endpoints
• The ratio of division is 1:1
Properties:
• Divides the line segment into two equal parts
• Is equidistant from both endpoints
• The ratio of division is 1:1
Midpoint Formula:
If $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a line segment, the midpoint $M$ is:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
In words:
• x-coordinate of midpoint = average of x-coordinates
• y-coordinate of midpoint = average of y-coordinates
If $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a line segment, the midpoint $M$ is:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
In words:
• x-coordinate of midpoint = average of x-coordinates
• y-coordinate of midpoint = average of y-coordinates
Steps to Find the Midpoint:
Step 1: Identify the coordinates of both endpoints $(x_1, y_1)$ and $(x_2, y_2)$
Step 2: Add the x-coordinates: $x_1 + x_2$
Step 3: Divide by 2 to get the x-coordinate of the midpoint: $\frac{x_1 + x_2}{2}$
Step 4: Add the y-coordinates: $y_1 + y_2$
Step 5: Divide by 2 to get the y-coordinate of the midpoint: $\frac{y_1 + y_2}{2}$
Step 6: Write the midpoint as an ordered pair
Step 1: Identify the coordinates of both endpoints $(x_1, y_1)$ and $(x_2, y_2)$
Step 2: Add the x-coordinates: $x_1 + x_2$
Step 3: Divide by 2 to get the x-coordinate of the midpoint: $\frac{x_1 + x_2}{2}$
Step 4: Add the y-coordinates: $y_1 + y_2$
Step 5: Divide by 2 to get the y-coordinate of the midpoint: $\frac{y_1 + y_2}{2}$
Step 6: Write the midpoint as an ordered pair
Example 1: Find the midpoint of the line segment with endpoints $(2, 6)$ and $(8, 10)$
Given: $(x_1, y_1) = (2, 6)$ and $(x_2, y_2) = (8, 10)$
Using the formula:
$M = \left(\frac{2 + 8}{2}, \frac{6 + 10}{2}\right)$
$M = \left(\frac{10}{2}, \frac{16}{2}\right)$
$M = (5, 8)$
Answer: The midpoint is $(5, 8)$
Given: $(x_1, y_1) = (2, 6)$ and $(x_2, y_2) = (8, 10)$
Using the formula:
$M = \left(\frac{2 + 8}{2}, \frac{6 + 10}{2}\right)$
$M = \left(\frac{10}{2}, \frac{16}{2}\right)$
$M = (5, 8)$
Answer: The midpoint is $(5, 8)$
Example 2: Find the midpoint between $(−3, 4)$ and $(5, −2)$
Solution:
$M = \left(\frac{-3 + 5}{2}, \frac{4 + (-2)}{2}\right)$
$M = \left(\frac{2}{2}, \frac{2}{2}\right)$
$M = (1, 1)$
Answer: $(1, 1)$
Solution:
$M = \left(\frac{-3 + 5}{2}, \frac{4 + (-2)}{2}\right)$
$M = \left(\frac{2}{2}, \frac{2}{2}\right)$
$M = (1, 1)$
Answer: $(1, 1)$
Example 3: Find the midpoint of the segment joining $(−6, −8)$ and $(−2, 4)$
Solution:
$M = \left(\frac{-6 + (-2)}{2}, \frac{-8 + 4}{2}\right)$
$M = \left(\frac{-8}{2}, \frac{-4}{2}\right)$
$M = (-4, -2)$
Answer: $(-4, -2)$
Solution:
$M = \left(\frac{-6 + (-2)}{2}, \frac{-8 + 4}{2}\right)$
$M = \left(\frac{-8}{2}, \frac{-4}{2}\right)$
$M = (-4, -2)$
Answer: $(-4, -2)$
Example 4: The endpoints are $(0, 5)$ and $(4, 0)$. Find the midpoint.
Solution:
$M = \left(\frac{0 + 4}{2}, \frac{5 + 0}{2}\right)$
$M = \left(\frac{4}{2}, \frac{5}{2}\right)$
$M = \left(2, 2.5\right)$ or $\left(2, \frac{5}{2}\right)$
Answer: $(2, 2.5)$ or $\left(2, \frac{5}{2}\right)$
Solution:
$M = \left(\frac{0 + 4}{2}, \frac{5 + 0}{2}\right)$
$M = \left(\frac{4}{2}, \frac{5}{2}\right)$
$M = \left(2, 2.5\right)$ or $\left(2, \frac{5}{2}\right)$
Answer: $(2, 2.5)$ or $\left(2, \frac{5}{2}\right)$
Key Points:
• The order of the points doesn't matter
• The midpoint formula works with positive and negative coordinates
• The result may be a fraction or decimal
• Always simplify your answer
• The order of the points doesn't matter
• The midpoint formula works with positive and negative coordinates
• The result may be a fraction or decimal
• Always simplify your answer
3. Midpoint Formula: Find the Endpoint
Finding an Endpoint: When you know the midpoint and one endpoint, you can find the other endpoint
Concept: Work backwards from the midpoint formula
Concept: Work backwards from the midpoint formula
Formula to Find the Unknown Endpoint:
If one endpoint is $(x_1, y_1)$, the midpoint is $(m_x, m_y)$, and the unknown endpoint is $(x_2, y_2)$, then:
$$x_2 = 2m_x - x_1$$
$$y_2 = 2m_y - y_1$$
Or as an ordered pair:
$$(x_2, y_2) = (2m_x - x_1, 2m_y - y_1)$$
In words: Multiply the midpoint coordinates by 2, then subtract the known endpoint coordinates
If one endpoint is $(x_1, y_1)$, the midpoint is $(m_x, m_y)$, and the unknown endpoint is $(x_2, y_2)$, then:
$$x_2 = 2m_x - x_1$$
$$y_2 = 2m_y - y_1$$
Or as an ordered pair:
$$(x_2, y_2) = (2m_x - x_1, 2m_y - y_1)$$
In words: Multiply the midpoint coordinates by 2, then subtract the known endpoint coordinates
Derivation from Midpoint Formula:
Starting with: $m_x = \frac{x_1 + x_2}{2}$
Multiply both sides by 2: $2m_x = x_1 + x_2$
Solve for $x_2$: $x_2 = 2m_x - x_1$
Same process for y-coordinate: $y_2 = 2m_y - y_1$
Starting with: $m_x = \frac{x_1 + x_2}{2}$
Multiply both sides by 2: $2m_x = x_1 + x_2$
Solve for $x_2$: $x_2 = 2m_x - x_1$
Same process for y-coordinate: $y_2 = 2m_y - y_1$
Steps to Find the Unknown Endpoint:
Step 1: Identify the given endpoint $(x_1, y_1)$ and midpoint $(m_x, m_y)$
Step 2: For the x-coordinate: Multiply midpoint's x-coordinate by 2, then subtract the known endpoint's x-coordinate
$x_2 = 2m_x - x_1$
Step 3: For the y-coordinate: Multiply midpoint's y-coordinate by 2, then subtract the known endpoint's y-coordinate
$y_2 = 2m_y - y_1$
Step 4: Write the unknown endpoint as $(x_2, y_2)$
Step 5: Check your answer by finding the midpoint of both endpoints
Step 1: Identify the given endpoint $(x_1, y_1)$ and midpoint $(m_x, m_y)$
Step 2: For the x-coordinate: Multiply midpoint's x-coordinate by 2, then subtract the known endpoint's x-coordinate
$x_2 = 2m_x - x_1$
Step 3: For the y-coordinate: Multiply midpoint's y-coordinate by 2, then subtract the known endpoint's y-coordinate
$y_2 = 2m_y - y_1$
Step 4: Write the unknown endpoint as $(x_2, y_2)$
Step 5: Check your answer by finding the midpoint of both endpoints
Example 1: One endpoint is $(3, 7)$ and the midpoint is $(5, 9)$. Find the other endpoint.
Given: Endpoint = $(3, 7)$, Midpoint = $(5, 9)$
Find: Other endpoint $(x_2, y_2)$
Solution:
$x_2 = 2(5) - 3 = 10 - 3 = 7$
$y_2 = 2(9) - 7 = 18 - 7 = 11$
Answer: The other endpoint is $(7, 11)$
Check: Midpoint of $(3, 7)$ and $(7, 11)$ = $\left(\frac{3+7}{2}, \frac{7+11}{2}\right) = (5, 9)$ ✓
Given: Endpoint = $(3, 7)$, Midpoint = $(5, 9)$
Find: Other endpoint $(x_2, y_2)$
Solution:
$x_2 = 2(5) - 3 = 10 - 3 = 7$
$y_2 = 2(9) - 7 = 18 - 7 = 11$
Answer: The other endpoint is $(7, 11)$
Check: Midpoint of $(3, 7)$ and $(7, 11)$ = $\left(\frac{3+7}{2}, \frac{7+11}{2}\right) = (5, 9)$ ✓
Example 2: The midpoint of a segment is $(4, 2)$ and one endpoint is $(1, 6)$. Find the other endpoint.
Solution:
$x_2 = 2(4) - 1 = 8 - 1 = 7$
$y_2 = 2(2) - 6 = 4 - 6 = -2$
Answer: $(7, -2)$
Check: Midpoint of $(1, 6)$ and $(7, -2)$ = $\left(\frac{1+7}{2}, \frac{6+(-2)}{2}\right) = (4, 2)$ ✓
Solution:
$x_2 = 2(4) - 1 = 8 - 1 = 7$
$y_2 = 2(2) - 6 = 4 - 6 = -2$
Answer: $(7, -2)$
Check: Midpoint of $(1, 6)$ and $(7, -2)$ = $\left(\frac{1+7}{2}, \frac{6+(-2)}{2}\right) = (4, 2)$ ✓
Example 3: One endpoint is $(-2, -5)$ and the midpoint is $(0, 1)$. Find the other endpoint.
Solution:
$x_2 = 2(0) - (-2) = 0 + 2 = 2$
$y_2 = 2(1) - (-5) = 2 + 5 = 7$
Answer: $(2, 7)$
Solution:
$x_2 = 2(0) - (-2) = 0 + 2 = 2$
$y_2 = 2(1) - (-5) = 2 + 5 = 7$
Answer: $(2, 7)$
Example 4: The midpoint is $(-1, 3)$ and one endpoint is $(-4, 8)$. Find the missing endpoint.
Solution:
$x_2 = 2(-1) - (-4) = -2 + 4 = 2$
$y_2 = 2(3) - 8 = 6 - 8 = -2$
Answer: $(2, -2)$
Solution:
$x_2 = 2(-1) - (-4) = -2 + 4 = 2$
$y_2 = 2(3) - 8 = 6 - 8 = -2$
Answer: $(2, -2)$
Tips for Finding Endpoints:
• Be careful with negative numbers when subtracting
• Remember: subtracting a negative is the same as adding
• Always check your answer by verifying the midpoint
• The formula works for all quadrants
• Be careful with negative numbers when subtracting
• Remember: subtracting a negative is the same as adding
• Always check your answer by verifying the midpoint
• The formula works for all quadrants
4. Distance Between Two Points
Distance: The length of the straight line segment connecting two points
Distance Formula: Based on the Pythagorean Theorem
Key Concept: Distance is always a positive value (or zero if points are the same)
Distance Formula: Based on the Pythagorean Theorem
Key Concept: Distance is always a positive value (or zero if points are the same)
Distance Formula:
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Alternative notation:
$$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$
where $\Delta x = x_2 - x_1$ and $\Delta y = y_2 - y_1$
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Alternative notation:
$$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$
where $\Delta x = x_2 - x_1$ and $\Delta y = y_2 - y_1$
Why This Formula Works (Pythagorean Theorem):
When you connect two points, you can form a right triangle where:
• Horizontal leg = $|x_2 - x_1|$ (horizontal distance)
• Vertical leg = $|y_2 - y_1|$ (vertical distance)
• Hypotenuse = distance between the points
Using Pythagorean Theorem: $c^2 = a^2 + b^2$
$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
When you connect two points, you can form a right triangle where:
• Horizontal leg = $|x_2 - x_1|$ (horizontal distance)
• Vertical leg = $|y_2 - y_1|$ (vertical distance)
• Hypotenuse = distance between the points
Using Pythagorean Theorem: $c^2 = a^2 + b^2$
$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Steps to Find Distance Between Two Points:
Step 1: Identify the coordinates: $(x_1, y_1)$ and $(x_2, y_2)$
Step 2: Find the horizontal distance: $x_2 - x_1$
Step 3: Square it: $(x_2 - x_1)^2$
Step 4: Find the vertical distance: $y_2 - y_1$
Step 5: Square it: $(y_2 - y_1)^2$
Step 6: Add the squares: $(x_2 - x_1)^2 + (y_2 - y_1)^2$
Step 7: Take the square root of the sum
Step 8: Simplify if possible
Step 1: Identify the coordinates: $(x_1, y_1)$ and $(x_2, y_2)$
Step 2: Find the horizontal distance: $x_2 - x_1$
Step 3: Square it: $(x_2 - x_1)^2$
Step 4: Find the vertical distance: $y_2 - y_1$
Step 5: Square it: $(y_2 - y_1)^2$
Step 6: Add the squares: $(x_2 - x_1)^2 + (y_2 - y_1)^2$
Step 7: Take the square root of the sum
Step 8: Simplify if possible
Example 1: Find the distance between $(1, 2)$ and $(4, 6)$
Given: $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (4, 6)$
Solution:
$d = \sqrt{(4 - 1)^2 + (6 - 2)^2}$
$d = \sqrt{(3)^2 + (4)^2}$
$d = \sqrt{9 + 16}$
$d = \sqrt{25}$
$d = 5$ units
Answer: 5 units
Given: $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (4, 6)$
Solution:
$d = \sqrt{(4 - 1)^2 + (6 - 2)^2}$
$d = \sqrt{(3)^2 + (4)^2}$
$d = \sqrt{9 + 16}$
$d = \sqrt{25}$
$d = 5$ units
Answer: 5 units
Example 2: Find the distance between $(−3, 2)$ and $(5, 8)$
Solution:
$d = \sqrt{(5 - (-3))^2 + (8 - 2)^2}$
$d = \sqrt{(5 + 3)^2 + (6)^2}$
$d = \sqrt{(8)^2 + (6)^2}$
$d = \sqrt{64 + 36}$
$d = \sqrt{100}$
$d = 10$ units
Answer: 10 units
Solution:
$d = \sqrt{(5 - (-3))^2 + (8 - 2)^2}$
$d = \sqrt{(5 + 3)^2 + (6)^2}$
$d = \sqrt{(8)^2 + (6)^2}$
$d = \sqrt{64 + 36}$
$d = \sqrt{100}$
$d = 10$ units
Answer: 10 units
Example 3: Find the distance between $(0, 0)$ and $(6, 8)$
Solution:
$d = \sqrt{(6 - 0)^2 + (8 - 0)^2}$
$d = \sqrt{6^2 + 8^2}$
$d = \sqrt{36 + 64}$
$d = \sqrt{100} = 10$ units
Answer: 10 units
Solution:
$d = \sqrt{(6 - 0)^2 + (8 - 0)^2}$
$d = \sqrt{6^2 + 8^2}$
$d = \sqrt{36 + 64}$
$d = \sqrt{100} = 10$ units
Answer: 10 units
Example 4: Find the distance between $(−4, −1)$ and $(2, 3)$
Solution:
$d = \sqrt{(2 - (-4))^2 + (3 - (-1))^2}$
$d = \sqrt{(6)^2 + (4)^2}$
$d = \sqrt{36 + 16}$
$d = \sqrt{52}$
$d = \sqrt{4 \times 13} = 2\sqrt{13}$ units
$d \approx 7.21$ units (rounded to 2 decimal places)
Answer: $2\sqrt{13}$ units or approximately 7.21 units
Solution:
$d = \sqrt{(2 - (-4))^2 + (3 - (-1))^2}$
$d = \sqrt{(6)^2 + (4)^2}$
$d = \sqrt{36 + 16}$
$d = \sqrt{52}$
$d = \sqrt{4 \times 13} = 2\sqrt{13}$ units
$d \approx 7.21$ units (rounded to 2 decimal places)
Answer: $2\sqrt{13}$ units or approximately 7.21 units
Example 5: Find the distance between $(3, 7)$ and $(3, -2)$
Solution:
$d = \sqrt{(3 - 3)^2 + (-2 - 7)^2}$
$d = \sqrt{(0)^2 + (-9)^2}$
$d = \sqrt{0 + 81}$
$d = \sqrt{81} = 9$ units
Answer: 9 units
Note: Points have the same x-coordinate, so they lie on a vertical line
Solution:
$d = \sqrt{(3 - 3)^2 + (-2 - 7)^2}$
$d = \sqrt{(0)^2 + (-9)^2}$
$d = \sqrt{0 + 81}$
$d = \sqrt{81} = 9$ units
Answer: 9 units
Note: Points have the same x-coordinate, so they lie on a vertical line
Special Cases:
Horizontal Distance (same y-coordinate):
If $(x_1, y)$ and $(x_2, y)$, then $d = |x_2 - x_1|$
Vertical Distance (same x-coordinate):
If $(x, y_1)$ and $(x, y_2)$, then $d = |y_2 - y_1|$
Distance from Origin:
Distance from $(0, 0)$ to $(x, y)$ is $d = \sqrt{x^2 + y^2}$
Horizontal Distance (same y-coordinate):
If $(x_1, y)$ and $(x_2, y)$, then $d = |x_2 - x_1|$
Vertical Distance (same x-coordinate):
If $(x, y_1)$ and $(x, y_2)$, then $d = |y_2 - y_1|$
Distance from Origin:
Distance from $(0, 0)$ to $(x, y)$ is $d = \sqrt{x^2 + y^2}$
Important Points About Distance:
• Distance is always positive (or zero)
• The order of points doesn't matter: $d(A, B) = d(B, A)$
• Distance can be an integer, fraction, or irrational number
• Leave answer in simplest radical form unless asked to round
• Squaring eliminates the need for absolute values
• Distance is always positive (or zero)
• The order of points doesn't matter: $d(A, B) = d(B, A)$
• Distance can be an integer, fraction, or irrational number
• Leave answer in simplest radical form unless asked to round
• Squaring eliminates the need for absolute values
Combined Application Problem
Example 6 (Combined): Points $A(2, 3)$ and $B(8, 11)$ are endpoints of a line segment.
Find: a) The midpoint M, b) The distance AB, c) The distance from A to M
Solution:
a) Midpoint M:
$M = \left(\frac{2 + 8}{2}, \frac{3 + 11}{2}\right) = \left(\frac{10}{2}, \frac{14}{2}\right) = (5, 7)$
b) Distance AB:
$d_{AB} = \sqrt{(8 - 2)^2 + (11 - 3)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ units
c) Distance from A to M:
$d_{AM} = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units
Note: The distance from A to M is exactly half of AB, as expected! $d_{AM} = \frac{1}{2}d_{AB}$
Find: a) The midpoint M, b) The distance AB, c) The distance from A to M
Solution:
a) Midpoint M:
$M = \left(\frac{2 + 8}{2}, \frac{3 + 11}{2}\right) = \left(\frac{10}{2}, \frac{14}{2}\right) = (5, 7)$
b) Distance AB:
$d_{AB} = \sqrt{(8 - 2)^2 + (11 - 3)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ units
c) Distance from A to M:
$d_{AM} = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units
Note: The distance from A to M is exactly half of AB, as expected! $d_{AM} = \frac{1}{2}d_{AB}$
Quick Reference Guide
Coordinate Plane Basics:
• Origin: $(0, 0)$
• Ordered Pair: $(x, y)$ where $x$ = horizontal, $y$ = vertical
• X-axis: Horizontal line, $y = 0$
• Y-axis: Vertical line, $x = 0$
• Origin: $(0, 0)$
• Ordered Pair: $(x, y)$ where $x$ = horizontal, $y$ = vertical
• X-axis: Horizontal line, $y = 0$
• Y-axis: Vertical line, $x = 0$
Quadrant Signs:
• Quadrant I: $(+, +)$ - Upper Right
• Quadrant II: $(-, +)$ - Upper Left
• Quadrant III: $(-, -)$ - Lower Left
• Quadrant IV: $(+, -)$ - Lower Right
• Quadrant I: $(+, +)$ - Upper Right
• Quadrant II: $(-, +)$ - Upper Left
• Quadrant III: $(-, -)$ - Lower Left
• Quadrant IV: $(+, -)$ - Lower Right
Midpoint Formula:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
Average the x-coordinates, average the y-coordinates
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
Average the x-coordinates, average the y-coordinates
Find Endpoint (given midpoint and one endpoint):
$$x_2 = 2m_x - x_1$$
$$y_2 = 2m_y - y_1$$
Multiply midpoint by 2, then subtract known endpoint
$$x_2 = 2m_x - x_1$$
$$y_2 = 2m_y - y_1$$
Multiply midpoint by 2, then subtract known endpoint
Distance Formula:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Square the differences, add them, take the square root
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Square the differences, add them, take the square root
| Formula Type | Formula | What It Finds |
|---|---|---|
| Midpoint | $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ | Point halfway between two endpoints |
| Endpoint | $(x_2, y_2) = (2m_x - x_1, 2m_y - y_1)$ | Missing endpoint given midpoint |
| Distance | $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ | Length between two points |
Special Distance Cases:
• Horizontal line: $d = |x_2 - x_1|$ (same y)
• Vertical line: $d = |y_2 - y_1|$ (same x)
• From origin: $d = \sqrt{x^2 + y^2}$
• Horizontal line: $d = |x_2 - x_1|$ (same y)
• Vertical line: $d = |y_2 - y_1|$ (same x)
• From origin: $d = \sqrt{x^2 + y^2}$
Success Tips for Coordinate Plane:
✓ Always write coordinates in order: $(x, y)$ not $(y, x)$
✓ Move horizontally first (x), then vertically (y)
✓ Be careful with negative signs when subtracting
✓ For midpoint: think "average" of coordinates
✓ For distance: remember it comes from Pythagorean Theorem
✓ Always simplify radicals when possible
✓ Check your work by plotting points when possible
✓ Remember: distance is always positive or zero
✓ Always write coordinates in order: $(x, y)$ not $(y, x)$
✓ Move horizontally first (x), then vertically (y)
✓ Be careful with negative signs when subtracting
✓ For midpoint: think "average" of coordinates
✓ For distance: remember it comes from Pythagorean Theorem
✓ Always simplify radicals when possible
✓ Check your work by plotting points when possible
✓ Remember: distance is always positive or zero