Distance Formula
Find distance between two points
Distance Formula
How do you find the distance between two points on a coordinate plane? The distance formula gives you the exact answer every time! This powerful tool combines the Pythagorean Theorem with coordinate geometry.
What Is the Distance Formula?
The distance formula calculates the straight-line distance between two points on a coordinate plane.
Given two points: (x₁, y₁) and (x₂, y₂)
Distance formula:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Where d represents the distance between the points.
Where Does It Come From?
The distance formula comes from the Pythagorean Theorem!
Imagine: Two points (x₁, y₁) and (x₂, y₂) on a graph.
Create a right triangle:
- Horizontal leg: |x₂ - x₁| (difference in x-coordinates)
- Vertical leg: |y₂ - y₁| (difference in y-coordinates)
- Hypotenuse: distance between the points
Apply Pythagorean Theorem: d² = (x₂ - x₁)² + (y₂ - y₁)²
Solve for d: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
This is the distance formula!
Using the Distance Formula
Example 1: Find the distance between (1, 2) and (4, 6)
Solution: Identify coordinates: (x₁, y₁) = (1, 2) (x₂, y₂) = (4, 6)
Apply formula: d = √[(4 - 1)² + (6 - 2)²] d = √[(3)² + (4)²] d = √[9 + 16] d = √25 d = 5
Answer: 5 units
Example 2: Find the distance between (-2, 3) and (4, -1)
Solution: (x₁, y₁) = (-2, 3) (x₂, y₂) = (4, -1)
d = √[(4 - (-2))² + (-1 - 3)²] d = √[(4 + 2)² + (-4)²] d = √[(6)² + (-4)²] d = √[36 + 16] d = √52 d = √(4 × 13) d = 2√13 ≈ 7.21
Answer: 2√13 units (or approximately 7.21 units)
Example 3: Find the distance between (0, 0) and (3, 4)
Solution: d = √[(3 - 0)² + (4 - 0)²] d = √[9 + 16] d = √25 d = 5
Answer: 5 units
This is a 3-4-5 right triangle!
Special Cases
Horizontal Distance (same y-coordinate):
Points (1, 3) and (7, 3): d = √[(7-1)² + (3-3)²] d = √[36 + 0] d = √36 = 6
Shortcut: Just find |x₂ - x₁| = |7 - 1| = 6
Vertical Distance (same x-coordinate):
Points (2, 1) and (2, 8): d = √[(2-2)² + (8-1)²] d = √[0 + 49] d = √49 = 7
Shortcut: Just find |y₂ - y₁| = |8 - 1| = 7
Distance from Origin:
Point (5, 12) to origin (0, 0): d = √[(5-0)² + (12-0)²] d = √[25 + 144] d = √169 = 13
Shortcut: d = √(x² + y²)
Working with Negative Coordinates
Negative coordinates work the same way!
Example: Find distance between (-5, -2) and (3, 4)
Solution: d = √[(3 - (-5))² + (4 - (-2))²] d = √[(3 + 5)² + (4 + 2)²] d = √[(8)² + (6)²] d = √[64 + 36] d = √100 d = 10
Answer: 10 units
Remember: Squaring eliminates negative signs!
- (-3)² = 9 (positive!)
- (3)² = 9 (positive!)
Simplifying Radical Answers
Sometimes the answer is not a perfect square.
Example: √50
Simplify: √50 = √(25 × 2) = √25 × √2 = 5√2
Common simplifications:
- √8 = 2√2
- √12 = 2√3
- √18 = 3√2
- √20 = 2√5
- √27 = 3√3
- √32 = 4√2
- √45 = 3√5
- √48 = 4√3
- √50 = 5√2
- √72 = 6√2
When to use decimals vs. radicals:
- Exact answer: Leave as radical (5√2)
- Approximate: Use calculator (≈ 7.07)
Finding a Missing Coordinate
Sometimes you know the distance and need to find a coordinate!
Example: Point A is at (2, 3) and point B is at (x, 7). The distance is 5 units. Find x.
Solution: Use distance formula: 5 = √[(x - 2)² + (7 - 3)²] 5 = √[(x - 2)² + 16]
Square both sides: 25 = (x - 2)² + 16
Subtract 16: 9 = (x - 2)²
Take square root: ±3 = x - 2
Solve: x - 2 = 3 or x - 2 = -3 x = 5 or x = -1
Answer: x = 5 or x = -1
Both work! There are two points 5 units from (2, 3) with y-coordinate 7.
Real-World Applications
Mapping and GPS:
- Distance between two cities on a coordinate map
- Shortest path "as the crow flies"
- GPS calculates distance between coordinates
Sports:
- Distance a player ran on a field
- Finding shortest route to a target
Aviation:
- Distance between aircraft on radar
- Flight path calculations
Video Games:
- Calculating distance between characters
- Detecting when objects are close enough to interact
Architecture:
- Diagonal measurements in construction
- Room layout planning
The Midpoint Formula (Related Concept)
The midpoint is the point exactly halfway between two points.
Midpoint Formula:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Example: Find the midpoint between (2, 5) and (8, 11)
Solution: M = ((2 + 8)/2, (5 + 11)/2) M = (10/2, 16/2) M = (5, 8)
Answer: (5, 8)
Note: Average the x-coordinates, average the y-coordinates!
Combining Distance and Midpoint
Example: Points A(1, 3) and B(7, 11) are endpoints of a diameter of a circle. Find the center and radius.
Solution:
Center = Midpoint of AB C = ((1 + 7)/2, (3 + 11)/2) = (4, 7)
Radius = Distance from center to either endpoint r = distance from (4, 7) to (1, 3) r = √[(4 - 1)² + (7 - 3)²] r = √[9 + 16] r = √25 = 5
Answer: Center (4, 7), Radius = 5 units
Verifying Geometric Shapes
Use the distance formula to verify properties of shapes!
Example: Verify that points A(0, 0), B(3, 4), and C(6, 0) form an isosceles triangle.
Solution: Find all three side lengths:
AB = √[(3-0)² + (4-0)²] = √[9 + 16] = √25 = 5
AC = √[(6-0)² + (0-0)²] = √36 = 6
BC = √[(6-3)² + (0-4)²] = √[9 + 16] = √25 = 5
AB = BC = 5 (two sides equal)
Answer: Yes, it's isosceles! (Two sides have equal length)
Common Mistakes to Avoid
❌ Mistake 1: Forgetting to square the differences
- Wrong: d = √[(x₂ - x₁) + (y₂ - y₁)]
- Right: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
❌ Mistake 2: Forgetting the square root
- Wrong: d = (x₂ - x₁)² + (y₂ - y₁)²
- Right: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
❌ Mistake 3: Sign errors with negatives
- Wrong: 5 - (-3) = 2
- Right: 5 - (-3) = 5 + 3 = 8
❌ Mistake 4: Mixing up x and y
- Wrong: d = √[(y₂ - y₁)² + (x₂ - x₁)²]
- Right: Either order works! (x₂-x₁)² + (y₂-y₁)² is same as (y₂-y₁)² + (x₂-x₁)²
❌ Mistake 5: Not simplifying radicals
- Not simplified: √50
- Simplified: 5√2
Step-by-Step Strategy
Step 1: Label your points clearly
- (x₁, y₁) and (x₂, y₂)
Step 2: Write the distance formula
- d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Step 3: Substitute the coordinates
- Watch negative signs!
Step 4: Calculate inside the parentheses
- x₂ - x₁ and y₂ - y₁
Step 5: Square each difference
- Remember: negatives become positive
Step 6: Add the squares
Step 7: Take the square root
- Simplify if possible
Step 8: Round if needed (or leave exact)
Quick Reference
Distance Formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Midpoint Formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Special Cases:
- Horizontal: d = |x₂ - x₁|
- Vertical: d = |y₂ - y₁|
- From origin: d = √(x² + y²)
Remember:
- Squaring eliminates negatives
- Order doesn't matter
- Simplify radicals when possible
Practice Tips
Tip 1: Draw a picture
- Plot the points on a coordinate plane
- Visualize the right triangle
- Check if your answer makes sense
Tip 2: Check your arithmetic
- Subtraction with negatives is tricky!
- Use parentheses carefully
- Verify by using the Pythagorean Theorem visually
Tip 3: Simplify radicals
- Factor out perfect squares
- √(a²·b) = a√b
Tip 4: Use a calculator wisely
- For exact answers, leave as radicals
- For approximate answers, round appropriately
Connecting to Other Concepts
Pythagorean Theorem: Distance formula IS the Pythagorean Theorem applied to coordinate plane!
Slope: Slope uses differences too: m = (y₂ - y₁)/(x₂ - x₁)
Circles: Circle equation uses distance from center: (x - h)² + (y - k)² = r²
3D Distance (Preview): In 3D: d = √[(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²]
Summary
The distance formula calculates the straight-line distance between two points:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Key concepts:
- Based on the Pythagorean Theorem
- Works with any coordinates (positive, negative, zero)
- Answers can be exact (radicals) or approximate (decimals)
- Related to midpoint formula
Applications:
- Navigation and mapping
- Geometry (verifying shapes)
- Real-world distance calculations
- Computer graphics and game design
Mastering the distance formula connects algebra, geometry, and real-world problem solving!
📚 Practice Problems
1Problem 1easy
❓ Question:
Find the distance between points (1, 2) and (4, 6).
💡 Show Solution
Use the distance formula: d = √[(x₂-x₁)² + (y₂-y₁)²]
d = √[(4-1)² + (6-2)²]
d = √[3² + 4²]
d = √[9 + 16]
d = √25 = 5
Answer: 5 units
2Problem 2easy
❓ Question:
Find the distance between points (-2, 3) and (1, -1).
💡 Show Solution
Use the distance formula: d = √[(x₂-x₁)² + (y₂-y₁)²]
d = √[(1-(-2))² + (-1-3)²]
d = √[3² + (-4)²]
d = √[9 + 16]
d = √25 = 5
Answer: 5 units
3Problem 3medium
❓ Question:
Find the distance from the origin to point (6, 8).
💡 Show Solution
Origin is (0, 0). Use distance formula:
d = √[(6-0)² + (8-0)²]
d = √[6² + 8²]
d = √[36 + 64]
d = √100 = 10
Answer: 10 units
4Problem 4medium
❓ Question:
Points A(2, 5) and B(x, 1) are 5 units apart. Find the possible values of x.
💡 Show Solution
Use distance formula and set d = 5:
5 = √[(x-2)² + (1-5)²]
5 = √[(x-2)² + 16]
25 = (x-2)² + 16
9 = (x-2)²
±3 = x-2
x = 2+3 = 5 or x = 2-3 = -1
Answer: x = 5 or x = -1
5Problem 5hard
❓ Question:
Triangle ABC has vertices A(0, 0), B(5, 0), and C(5, 12). Determine if it is a right triangle.
💡 Show Solution
Find all three side lengths:
AB = √[(5-0)² + (0-0)²] = √25 = 5
BC = √[(5-5)² + (12-0)²] = √144 = 12
AC = √[(5-0)² + (12-0)²] = √[25+144] = √169 = 13
Check Pythagorean theorem: 5² + 12² = 25 + 144 = 169 = 13²
Yes! AB² + BC² = AC²
Answer: Yes, it is a right triangle (with right angle at B)
Practice with Flashcards
Review key concepts with our flashcard system
Browse All Topics
Explore other calculus topics