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!