Distance Between Points

For Elementary Students

What Is Distance Between Points?

Distance tells you how far apart two points are on the coordinate plane!

Think about it like this: If you put your finger on one point and want to walk to another point, the distance tells you how far you need to travel!

Easy Case: Same Row or Same Column

Sometimes finding distance is SUPER easy!

Points in the Same Row (Same y)

When two points are in the same horizontal row, they have the same y-coordinate!

Example: Points (2, 5) and (7, 5)

`(2,5)`────────`(7,5)`
  │   5 units │

To find distance: Just subtract the x-coordinates!

Distance = 7 − 2 = 5 units

Another example: (1, 3) and (8, 3)

Distance = 8 − 1 = 7 units ✓

Points in the Same Column (Same x)

When two points are in the same vertical column, they have the same x-coordinate!

Example: Points (4, 2) and (4, 9)

`(4,9)`
  │
  │ 7 units
  │
`(4,2)`

To find distance: Subtract the y-coordinates!

Distance = 9 − 2 = 7 units

Use Absolute Value

Since distance is always positive, use absolute value if you're not sure which number is bigger!

|7 − 2| = 5
|2 − 7| = 5  (same answer!)

Diagonal Points: The Tricky Case

When points are diagonal (different x AND different y), we need a special formula!

Think about it like this: Imagine making a right triangle!

Example: From (1, 1) to (4, 5)

    `(4,5)`
     /│
    / │ 4 up
   /  │
  /   │
`(1,1)`──┘
  3 right

• Go 3 units right (from x=1 to x=4)
• Go 4 units up (from y=1 to y=5)
• The diagonal is the distance!

The Distance Formula (Magic Formula!)

For diagonal points, use this formula:

Distance = √(horizontal² + vertical²)

Step by step:

Step 1: Find horizontal distance (change in x) Step 2: Find vertical distance (change in y) Step 3: Square both numbers Step 4: Add them Step 5: Take the square root!

Example: Find Distance Between (1, 2) and (4, 6)

Step 1: Horizontal change

4 − 1 = 3

Step 2: Vertical change

6 − 2 = 4

Step 3: Square both

3² = 9
4² = 16

Step 4: Add

9 + 16 = 25

Step 5: Square root

√25 = 5

Answer: 5 units! ✓

Why Does This Work?

It's actually the Pythagorean Theorem in disguise!

a² + b² = c²

Horizontal² + Vertical² = Distance²

3² + 4² = 5²
9 + 16 = 25
25 = 25 ✓

The diagonal makes the hypotenuse (longest side) of a right triangle!

Another Example: (0, 0) to (3, 4)

Horizontal: 3 − 0 = 3
Vertical: 4 − 0 = 4

Distance = √(3² + 4²)
        = √(9 + 16)
        = √25
        = 5 units

This is the famous 3-4-5 triangle!

When the Answer Isn't a Whole Number

Sometimes the distance is not a nice whole number!

Example: From (0, 0) to (1, 1)

Distance = √(1² + 1²)
        = √(1 + 1)
        = √2
        = about 1.41 units

That's okay! Just leave it as √2 or use the decimal.

Quick Reference

Same row (same y)? → Subtract x-coordinates Same column (same x)? → Subtract y-coordinates Diagonal? → Use √(horizontal² + vertical²)

For Junior High Students

Understanding Distance on the Coordinate Plane

The distance between two points is the length of the straight-line segment connecting them.

Context: This concept extends the idea of distance on a number line to two dimensions.

Special Cases: Horizontal and Vertical Distance

Horizontal distance (points on same horizontal line):

When two points share the same y-coordinate: (x₁, y) and (x₂, y)

Formula: d = |x₂ − x₁|

Example: Distance between (1, 5) and (9, 5)

d = |9 − 1| = 8 units

Why absolute value? Distance is always non-negative, regardless of which point is listed first.

Vertical distance (points on same vertical line):

When two points share the same x-coordinate: (x, y₁) and (x, y₂)

Formula: d = |y₂ − y₁|

Example: Distance between (3, 2) and (3, 8)

d = |8 − 2| = 6 units

The Distance Formula

For any two points (x₁, y₁) and (x₂, y₂) in the coordinate plane:

Distance Formula:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Derivation from Pythagorean Theorem:

Consider points A(x₁, y₁) and B(x₂, y₂). The horizontal distance is |x₂ − x₁| and the vertical distance is |y₂ − y₁|.

These form the legs of a right triangle, with the distance d as the hypotenuse:

d² = (x₂ − x₁)² + (y₂ − y₁)²
d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Note: Since we're squaring the differences, we don't need absolute value bars—squaring automatically makes negative values positive.

Applying the Distance Formula

Example 1: Find the distance between (1, 2) and (4, 6)

Given: (x₁, y₁) = `(1, 2)` and (x₂, y₂) = `(4, 6)`

d = √[(4 − 1)² + (6 − 2)²]
  = √[3² + 4²]
  = √[9 + 16]
  = √25
  = 5

Distance: 5 units

Verification: The legs are 3 and 4, so by the Pythagorean theorem, the hypotenuse is 5 (a 3-4-5 right triangle).

Example 2: Distance between (-2, 1) and (1, 5)

d = √[(1 − (−2))² + (5 − 1)²]
  = √[(1 + 2)² + 4²]
  = √[3² + 4²]
  = √[9 + 16]
  = √25
  = 5 units

Example 3: Distance between (-3, -1) and (2, 3)

d = √[(2 − (−3))² + (3 − (−1))²]
  = √[5² + 4²]
  = √[25 + 16]
  = √41
  ≈ 6.40 units

Note: Not all distances simplify to whole numbers. √41 is the exact answer; 6.40 is approximate.

Order Doesn't Matter

The distance formula gives the same result regardless of which point is labeled first.

Proof:

d₁ = √[(x₂ − x₁)² + (y₂ − y₁)²]
d₂ = √[(x₁ − x₂)² + (y₁ − y₂)²]

Since (x₂ − x₁)² = (x₁ − x₂)² (squaring eliminates sign)
And (y₂ − y₁)² = (y₁ − y₂)²

Therefore: d₁ = d₂

Irrational Distances

Many distances are irrational numbers (cannot be expressed as fractions).

Example: Distance from (0, 0) to (1, 1)

d = √[(1 − 0)² + (1 − 0)²]
  = √[1 + 1]
  = √2
  ≈ 1.414...

√2 is irrational; leave the answer in radical form unless asked for a decimal approximation.

Connection to the Pythagorean Theorem

The distance formula is an application of the Pythagorean theorem: a² + b² = c²

Components:

• a = |x₂ − x₁| (horizontal leg)
• b = |y₂ − y₁| (vertical leg)
• c = d (distance, the hypotenuse)

Formula: d² = (x₂ − x₁)² + (y₂ − y₁)²

Solving for d: d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Distance in Different Quadrants

The formula works for points in any quadrant, including negative coordinates.

Example: Distance from (-4, -3) to (2, 5)

d = √[(2 − (−4))² + (5 − (−3))²]
  = √[(2 + 4)² + (5 + 3)²]
  = √[6² + 8²]
  = √[36 + 64]
  = √100
  = 10 units

Using Distance to Verify Geometric Properties

Example: Verify that (0, 0), (4, 0), (4, 3), (0, 3) form a rectangle.

Solution: Check that opposite sides are equal:

Side 1: `(0,0)` to `(4,0)`: d = 4
Side 2: `(4,0)` to `(4,3)`: d = 3
Side 3: `(4,3)` to `(0,3)`: d = 4
Side 4: `(0,3)` to `(0,0)`: d = 3

Opposite sides equal: 4 = 4, 3 = 3 ✓

Diagonal 1: `(0,0)` to `(4,3)`: √(16 + 9) = 5
Diagonal 2: `(4,0)` to `(0,3)`: √(16 + 9) = 5

Diagonals equal: 5 = 5 ✓

Conclusion: It's a rectangle

Real-Life Applications

Navigation: Distance between two locations on a map

Physics: Displacement between two positions

Computer graphics: Distance calculations for rendering and collision detection

Engineering: Measuring distances in 2D plans

Robotics: Path planning and navigation

Common Mistakes

Mistake 1: Forgetting to square the differences

❌ d = √[(x₂ − x₁) + (y₂ − y₁)] ✓ d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Mistake 2: Squaring after adding

❌ d = √(x₂ − x₁ + y₂ − y₁)² ✓ d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Mistake 3: Sign errors with negative coordinates

❌ (3 − (−2)) = 1 ✓ (3 − (−2)) = 3 + 2 = 5

Mistake 4: Using subtraction instead of Pythagorean formula

❌ For diagonal points: d = (x₂ − x₁) + (y₂ − y₁) ✓ Must use: d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Mistake 5: Forgetting the square root

❌ d = (x₂ − x₁)² + (y₂ − y₁)² ✓ d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Tips for Success

Tip 1: Draw a diagram showing the two points and the right triangle

Tip 2: Label horizontal and vertical distances before using the formula

Tip 3: Check if points share the same x or y coordinate for simpler calculation

Tip 4: Verify with Pythagorean theorem when possible (3-4-5, 5-12-13 triangles)

Tip 5: Leave answers in radical form unless decimal approximation requested

Tip 6: Use parentheses carefully with negative coordinates

Special Right Triangles

Recognize common distance patterns:

Knowing these saves time in calculations.

Formula Summary

For any two points:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Horizontal only (y₁ = y₂):

d = |x₂ − x₁|

Vertical only (x₁ = x₂):

d = |y₂ − y₁|

Practice