Lesson 3: The Distance Formula

Learning Objectives

By the end of this lesson, students will be able to:

  • Understand the connection between the Pythagorean theorem and the distance formula
  • Apply the distance formula to calculate the distance between any two points
  • Simplify radical expressions in distance calculations
  • Solve real-world problems involving distance on coordinate planes
  • Find unknown coordinates when given distance information

Introduction

One of the most important applications of coordinate geometry is finding the distance between two points. Whether you’re calculating how far apart two cities are on a map, finding the straight-line distance a delivery drone needs to travel, or determining if two locations are within wireless signal range, the distance formula is essential.

The Pythagorean Theorem Review

Before we explore the distance formula, let’s recall the Pythagorean theorem:

For a right triangle with legs of length \(a\) and \(b\), and hypotenuse of length \(c\):

\[a^2 + b^2 = c^2\]

Example: If a right triangle has legs of 3 and 4 units, find the hypotenuse.

\[c^2 = 3^2 + 4^2 = 9 + 16 = 25\] \[c = \sqrt{25} = 5\]

Deriving the Distance Formula

When we plot two points on a coordinate plane, we can create a right triangle:

Consider points A\((x_1, y_1)\) and B\((x_2, y_2)\): - The horizontal leg has length \(|x_2 - x_1|\) - The vertical leg has length \(|y_2 - y_1|\) - The distance between A and B is the hypotenuse

Using the Pythagorean theorem: \[d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\]

Taking the square root of both sides:

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

This is the distance formula.

Using the Distance Formula

Example 1: Basic Distance Calculation

Find the distance between points A(1, 2) and B(4, 6).

Solution: Identify the coordinates: \((x_1, y_1) = (1, 2)\) and \((x_2, y_2) = (4, 6)\)

\[d = \sqrt{(4-1)^2 + (6-2)^2}\] \[d = \sqrt{3^2 + 4^2}\] \[d = \sqrt{9 + 16}\] \[d = \sqrt{25} = 5\]

The distance is exactly 5 units.

Example 2: Distance with Negative Coordinates

Find the distance between C(-2, 3) and D(4, -1).

Solution: \[d = \sqrt{(4-(-2))^2 + (-1-3)^2}\] \[d = \sqrt{(4+2)^2 + (-4)^2}\] \[d = \sqrt{6^2 + 16}\] \[d = \sqrt{36 + 16}\] \[d = \sqrt{52}\]

To simplify: \(\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}\)

Approximately: \(d \approx 7.21\) units

Example 3: Points on the Same Horizontal or Vertical Line

Find the distance between E(2, 5) and F(2, 9).

Solution: \[d = \sqrt{(2-2)^2 + (9-5)^2}\] \[d = \sqrt{0 + 4^2}\] \[d = \sqrt{16} = 4\]

Note: When points share the same x-coordinate (vertical line) or y-coordinate (horizontal line), the distance is simply the difference in the other coordinate.

Example 4: Distance from the Origin

Find the distance from the origin O(0, 0) to point P(6, 8).

Solution: \[d = \sqrt{(6-0)^2 + (8-0)^2}\] \[d = \sqrt{36 + 64}\] \[d = \sqrt{100} = 10\]

The distance from the origin to any point \((x, y)\) is \(\sqrt{x^2 + y^2}\).

Simplifying Radicals

When the distance is not a perfect square, we should simplify:

Example 5: Simplifying Radical Answers

Find the distance between G(-1, 2) and H(5, 4), leaving your answer in simplest radical form.

Solution: \[d = \sqrt{(5-(-1))^2 + (4-2)^2}\] \[d = \sqrt{6^2 + 2^2}\] \[d = \sqrt{36 + 4}\] \[d = \sqrt{40}\]

Factor out perfect squares: \(\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}\)

Decimal approximation: \(d \approx 6.32\) units

Real-World Applications

Example 6: Map Distances

A park ranger at point P(2, 5) needs to reach a hiker in distress at point H(8, 13). Each coordinate unit represents 100 meters. How far must the ranger travel in a straight line?

Solution: \[d = \sqrt{(8-2)^2 + (13-5)^2}\] \[d = \sqrt{6^2 + 8^2}\] \[d = \sqrt{36 + 64}\] \[d = \sqrt{100} = 10 \text{ units}\]

Real distance = 10 × 100 = 1000 meters = 1 km

Example 7: Cell Tower Coverage

A cell tower is located at coordinates (3, 4) and has a signal range of 10 km. A house is located at (11, 10). Is the house within range? (Units are in kilometers)

Solution: Distance from tower to house: \[d = \sqrt{(11-3)^2 + (10-4)^2}\] \[d = \sqrt{8^2 + 6^2}\] \[d = \sqrt{64 + 36}\] \[d = \sqrt{100} = 10 \text{ km}\]

The house is exactly at the edge of the signal range (10 km).

Example 8: Comparing Distances

A delivery company needs to determine which warehouse is closer to a customer. - Warehouse A is at (2, 3) - Warehouse B is at (-1, 7) - Customer is at (5, 6)

Which warehouse should make the delivery?

Solution:

Distance from A to Customer: \[d_A = \sqrt{(5-2)^2 + (6-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \approx 4.24\]

Distance from B to Customer: \[d_B = \sqrt{(5-(-1))^2 + (6-7)^2} = \sqrt{36 + 1} = \sqrt{37} \approx 6.08\]

Warehouse A is closer (4.24 units vs 6.08 units).

Working Backwards: Finding Unknown Coordinates

Example 9: Finding an Unknown x-coordinate

Point A is at (3, 4) and point B is at (x, 10). The distance between them is 10 units. Find all possible values of x.

Solution: \[10 = \sqrt{(x-3)^2 + (10-4)^2}\] \[10 = \sqrt{(x-3)^2 + 36}\]

Square both sides: \[100 = (x-3)^2 + 36\] \[(x-3)^2 = 64\] \[x - 3 = \pm 8\]

Therefore: \(x = 3 + 8 = 11\) or \(x = 3 - 8 = -5\)

Answer: B could be at (11, 10) or (-5, 10).

Example 10: Finding an Unknown y-coordinate

Points P(2, y) and Q(6, 3) are 5 units apart. Find y.

Solution: \[5 = \sqrt{(6-2)^2 + (3-y)^2}\] \[5 = \sqrt{16 + (3-y)^2}\] \[25 = 16 + (3-y)^2\] \[(3-y)^2 = 9\] \[3 - y = \pm 3\]

Therefore: \(y = 3 - 3 = 0\) or \(y = 3 + 3 = 6\)

Answer: P could be at (2, 0) or (2, 6).

Connection to Lesson 2: Maps and Scale

Remember from Lesson 2 that maps use scales. When using the distance formula with map coordinates, don’t forget to convert to real distances:

Example 11: On a map with scale 1:50,000, two landmarks are at grid coordinates (120, 80) and (180, 140). Each grid unit represents 1 meter. What is the real distance between the landmarks?

Solution: Map distance (in grid units): \[d = \sqrt{(180-120)^2 + (140-80)^2}\] \[d = \sqrt{60^2 + 60^2}\] \[d = \sqrt{3600 + 3600}\] \[d = \sqrt{7200} = 60\sqrt{2} \approx 84.85 \text{ grid units}\]

Since each unit = 1 meter, map distance = 84.85 m

Real distance = 84.85 m × scale = 84.85 × 1 = 84.85 m (the scale tells us about map:real ratio, but here we’re working with grid units that are already scaled)

Practice Exercises

Worksheet 1B: Distance Practice

  1. Calculate the distance between:
    • A(1, 1) and B(4, 5)
    • C(-2, 3) and D(3, -9)
    • E(0, 0) and F(-5, 12)

Set 1: Basic Calculations

  1. Find the distance between:
    1. A(0, 0) and B(5, 12)
    2. C(2, 3) and D(8, 11)
    3. E(-4, 1) and F(2, -7)
    4. G(-3, -2) and H(-7, -5)
  2. Simplify your answers in simplest radical form:
    1. Distance between (0, 0) and (3, 7)
    2. Distance between (1, 2) and (6, 5)
    3. Distance between (-2, 3) and (4, 8)

Set 2: Special Cases

  1. Find the distance between:
    1. Two points on a horizontal line: (2, 5) and (9, 5)
    2. Two points on a vertical line: (3, 1) and (3, 8)
    3. Two points where one is the origin: (0, 0) and (-6, -8)

Set 3: Applications

  1. A drone flies from point A(10, 20) to point B(40, 50). Each unit represents 10 meters.
    1. How far does the drone travel?
    2. Express your answer in kilometers.
  2. Which point is closer to (5, 5)?
    • Point A at (1, 2)
    • Point B at (9, 9)
  3. A circular running track has its center at (0, 0) and a radius of 50 meters. A water station is at point (30, 40). Is the water station inside, on, or outside the track?

Set 4: Unknown Coordinates

  1. Point M is at (x, 6) and N is at (4, 2). If MN = 5, find all possible values of x.

  2. Points P(3, y) and Q(-1, 5) are \(2\sqrt{5}\) units apart. Find y.

  3. A point on the x-axis is 13 units from point (5, 12). Find the coordinates of the point.

Set 5: Challenge Problems

  1. Three vertices of a square are at A(1, 1), B(4, 1), and C(4, 4). Find the coordinates of the fourth vertex D and verify your answer by showing all four sides have equal length.

  2. Prove that the triangle with vertices A(0, 0), B(4, 0), and C(2, 2√3) is equilateral by showing all three sides have the same length.

  3. A lighthouse at position (0, 0) has a light that can be seen up to 20 km away. A ship is traveling in a straight line from point (-15, 20) to point (25, 15). Will the ship be visible from the lighthouse at any point during its journey? (Hint: Find the closest point on the path to the lighthouse)

Additional Resources

Online Tools

  • Use graphing calculators or Desmos to visualize distances
  • GeoGebra has excellent tools for exploring coordinate geometry
  • Google Earth for real-world distance applications

Video Resources

  • Search for “distance formula derivation” to see animated proofs
  • “Pythagorean theorem applications” shows real-world uses

Extension Topics

  • 3D Distance Formula: For three dimensions: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\)
  • Taxicab Distance: In cities with grid layouts, you can’t travel “as the crow flies”
  • GPS and Great Circle Distance: How GPS calculates distance on the curved Earth

Homework

Complete Practice Exercises Sets 1 and 2 completely, and at least 3 problems from Set 3.

For additional practice, see the Resources section.

  1. AB = 5
  2. CD = 13
  3. EF = 13

Previous Lesson: Lesson 2: Map It - Scale, Coordinates, and Bearings
Next Lesson: Lesson 4: The Midpoint Formula