Solved Exercises on Distance Formula in Grade 8

Master distance formula: coordinate points, real-world applications, and geometric connections through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Distance Calculation
Exercise 1
Find the distance between points A(2, 3) and B(6, 7). Use the distance formula and round to the nearest hundredth.
Definition:

Distance Formula: The distance between two points (x₁, y₁) and (x₂, y₂) on the coordinate plane is d = √[(x₂ - x₁)² + (y₂ - y₁)²]. This formula is derived from the Pythagorean theorem.

Distance Calculation Method:
  1. Identify the coordinates: (x₁, y₁) and (x₂, y₂)
  2. Subtract the x-coordinates: (x₂ - x₁)
  3. Subtract the y-coordinates: (y₂ - y₁)
  4. Square both differences
  5. Add the squared differences
  6. Take the square root of the sum
Given Points
A(2, 3), B(6, 7)
Formula
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Result
d ≈ 5.66
Step 1: Identify coordinates

Point A: (x₁, y₁) = (2, 3)

Point B: (x₂, y₂) = (6, 7)

Step 2: Calculate differences

x₂ - x₁ = 6 - 2 = 4

y₂ - y₁ = 7 - 3 = 4

Step 3: Square the differences

(x₂ - x₁)² = 4² = 16

(y₂ - y₁)² = 4² = 16

Step 4: Add the squares

16 + 16 = 32

Step 5: Take the square root

d = √32 ≈ 5.66 units

d ≈ 5.66 units
Final answer:

The distance between points A(2, 3) and B(6, 7) is approximately 5.66 units.

Applied rules:

• Distance Formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]

• Order Matters: Subtract consistently (always 2nd - 1st)

• Pythagorean Connection: Distance formula is based on the Pythagorean theorem

2 Negative Coordinates
Exercise 2
Find the distance between points C(-3, 4) and D(2, -1). Use the distance formula and round to the nearest hundredth.
Definition:

Handling Negative Coordinates: When using the distance formula with negative coordinates, remember that subtracting a negative number is equivalent to adding its positive value. The squaring operation eliminates negative signs.

Given Points
C(-3, 4), D(2, -1)
Differences
x₂ - x₁ = 5, y₂ - y₁ = -5
Distance
d ≈ 7.07
Step 1: Identify coordinates

Point C: (x₁, y₁) = (-3, 4)

Point D: (x₂, y₂) = (2, -1)

Step 2: Calculate differences

x₂ - x₁ = 2 - (-3) = 2 + 3 = 5

y₂ - y₁ = -1 - 4 = -5

Step 3: Square the differences

(x₂ - x₁)² = 5² = 25

(y₂ - y₁)² = (-5)² = 25

Step 4: Add the squares

25 + 25 = 50

Step 5: Take the square root

d = √50 ≈ 7.07 units

d ≈ 7.07 units
Final answer:

The distance between points C(-3, 4) and D(2, -1) is approximately 7.07 units.

Applied rules:

• Subtracting Negatives: a - (-b) = a + b

• Squaring Eliminates Signs: (-x)² = x²

• Distance is Always Positive: The square root of a positive number is positive

3 Points on Same Horizontal Line
Exercise 3
Find the distance between points E(4, 5) and F(-2, 5). Explain why this is a special case and verify with the distance formula.
Definition:

Horizontal Line Distance: When two points have the same y-coordinate, they lie on a horizontal line. The distance is simply the absolute difference of their x-coordinates: |x₂ - x₁|.

Given Points
E(4, 5), F(-2, 5)
Same y-coordinate
y₁ = y₂ = 5
Distance
d = 6
Step 1: Identify coordinates

Point E: (x₁, y₁) = (4, 5)

Point F: (x₂, y₂) = (-2, 5)

Step 2: Notice y-coordinates are equal

y₁ = y₂ = 5, so points are on a horizontal line

Step 3: Calculate differences

x₂ - x₁ = -2 - 4 = -6

y₂ - y₁ = 5 - 5 = 0

Step 4: Apply distance formula

d = √[(-6)² + (0)²] = √[36 + 0] = √36 = 6 units

Step 5: Verify with shortcut

For horizontal lines: d = |x₂ - x₁| = |-2 - 4| = |-6| = 6 units

d = 6 units
Final answer:

The distance between points E(4, 5) and F(-2, 5) is 6 units.

Applied rules:

• Horizontal Line: When y₁ = y₂, distance = |x₂ - x₁|

• Vertical Line: When x₁ = x₂, distance = |y₂ - y₁|

• Shortcut Verification: The general formula confirms the shortcut

Rules and methods, laws,...
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Distance Formula
General Formula
d = √[(x₂-x₁)² + (y₂-y₁)²]
Works for any two points
Horizontal Line
d = |x₂ - x₁|
When y₁ = y₂
Vertical Line
d = |y₂ - y₁|
When x₁ = x₂
Pythagorean Theorem
a² + b² = c²
Foundation of distance formula
Midpoint Formula
M = ((x₁+x₂)/2, (y₁+y₂)/2)
Related concept
Distance is Always Positive: The square root of a positive number is positive
Order Independent: Distance from A to B equals distance from B to A
Key definitions:

Distance: The length of the shortest path between two points, measured in linear units.

Coordinate Plane: A two-dimensional surface formed by the intersection of two perpendicular number lines.

Pythagorean Theorem: In a right triangle, a² + b² = c², where a and b are the legs and c is the hypotenuse.

Horizontal Line: A line parallel to the x-axis where all points have the same y-coordinate.

Vertical Line: A line parallel to the y-axis where all points have the same x-coordinate.

Complete methodology:
  1. Identify coordinates: Determine (x₁, y₁) and (x₂, y₂)
  2. Calculate differences: Find (x₂ - x₁) and (y₂ - y₁)
  3. Square differences: Calculate (x₂ - x₁)² and (y₂ - y₁)²
  4. Sum squares: Add the squared differences
  5. Take square root: Find √[sum of squares]
  6. Round if needed: Round to the required precision
Tip 1: Always subtract in the same order: (2nd coordinate - 1st coordinate).
Tip 2: Remember that squaring eliminates negative signs: (-a)² = a².
Tip 3: Use shortcuts for horizontal/vertical lines to save time.

Solution: Exercises 4 to 5
4 Real-World Application
Exercise 4
Sarah is at point A(1, 2) and Tom is at point B(7, 8). If each unit represents 100 meters, how far apart are they in kilometers?
Definition:

Real-World Application: The distance formula can be used to calculate actual distances between locations when coordinates represent physical positions.

Given Points
A(1, 2), B(7, 8)
Scale Factor
1 unit = 100 meters
Distance
d = 0.85 km
Step 1: Calculate distance in coordinate units

d = √[(7-1)² + (8-2)²] = √[6² + 6²] = √[36 + 36] = √72 ≈ 8.49 units

Step 2: Convert to real-world units

Distance = 8.49 units × 100 meters/unit = 849 meters

Step 3: Convert to kilometers

Distance = 849 meters ÷ 1000 = 0.849 km ≈ 0.85 km

0.85 km
Final answer:

Sarah and Tom are approximately 0.85 kilometers apart.

Applied rules:

• Distance Formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]

• Unit Conversion: 1 km = 1000 m

• Scale Factor: Multiply coordinate distance by real-world scale

5 Geometric Connection
Exercise 5
Show that points P(0, 0), Q(3, 4), and R(0, 8) form a right triangle by calculating all three side lengths and verifying the Pythagorean theorem.
Definition:

Pythagorean Theorem Verification: For a right triangle with sides a, b, and hypotenuse c, the relationship a² + b² = c² must hold true.

Side PQ
d = 5
Side QR
d = 5
Side PR
d = 8
Step 1: Calculate distance PQ

PQ = √[(3-0)² + (4-0)²] = √[9 + 16] = √25 = 5

Step 2: Calculate distance QR

QR = √[(0-3)² + (8-4)²] = √[9 + 16] = √25 = 5

Step 3: Calculate distance PR

PR = √[(0-0)² + (8-0)²] = √[0 + 64] = √64 = 8

Step 4: Verify Pythagorean theorem

Check if (PQ)² + (QR)² = (PR)²

5² + 5² = 25 + 25 = 50

8² = 64

Since 50 ≠ 64, these points do not form a right triangle.

Step 5: Recalculate (corrected approach)

Actually, let's check if the longest side is the hypotenuse:

Let's verify: (PQ)² + (QR)² = 25 + 25 = 50 and (PR)² = 64

Since 50 ≠ 64, it's not a right triangle with PR as hypotenuse.

But 5² + 5² = 50 and 8² = 64, so 5² + 5² + 14 = 8², showing it's close to a right triangle but not exact.

Triangle PQR is not a right triangle
Final answer:

Points P(0, 0), Q(3, 4), and R(0, 8) do not form a right triangle since 5² + 5² ≠ 8².

Applied rules:

• Distance Formula: Calculate all three side lengths

• Pythagorean Theorem: a² + b² = c² for right triangles

• Verification: Check if the equation holds true

Distance Formula Laws, Methods, and Properties
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Distance Formula
Key definitions:

Distance Formula: A mathematical formula derived from the Pythagorean theorem that calculates the distance between any two points in a coordinate plane. It represents the length of the straight line connecting two points.

Pythagorean Theorem: The foundation of the distance formula, stating that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

Euclidean Distance: The "as-the-crow-flies" distance between two points in Euclidean space.

Complete methodology:
  1. Identify coordinates: Determine the exact coordinates of both points
  2. Organize values: Assign coordinates to x₁, y₁, x₂, y₂ consistently
  3. Calculate differences: Find horizontal and vertical distances
  4. Apply formula: Square differences, sum them, then take square root
  5. Simplify: Calculate the final distance value
  6. Verify: Check that the result makes sense in context
Tip 1: The distance is always positive regardless of coordinate signs.
Tip 2: Use mental math shortcuts for simple coordinate differences.

Tip 3: Remember that distance is the same in both directions (A to B equals B to A).

Tip 4: For points on the same axis, use the simpler absolute difference.

Common errors: Forgetting to square differences, subtracting in wrong order, miscounting coordinate values, forgetting that distance is always positive.
Exam preparation: Practice with mixed positive/negative coordinates, memorize the formula, understand geometric applications, work with real-world contexts.
Distance formula properties:

• Symmetry: d(A,B) = d(B,A) - distance is the same in both directions

• Non-negativity: d ≥ 0 - distance is always positive or zero

• Zero Distance: d(A,B) = 0 if and only if A and B are the same point

• Triangle Inequality: d(A,C) ≤ d(A,B) + d(B,C) for any three points

• Horizontal Distance: When y₁ = y₂, d = |x₂ - x₁|

• Vertical Distance: When x₁ = x₂, d = |y₂ - y₁|

• Pythagorean Foundation: Based on a² + b² = c² where a and b are coordinate differences

Exercise with Visualization: Distance Relationships
Exercise 6: Distance vs Coordinate Differences
Consider points with various coordinate differences:
Δx = 0: (0,0) to (0,3), (0,0) to (0,6), (0,0) to (0,9)
Δy = 0: (0,0) to (3,0), (0,0) to (6,0), (0,0) to (9,0)
Equal differences: (0,0) to (3,3), (0,0) to (4,4), (0,0) to (5,5)

Analysis: The chart shows how distance varies with coordinate differences.

  • Vertical/horizontal distances grow linearly with coordinate differences
  • Diagonal distances grow with the square root of the sum of squares
  • This demonstrates the relationship in the distance formula

Questions & Answers

Question: Why does the distance formula involve square roots? Can't we just add the differences?

Answer: The distance formula involves square roots because it's based on the Pythagorean theorem. When you connect two points, you form a right triangle where:

  • The horizontal difference (x₂ - x₁) is one leg
  • The vertical difference (y₂ - y₁) is the other leg
  • The distance between points is the hypotenuse

According to the Pythagorean theorem: (leg₁)² + (leg₂)² = (hypotenuse)²

So: (x₂ - x₁)² + (y₂ - y₁)² = d²

Therefore: d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Simply adding differences would give the "city block" distance, not the straight-line distance!

Question: How do I remember the order of subtraction in the distance formula? Does it matter?

Answer: The order of subtraction does NOT matter because of the squaring operation:

However, you must be consistent within each coordinate:

Correct: (x₂ - x₁)² + (y₂ - y₁)²

Incorrect: (x₂ - x₁)² + (y₁ - y₂)²

I recommend always subtracting "second point - first point" to maintain consistency: (x₂ - x₁)² + (y₂ - y₁)².

Question: What's the difference between distance and displacement? Are they the same thing?

Answer: While related, distance and displacement are different concepts:

For two points in coordinate geometry, the distance formula gives the displacement (straight-line distance). The distance is always the shortest path between two points in Euclidean geometry.

In our coordinate system, distance and displacement magnitude are the same!

Question: How can I quickly estimate the distance without calculating the exact square root?

Answer: Here are some estimation strategies:

  • Perfect Squares: Know common perfect squares (4, 9, 16, 25, 36, 49, 64, 81, 100)
  • Between Values: If the sum is between perfect squares, estimate between their square roots
  • Common Patterns: √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2
  • Factor Out: √50 = √(25×2) = 5√2 ≈ 5×1.4 = 7

For √32: It's between √25=5 and √36=6, closer to 6, so about 5.6-5.7.

This helps verify your exact calculations!

Question: Can the distance formula be extended to three dimensions? How would that work?

Answer: Yes! The distance formula extends naturally to three dimensions:

For points (x₁, y₁, z₁) and (x₂, y₂, z₂):

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

This adds a third coordinate difference under the square root. The principle remains the same - it's based on extending the Pythagorean theorem to three dimensions.

In n-dimensions: d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)² + ... + (n₂ - n₁)²]

This concept is fundamental in higher mathematics and physics!