Solved Exercises on Distance Between Points in Grade 8

Master distance between points: coordinate plane, real-world applications, and problem-solving through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Distance Formula
Exercise 1
Find the distance between points A(3, 4) and B(7, 9).
Definition:

Distance formula: The distance between points (x₁, y₁) and (x₂, y₂) is d = √[(x₂-x₁)² + (y₂-y₁)²]

Solution method:
  1. Identify coordinates: (x₁, y₁) and (x₂, y₂)
  2. Substitute into distance formula
  3. Calculate differences and squares
  4. Find the square root
Given
A(3,4), B(7,9)
Formula
d = √[(7-3)² + (9-4)²]
Solution
d = √41 ≈ 6.4
Step 1: Identify coordinates

Point A: (x₁, y₁) = (3, 4)

Point B: (x₂, y₂) = (7, 9)

Step 2: Apply distance formula

d = √[(x₂-x₁)² + (y₂-y₁)²]

d = √[(7-3)² + (9-4)²]

Step 3: Calculate differences

x₂ - x₁ = 7 - 3 = 4

y₂ - y₁ = 9 - 4 = 5

Step 4: Square and add

d = √[4² + 5²]

d = √[16 + 25]

d = √41 ≈ 6.40 units

Distance ≈ 6.40 units
Final answer:

The distance between A(3, 4) and B(7, 9) is √41 ≈ 6.40 units

Applied rules:

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

• Order matters: Subtract consistently (second point minus first)

• Verification: Check that (4)² + (5)² = 16 + 25 = 41 ✓

2 Distance with Negative Coordinates
Exercise 2
Find the distance between points P(-2, 5) and Q(4, -3).
Definition:

Handling negative coordinates: The distance formula works the same with negative values

Given
P(-2,5), Q(4,-3)
Formula
d = √[(4-(-2))² + (-3-5)²]
Solution
d = √100 = 10
Step 1: Identify coordinates

Point P: (x₁, y₁) = (-2, 5)

Point Q: (x₂, y₂) = (4, -3)

Step 2: Apply distance formula

d = √[(x₂-x₁)² + (y₂-y₁)²]

d = √[(4-(-2))² + (-3-5)²]

Step 3: Calculate differences

x₂ - x₁ = 4 - (-2) = 4 + 2 = 6

y₂ - y₁ = -3 - 5 = -8

Step 4: Square and add

d = √[6² + (-8)²]

d = √[36 + 64]

d = √100 = 10 units

Distance = 10 units
Final answer:

The distance between P(-2, 5) and Q(4, -3) is 10 units

Applied rules:

• Negative subtraction: Subtracting a negative is adding a positive

• Squaring negatives: (-a)² = a² (always positive)

• Distance is always positive: The result under the square root is always ≥ 0

3 Real-World Application
Exercise 3
On a city map, the library is at (-3, 2) and the park is at (5, 7). If each unit represents 100 meters, how far apart are they in kilometers?
Definition:

Scale conversion: Convert coordinate distance to real-world distance using the given scale

Given
Library(-3,2), Park(5,7)
Formula
d = √[(5-(-3))² + (7-2)²]
Conversion
890 meters = 0.89 km
Step 1: Identify coordinates

Library: (x₁, y₁) = (-3, 2)

Park: (x₂, y₂) = (5, 7)

Step 2: Apply distance formula

d = √[(x₂-x₁)² + (y₂-y₁)²]

d = √[(5-(-3))² + (7-2)²]

Step 3: Calculate differences

x₂ - x₁ = 5 - (-3) = 8

y₂ - y₁ = 7 - 2 = 5

Step 4: Square and add

d = √[8² + 5²]

d = √[64 + 25]

d = √89 ≈ 9.43 coordinate units

Step 5: Convert to real distance

Each unit = 100 meters

Distance = 9.43 × 100 = 943 meters

Distance = 943 ÷ 1000 = 0.943 kilometers

Distance ≈ 0.94 km
Final answer:

The library and park are approximately 0.94 kilometers apart

Applied rules:

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

• Scale conversion: Multiply coordinate distance by scale factor

• Unit conversion: 1 km = 1000 m

Rules and methods, laws,...
d = √[(x₂-x₁)² + (y₂-y₁)²]
Distance Formula
d = √[(x₂-x₁)²]
Horizontal Distance
d = √[(y₂-y₁)²]
Vertical Distance
Horizontal Line
d = |x₂ - x₁|
When y₁ = y₂
Vertical Line
d = |y₂ - y₁|
When x₁ = x₂
Origin Distance
d = √(x² + y²)
Distance from (0,0) to (x,y)
Key definitions:

Coordinate plane: A two-dimensional plane defined by x-axis and y-axis

Ordered pair: A point represented as (x, y) where x is horizontal and y is vertical

Distance: The length of the straight line connecting two points

Distance formula: d = √[(x₂-x₁)² + (y₂-y₁)²], derived from Pythagorean theorem

Horizontal distance: Distance between points with the same y-coordinate

Vertical distance: Distance between points with the same x-coordinate

Midpoint: The point exactly halfway between two points

Complete methodology:
  1. Identify the coordinates: Determine (x₁, y₁) and (x₂, y₂) for the two points
  2. Substitute into formula: Place coordinates into d = √[(x₂-x₁)² + (y₂-y₁)²]
  3. Calculate differences: Find x₂-x₁ and y₂-y₁
  4. Square each difference: Compute (x₂-x₁)² and (y₂-y₁)²
  5. Add the squares: Sum the two squared differences
  6. Take the square root: Find the final distance value
  7. Convert units if needed: Apply scale factors or unit conversions
Tip 1: The distance is always positive, regardless of coordinate signs.
Tip 2: Order doesn't matter since we're squaring the differences.
Tip 3: For horizontal lines (same y), distance is |x₂-x₁|.
Tip 4: For vertical lines (same x), distance is |y₂-y₁|.
Common errors: Forgetting to subtract coordinates properly, mixing up x and y values, making arithmetic errors with negative numbers, forgetting to take the square root.
Exam preparation: Practice with various coordinate combinations, memorize the formula, work on problems with different scales, understand when to use shortcuts for horizontal/vertical distances.
Formulas to know by heart:

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

• Horizontal distance: d = |x₂ - x₁| (when y₁ = y₂)

• Vertical distance: d = |y₂ - y₁| (when x₁ = x₂)

• Distance from origin: d = √(x² + y²)

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)

• Relationship to Pythagorean theorem: The distance formula is derived from a² + b² = c²

Solution: Exercises 4 to 5
4 Perimeter of a Triangle
Exercise 4
Find the perimeter of triangle ABC with vertices A(1, 2), B(4, 6), and C(7, 2).
Definition:

Perimeter: The sum of all side lengths of a polygon

Side AB
√[(4-1)² + (6-2)²] = 5
Side BC
√[(7-4)² + (2-6)²] = 5
Side AC
√[(7-1)² + (2-2)²] = 6
Step 1: Identify all vertices

A(1, 2), B(4, 6), C(7, 2)

Step 2: Calculate side AB

d_AB = √[(4-1)² + (6-2)²]

d_AB = √[3² + 4²] = √[9 + 16] = √25 = 5 units

Step 3: Calculate side BC

d_BC = √[(7-4)² + (2-6)²]

d_BC = √[3² + (-4)²] = √[9 + 16] = √25 = 5 units

Step 4: Calculate side AC

d_AC = √[(7-1)² + (2-2)²]

d_AC = √[6² + 0²] = √36 = 6 units

Step 5: Calculate perimeter

Perimeter = AB + BC + AC

Perimeter = 5 + 5 + 6 = 16 units

Perimeter = 16 units
Final answer:

The perimeter of triangle ABC is 16 units

Applied rules:

• Distance formula: Calculate each side separately

• Perimeter: Sum of all side lengths

• Verification: Check that all three distances are positive

5 Verifying Geometric Shapes
Exercise 5
Show that the points A(0, 0), B(4, 0), C(4, 3), and D(0, 3) form a rectangle by calculating all side lengths and diagonals.
Definition:

Rectangle verification: Opposite sides equal, all angles 90°, diagonals equal

AB & CD
4 units each
BC & AD
3 units each
Diagonals
5 units each
Step 1: Calculate all side lengths

AB = √[(4-0)² + (0-0)²] = √16 = 4 units

BC = √[(4-4)² + (3-0)²] = √9 = 3 units

CD = √[(0-4)² + (3-3)²] = √16 = 4 units

AD = √[(0-0)² + (3-0)²] = √9 = 3 units

Step 2: Calculate diagonals

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

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

Step 3: Verify rectangle properties

Opposite sides equal: AB = CD = 4, BC = AD = 3 ✓

Diagonals equal: AC = BD = 5 ✓

All angles are 90° (can be verified using slopes)

Rectangle confirmed
Final answer:

The points form a rectangle since opposite sides are equal and diagonals are equal.

Applied rules:

• Rectangle properties: Opposite sides equal, diagonals equal

• Distance formula: Calculate all sides and diagonals

• Verification: Check all necessary conditions

Key Concepts, Laws, Methods, and Formulas for Distance Between Points
d = √[(x₂-x₁)² + (y₂-y₁)²]
Distance Formula
Key definitions:

Coordinate system: A system that uses ordered pairs (x, y) to locate points on a plane

Distance: The length of the shortest path between two points, always a positive value

Distance formula: d = √[(x₂-x₁)² + (y₂-y₁)²], derived from the Pythagorean theorem

Coordinate plane: A two-dimensional plane with x-axis (horizontal) and y-axis (vertical)

Ordered pair: A set of coordinates (x, y) representing a point's location

Quadrants: Four regions of the coordinate plane divided by the axes

Horizontal distance: The distance between two points with the same y-coordinate: |x₂ - x₁|

Vertical distance: The distance between two points with the same x-coordinate: |y₂ - y₁|

Origin: The point (0, 0) where the x-axis and y-axis intersect

Pythagorean theorem connection: The distance formula is derived from a² + b² = c² where the legs are the differences in x and y coordinates

Complete methodology:
  1. Plot the points: If not already plotted, mark the locations on the coordinate plane to visualize the problem
  2. Identify coordinates: Carefully read and record the (x, y) coordinates of both points
  3. Assign variables: Designate one point as (x₁, y₁) and the other as (x₂, y₂)
  4. Substitute into formula: Place the coordinates into d = √[(x₂-x₁)² + (y₂-y₁)²]
  5. Calculate differences: Find x₂-x₁ and y₂-y₁, paying attention to signs
  6. Square the differences: Calculate (x₂-x₁)² and (y₂-y₁)², noting that squares are always positive
  7. Add the squares: Sum the two squared differences
  8. Take the square root: Find the principal (positive) square root of the sum
  9. Simplify if possible: Express the answer in simplest radical form or as a decimal approximation
  10. Check reasonableness: Verify that the calculated distance makes sense in the context of the problem
Tip 1: Distance is always positive - if you get a negative result, you made an error somewhere.
Tip 2: The order of points doesn't matter since we're squaring the differences (a² = (-a)²).
Tip 3: For horizontal lines (same y-values), distance is simply |x₂ - x₁|.
Tip 4: For vertical lines (same x-values), distance is simply |y₂ - y₁|.
Tip 5: Always double-check your subtraction to avoid sign errors.
Tip 6: When dealing with real-world problems, pay attention to the scale of the coordinate system.
Common errors: Mixing up x and y coordinates, making subtraction errors with negative numbers, forgetting to take the square root, confusing the formula with midpoint formula, arithmetic mistakes when squaring numbers.
Memory aids: "Difference in x, square it; difference in y, square it; add them, square root it", "Distance is always positive", "The formula comes from the Pythagorean theorem".
Problem-solving strategies: Draw a rough sketch to visualize the points, label coordinates clearly, substitute carefully into the formula, check your answer by estimating the distance visually.
Essential formulas and theorems:

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

• Horizontal distance: d = |x₂ - x₁| (when y₁ = y₂)

• Vertical distance: d = |y₂ - y₁| (when x₁ = x₂)

• Distance from origin: d = √(x² + y²)

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)

• Pythagorean theorem: a² + b² = c² (distance formula derivation)

• Properties of geometric shapes: Use distances to verify squares, rectangles, triangles

Visual Representation: Coordinate Distances
Exercise 6: Distance vs Coordinate Differences
Visual representation of how coordinate differences affect distance calculations:
- Horizontal distances
- Vertical distances
- Diagonal distances
- Scale factors

Analysis: The chart illustrates how different coordinate differences result in various distance values.

  • Equal x-differences with varying y-differences show increasing distance
  • The relationship follows the Pythagorean theorem pattern
  • Small coordinate changes can result in significant distance differences
  • Distance grows proportionally to the square root of the sum of squares

Questions & Answers

Question: I sometimes get confused about which point should be (x₁, y₁) and which should be (x₂, y₂). Does it matter?

Answer: No, it doesn't matter which point you assign as (x₁, y₁) and which as (x₂, y₂). Here's why:

  • When you subtract coordinates, you might get different signs: (x₂-x₁) vs (x₁-x₂)
  • But when you square these differences, the result is the same: (x₂-x₁)² = (x₁-x₂)²
  • This is because (-a)² = a² for any number a

For example, if you have points A(2, 3) and B(5, 7):

  • Option 1: d = √[(5-2)² + (7-3)²] = √[3² + 4²] = √25 = 5
  • Option 2: d = √[(2-5)² + (3-7)²] = √[(-3)² + (-4)²] = √25 = 5

Both approaches give the same result. Just be consistent with which point is first and which is second throughout the calculation.

Question: Can the distance between two points ever be negative?

Answer: No, distance can never be negative! Here's why:

  • Physical meaning: Distance represents the length between two points, which is always a positive quantity
  • Mathematical reason: The distance formula involves squaring differences, which always gives positive results, then taking the positive square root
  • Formula structure: d = √[(x₂-x₁)² + (y₂-y₁)²] - the sum of squares is always ≥ 0, and the square root symbol indicates the principal (positive) root

Even when dealing with negative coordinates, the distance will still be positive. For example, the distance between (-3, -4) and (0, 0) is √[(-3)² + (-4)²] = √[9 + 16] = √25 = 5 units.

If you get a negative result, you've made an error in your calculation.

Question: When should I use shortcuts for horizontal or vertical distances instead of the full distance formula?

Answer: Use shortcuts when possible to save time:

Horizontal distance shortcut: When y-coordinates are the same

  • Points A(x₁, k) and B(x₂, k) have distance = |x₂ - x₁|
  • Example: Distance between (3, 5) and (8, 5) is |8 - 3| = 5

Vertical distance shortcut: When x-coordinates are the same

  • Points A(k, y₁) and B(k, y₂) have distance = |y₂ - y₁|
  • Example: Distance between (4, 2) and (4, 9) is |9 - 2| = 7

These shortcuts come from the full distance formula:

  • Horizontal: √[(x₂-x₁)² + (k-k)²] = √[(x₂-x₁)² + 0] = |x₂-x₁|
  • Vertical: √[(k-k)² + (y₂-y₁)²] = √[0 + (y₂-y₁)²] = |y₂-y₁|

Using shortcuts reduces the chance of calculation errors and saves time.

Question: How does the distance formula connect to the Pythagorean theorem?

Answer: The distance formula is derived directly from the Pythagorean theorem! Here's the connection:

  • When you have two points on a coordinate plane, you can form a right triangle by drawing horizontal and vertical lines
  • The horizontal distance (x₂-x₁) forms one leg of the right triangle
  • The vertical distance (y₂-y₁) forms the other leg of the right triangle
  • The direct distance between the points is the hypotenuse of the right triangle

Applying the Pythagorean theorem: a² + b² = c²

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

This is exactly the distance formula! So the distance formula is just the Pythagorean theorem applied to coordinate geometry.

Question: How can I check if my distance calculation is reasonable?

Answer: Here are several ways to verify your distance calculation:

  1. Visual estimation: Look at the coordinate plane and estimate if your answer seems reasonable
  2. Triangle inequality: Each side of a triangle should be shorter than the sum of the other two sides
  3. Component comparison: The distance should be greater than or equal to the absolute difference in x-coordinates and y-coordinates
  4. Reverse calculation: Square your answer and see if it equals (x₂-x₁)² + (y₂-y₁)²
  5. Calculator verification: Use a calculator to double-check your arithmetic

For example, if points are at (1, 2) and (4, 6), the distance should be more than |4-1| = 3 and more than |6-2| = 4, which it is (√25 = 5).

Always consider whether your answer makes sense in the context of the problem.