Solved Exercises on Distance on the Coordinate Plane in Grade 7

Master distance calculations on the coordinate plane: distance formula, Pythagorean theorem, and geometric applications through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Distance Between Two Points
Exercise 1
Find the distance between points A(3, 4) and B(-1, 2) on the coordinate plane. Show your work using the distance formula.
Definition:

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

Distance calculation method:
  1. Identify coordinates: (x₁, y₁) and (x₂, y₂)
  2. Substitute values into the distance formula
  3. Calculate the differences and squares
  4. Take the square root of the sum
Points
A(3,4), B(-1,2)
Distance Formula
d = √[(x₂-x₁)² + (y₂-y₁)²]
Calculated Distance
d = √20 = 2√5
Step 1: Identify coordinates

A(3, 4): x₁ = 3, y₁ = 4

B(-1, 2): x₂ = -1, y₂ = 2

Step 2: Apply distance formula

d = √[(x₂-x₁)² + (y₂-y₁)²]

d = √[(-1-3)² + (2-4)²]

Step 3: Simplify inside brackets

d = √[(-4)² + (-2)²]

d = √[16 + 4]

Step 4: Complete the calculation

d = √20 = √(4×5) = 2√5 ≈ 4.47 units

Distance = 2√5 ≈ 4.47 units
Final answer:

The distance between points A(3, 4) and B(-1, 2) is 2√5 ≈ 4.47 units

Applied rules:

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

• Subtraction with negatives: Be careful with signs

• Squaring: Squaring eliminates negative signs

• Square root simplification: Factor out perfect squares when possible

2 Distance from a Point to the Origin
Exercise 2
Find the distance from point P(-4, 3) to the origin (0, 0). Then determine if point P is closer to the origin than point Q(2, -5).
Definition:

Distance from origin: A special case of the distance formula where one point is (0, 0). The distance from (x, y) to origin is: d = √(x² + y²)

Point P
P(-4, 3)
Point Q
Q(2, -5)
Distances
P: 5, Q: √29
Step 1: Calculate distance from P to origin

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

Step 2: Calculate distance from Q to origin

d_Q = √[(2)² + (-5)²] = √[4 + 25] = √29 ≈ 5.39 units

Step 3: Compare the distances

Since 5 < √29 (or 5 < 5.39), point P is closer to the origin than point Q

Step 4: Verify by comparing squared distances

To avoid square roots: compare 25 (P's squared distance) vs 29 (Q's squared distance)

P is 5 units from origin, Q is √29 ≈ 5.39 units from origin, so P is closer
Final answer:

Point P(-4, 3) is 5 units from the origin, point Q(2, -5) is √29 ≈ 5.39 units from the origin, so P is closer to the origin than Q.

Applied rules:

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

• Comparison shortcut: Compare squared distances instead of actual distances

• Pythagorean theorem: Distance formula is based on this theorem

3 Perimeter of a Triangle
Exercise 3
Triangle ABC has vertices A(1, 2), B(4, 6), and C(7, 2). Find the perimeter of triangle ABC by calculating the distance between each pair of vertices.
Definition:

Perimeter: The total distance around a geometric figure. For a triangle, perimeter = sum of all three side lengths.

Vertices
A(1,2), B(4,6), C(7,2)
Side Lengths
AB = 5, BC = 5, AC = 6
Perimeter
P = AB + BC + AC
Step 1: Calculate distance AB

A(1, 2) to B(4, 6): d_AB = √[(4-1)² + (6-2)²] = √[9 + 16] = √25 = 5 units

Step 2: Calculate distance BC

B(4, 6) to C(7, 2): d_BC = √[(7-4)² + (2-6)²] = √[9 + 16] = √25 = 5 units

Step 3: Calculate distance AC

A(1, 2) to C(7, 2): d_AC = √[(7-1)² + (2-2)²] = √[36 + 0] = √36 = 6 units

Step 4: Calculate the perimeter

Perimeter = AB + BC + AC = 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 definition: Sum of all side lengths

• Special case: When y-coordinates are equal, distance is simply |x₂-x₁|

Rules and methods, laws,...
d = √[(x₂-x₁)² + (y₂-y₁)²]
Distance Formula
Distance from Origin
d = √(x² + y²)
Special case of distance formula
Horizontal Distance
d = |x₂ - x₁|
When y-coordinates are equal
Vertical Distance
d = |y₂ - y₁|
When x-coordinates are equal
Pythagorean Theorem
a² + b² = c²
Base of distance formula
Key definitions:

Distance on coordinate plane: The straight-line measurement between two points (x₁, y₁) and (x₂, y₂)

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

Perimeter: The total distance around a geometric figure

Euclidean distance: The shortest path between two points in a plane

Complete methodology:
  1. Identify coordinates: Determine the x and y values of both points
  2. Subtract coordinates: Calculate (x₂-x₁) and (y₂-y₁)
  3. Square differences: Square both differences
  4. Sum squares: Add the squared differences
  5. Take square root: Find the square root of the sum
Tip 1: When comparing distances, you can compare squared distances to avoid square roots.
Tip 2: For horizontal lines (same y-coordinates), distance is |x₂ - x₁|.
Tip 3: For vertical lines (same x-coordinates), distance is |y₂ - y₁|.
Tip 4: Always double-check your subtraction to avoid sign errors.
Common errors: Forgetting to subtract coordinates correctly, making sign errors when subtracting negatives, forgetting to take the square root, mixing up x and y coordinates.
Exam preparation: Practice with various point combinations, memorize the distance formula, understand when to use shortcuts for horizontal/vertical distances, master Pythagorean theorem applications.
Solution: Exercises 4 to 5
4 Distance and Scale Applications
Exercise 4
On a map with a scale of 1 unit = 10 miles, two towns are located at T₁(2, 5) and T₂(8, 9). Find the actual distance between the towns in miles.
Definition:

Scale factor: A ratio that compares the measurements on a map to the actual measurements. Actual distance = Map distance × Scale factor.

Map Coordinates
T₁(2,5), T₂(8,9)
Scale Factor
1 unit = 10 miles
Actual Distance
72.1 miles
Step 1: Calculate map distance between towns

d_map = √[(8-2)² + (9-5)²] = √[36 + 16] = √52 = √(4×13) = 2√13 ≈ 7.21 units

Step 2: Apply the scale factor

Scale: 1 unit on map = 10 miles in reality

Actual distance = Map distance × Scale factor

Actual distance = 2√13 × 10 = 20√13 ≈ 72.1 miles

Step 3: Verify the calculation

2√13 ≈ 7.21, and 7.21 × 10 = 72.1 miles ✓

Step 4: Round appropriately

Since we're dealing with distance, round to reasonable precision: approximately 72.1 miles

Actual distance = 20√13 ≈ 72.1 miles
Final answer:

The actual distance between the towns is 20√13 ≈ 72.1 miles

Applied rules:

• Distance formula: Calculate map distance first

• Scale conversion: Multiply map distance by scale factor

• Unit conversion: Apply scale to convert to real-world units

5 Real-World Application Problem
Exercise 5
A robot starts at point R(0, 0) and needs to travel to point S(6, 8). If the robot can only move at 2 units per minute, how long will it take to reach point S?
Definition:

Time-Speed-Distance relationship: Time = Distance ÷ Speed. This is fundamental for motion problems.

Start Point
R(0, 0)
End Point
S(6, 8)
Speed
2 units/min
Step 1: Calculate the distance from R to S

d = √[(6-0)² + (8-0)²] = √[36 + 64] = √100 = 10 units

Step 2: Apply the time formula

Time = Distance ÷ Speed

Time = 10 units ÷ 2 units/minute = 5 minutes

Step 3: Verify the calculation

Robot travels 2 units per minute for 5 minutes = 10 units ✓

Step 4: State the final answer

The robot will take exactly 5 minutes to reach point S

Time = 5 minutes
Final answer:

The robot will take 5 minutes to travel from point R(0, 0) to point S(6, 8)

Applied rules:

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

• Time-Speed-Distance: Time = Distance ÷ Speed

• Unit consistency: Ensure units match in calculations

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

Coordinate plane distance: The Euclidean distance between two points (x₁, y₁) and (x₂, y₂)

Pythagorean theorem: The foundation of the distance formula: a² + b² = c²

Cartesian distance: The straight-line distance in a coordinate system

Scale factor: The ratio between map distance and actual distance

Euclidean metric: The standard way to measure distance in a plane

Complete methodology:
  1. Identify coordinates: Determine the exact coordinates of both points
  2. Apply distance formula: Substitute into d = √[(x₂-x₁)² + (y₂-y₁)²]
  3. Perform calculations: Carefully compute differences and squares
  4. Simplify: Reduce radicals when possible
  5. Apply context: Convert units or apply to real-world scenarios as needed
Tip 1: When working with real-world problems, always consider the units of measurement.
Tip 2: For problems involving time and speed, use the relationship: Time = Distance ÷ Speed.
Tip 3: When comparing distances, compare squared distances to avoid computing square roots.
Tip 4: Always verify your calculations by checking if the answer makes sense in the context of the problem.
Common errors: Sign errors in coordinate subtraction, incorrect application of the distance formula, unit conversion mistakes, arithmetic errors with radicals.
Exam preparation: Master the distance formula, practice with various coordinate combinations, understand real-world applications, work on speed and accuracy with calculations, learn shortcuts for special cases.
Distance Rules:

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

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

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

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

• Squared distance: d² = (x₂-x₁)² + (y₂-y₁)²

• Triangle inequality: For three points, sum of any two sides ≥ third side

Questions & Answers

Question: Why do we use the distance formula instead of just counting the spaces between points on the coordinate plane?

Answer: Great question! Here's why the distance formula is essential:

  • Precision: Counting spaces only works for integer coordinates and gives approximate results
  • Diagonal distances: Points rarely align horizontally or vertically, so counting doesn't work for diagonal distances
  • Real-world applications: We often deal with decimal coordinates that require precise calculations
  • Mathematical rigor: The distance formula provides exact values based on the Pythagorean theorem

Example: For points (1, 2) and (4, 6), counting doesn't work directly. Using the formula: d = √[(4-1)² + (6-2)²] = √[9+16] = √25 = 5 units.

Question: How do I simplify square roots when finding distances? Sometimes I get expressions like √50 or √72.

Answer: Here's the systematic approach to simplify square roots:

  1. Factor out perfect squares: Find the largest perfect square that divides the number
  2. Separate the square root: √(a×b) = √a × √b
  3. Simplify: Take the square root of the perfect square

Examples:

  • √50 = √(25×2) = √25 × √2 = 5√2
  • √72 = √(36×2) = √36 × √2 = 6√2
  • √20 = √(4×5) = √4 × √5 = 2√5

Memorize perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81, 100) to make factoring easier!

Question: When solving problems with speed and time, how do I know which values to use in the distance formula versus the time formula?

Answer: Here's how to organize these multi-step problems:

  1. First: Use the distance formula to find the distance between coordinates
  2. Then: Apply the time-speed-distance relationship: Time = Distance ÷ Speed
  3. Units: Make sure all units are consistent (time, distance, speed)

Example: Robot moving from (0,0) to (3,4) at 1 unit/minute

  • Step 1: Distance = √[(3-0)² + (4-0)²] = √[9+16] = √25 = 5 units
  • Step 2: Time = Distance ÷ Speed = 5 units ÷ 1 unit/minute = 5 minutes

The distance formula finds the spatial distance; the time formula uses that distance to find temporal duration.