Instructions
Welcome, brave adventurer! The Cartesian plane is a vast land of coordinates, and you must navigate it. The shortest distance between two points is a straight line, but to find its length, you must use the wisdom of the ancient Pythagorean Theorem: a² + b² = c².
For each problem, you will be given two points. Your task is to find the direct distance between them. You can do this by forming a right-angled triangle where the distance between the points is the hypotenuse (side 'c'). The lengths of the other two sides ('a' and 'b') will be the horizontal and vertical distances between the points.
Follow these steps for each path:
- Plot the two points on the grid provided.
- Draw a straight line connecting the points. This is your hypotenuse (c).
- Draw the horizontal (run) and vertical (rise) lines to form a right-angled triangle.
- Count the units for the rise (a) and the run (b).
- Use the Pythagorean Theorem to calculate the length of the hypotenuse (c).
- Write your answer in two ways: in its simplest radical form and rounded to one decimal place.
Part 1: Charting the Basic Paths
Example Path: From Point A (1, 2) to Point B (4, 6)
- Horizontal distance (run): 4 - 1 = 3 units. So, a = 3.
- Vertical distance (rise): 6 - 2 = 4 units. So, b = 4.
- a² + b² = c²
- 3² + 4² = c²
- 9 + 16 = c²
- 25 = c²
- c = √25 = 5
- Distance: 5 units.
1. Path from Point P (-2, 1) to Point Q (6, 7)
Use the grid below to plot your points and draw your triangle.
Horizontal Distance (a): __________
Vertical Distance (b): __________
Calculation:
Final Distance (simplest radical form): __________
Final Distance (rounded to one decimal place): __________
2. Path from Point R (5, -1) to Point S (-1, -5)
Use the grid below to plot your points and draw your triangle.
Horizontal Distance (a): __________
Vertical Distance (b): __________
Calculation:
Final Distance (simplest radical form): __________
Final Distance (rounded to one decimal place): __________
Part 2: The Explorer's Challenge
Read the scenarios and calculate the shortest possible distance ("as the crow flies"). Show your work.
3. The Hidden Treasure
An old map shows the location of a pirate's treasure chest. You are standing at the old oak tree, located at (-5, -3). The treasure is buried near the rocky outcrop at (2, 8). What is the shortest distance, in map units, you must travel to reach the treasure?
Work Space:
Final Distance (simplest radical form): __________
Final Distance (rounded to one decimal place): __________
4. The Drone Delivery
A delivery drone takes off from its warehouse at (-8, 9). It must fly in a straight line to deliver a package to a house located at (10, -2). How far does the drone have to fly?
Work Space:
Final Distance (simplest radical form): __________
Final Distance (rounded to one decimal place): __________
Part 3: Enrichment - A Multi-Point Journey
A hiker is planning a trip between three scenic viewpoints: The Ridge (A), The Basin (B), and The Pinnacle (C). Their coordinates on the park map are:
- A = (-6, 2)
- B = (2, -4)
- C = (5, 0)
5. First, calculate the distance of the path from The Ridge (A) to The Basin (B).
Distance A to B: __________ units
6. Next, calculate the distance of the path from The Basin (B) to The Pinnacle (C).
Distance B to C: __________ units
7. Challenge: When viewed on the map, does the turn the hiker makes at The Basin (B) form a perfect right angle? Use the Converse of the Pythagorean Theorem to justify your answer. (Hint: Calculate the distance from A to C and see if the three path lengths form a Pythagorean triple.)
Justification:
Answer Key
1. Path from Point P (-2, 1) to Point Q (6, 7)
- Horizontal Distance (a): 6 - (-2) = 8 units
- Vertical Distance (b): 7 - 1 = 6 units
- Calculation: 8² + 6² = c² => 64 + 36 = c² => 100 = c² => c = 10
- Final Distance (simplest radical form): 10
- Final Distance (rounded to one decimal place): 10.0
2. Path from Point R (5, -1) to Point S (-1, -5)
- Horizontal Distance (a): 5 - (-1) = 6 units
- Vertical Distance (b): -1 - (-5) = 4 units
- Calculation: 6² + 4² = c² => 36 + 16 = c² => 52 = c² => c = √52
- Final Distance (simplest radical form): √52 = √(4 * 13) = 2√13
- Final Distance (rounded to one decimal place): 7.2
3. The Hidden Treasure
- Points: (-5, -3) and (2, 8)
- Horizontal Distance (a): 2 - (-5) = 7 units
- Vertical Distance (b): 8 - (-3) = 11 units
- Calculation: 7² + 11² = c² => 49 + 121 = c² => 170 = c² => c = √170
- Final Distance (simplest radical form): √170
- Final Distance (rounded to one decimal place): 13.0
4. The Drone Delivery
- Points: (-8, 9) and (10, -2)
- Horizontal Distance (a): 10 - (-8) = 18 units
- Vertical Distance (b): 9 - (-2) = 11 units
- Calculation: 18² + 11² = c² => 324 + 121 = c² => 445 = c² => c = √445
- Final Distance (simplest radical form): √445
- Final Distance (rounded to one decimal place): 21.1
5. Distance A to B
- Points: A(-6, 2) and B(2, -4)
- Horizontal Distance: 2 - (-6) = 8
- Vertical Distance: 2 - (-4) = 6
- Calculation: 8² + 6² = 64 + 36 = 100. √100 = 10.
- Distance A to B: 10 units
6. Distance B to C
- Points: B(2, -4) and C(5, 0)
- Horizontal Distance: 5 - 2 = 3
- Vertical Distance: 0 - (-4) = 4
- Calculation: 3² + 4² = 9 + 16 = 25. √25 = 5.
- Distance B to C: 5 units
7. Challenge
- First, find the direct distance from A to C to serve as the potential hypotenuse. Points: A(-6, 2) and C(5, 0).
- Horizontal Distance: 5 - (-6) = 11
- Vertical Distance: 2 - 0 = 2
- Distance (AC)² = 11² + 2² = 121 + 4 = 125.
- Now, use the Converse of the Pythagorean Theorem with the three path lengths: (AB)², (BC)², and (AC)².
- Check if (AB)² + (BC)² = (AC)².
- 10² + 5² = (√125)²
- 100 + 25 = 125
- 125 = 125
- Justification: Yes, the path forms a perfect right angle at B. The squares of the lengths of the two shorter paths (A to B and B to C) sum to equal the square of the length of the direct path from A to C.
On the Assessment of a Scholar’s Mathematical Perspicacity
An Analytic Rubric for the Consideration of Tutors (Year Levels 8 through 12)
It is a truth universally acknowledged that a young mind in possession of a good problem must be in want of a method. The purpose of this document, therefore, is to provide the discerning educator with a proper framework by which to judge the merits of a student’s endeavours in the logical arts, as they pertain to the application of the venerable Pythagorean Theorem and its related Cartesian explorations, in accordance with the standards set forth by the Australian Curriculum.
An Exemplary Accomplishment (An 'A' Standard)
A student of this disposition demonstrates a command of the material so thorough and elegant as to be an object of admiration. Their calculations are executed with unimpeachable propriety, free from the slightest impropriety or error. The reasoning they present is not merely correct, but articulated with a clarity and precision that reveals a profound and intuitive grasp of geometric principles. Such a scholar not only arrives at the correct destination but can also recount every step of the journey with eloquence, justifying their chosen path with sound logic and, where appropriate, applying converse principles with commendable ingenuity. Their work is a model of intellectual refinement.
A Most Commendable Endeavour (A 'B' Standard)
Here we observe a student of sound sense and diligent application. The course of their logic is, for the most part, true and well-reasoned, leading them to a conclusion of considerable merit. They navigate the coordinate system with a steady hand, and their application of the theorem is both appropriate and effective. If a fault is to be found, it is but a trifling one—a minor slip in computation, or a moment's hesitation in simplifying a radical expression—which, though it may mar the final perfection, does not betray a fundamental want of understanding. Their arguments are solid, if not always of the most polished character.
A Sound, if Uninspired, Attempt (A 'C' Standard)
This student's performance is one of respectable adequacy. They possess a foundational knowledge of the required formulae and procedures and can apply them to situations of a familiar and straightforward nature. However, when faced with a problem of greater complexity or one that requires a measure of independent thought, their confidence may waver. The work may contain several small errors in calculation or a certain clumsiness in the setting out of their argument, suggesting that their acquaintance with the concepts is more mechanical than intimate. They have grasped the letter of the law, but not yet its spirit.
A Concerning Deficiency in Understanding (A 'D' Standard)
It is with some regret that one must observe the work of a student whose reasoning is yet in a state of disarray. Such a student may recognise the name of Pythagoras, and may even attempt to assemble the formula, but does so with a profound misunderstanding of its constituent parts or proper application. There may be significant and repeated errors in identifying the lengths of the triangle's legs or in the algebraic manipulation that follows. The path they trace upon the grid is often erratic, betraying a mind not yet capable of navigating the logical demands of the question. A great deal of careful instruction is henceforth required.