How to Solve: Graph Distance Problems

How to Solve: Graph Distance Problems

Introduction

"Imagine you’re planning a road trip—how do you find the shortest route between two cities on a map? Or, in your exam, how do you calculate the exact distance between two points on a coordinate grid? Mastering graph distance problems unlocks real-world navigation AND easy marks on your geometry test!

What You Need To Know First

Before tackling graph distance problems, you must already understand: 1. Plotting points on a coordinate plane – How to read (x, y) coordinates and locate them on a grid. 2. Pythagoras’ Theorem – The formula a² + b² = c² for right-angled triangles. 3. Absolute value – The distance between numbers on a number line, regardless of direction.

If any of these are shaky, review them first—this guide assumes you’re solid on them.

Key Vocabulary

Formulas To Know

1. Distance Formula

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

Variables: - ( d ) = distance between two points - ( (x_1, y_1) ) = coordinates of the first point - ( (x_2, y_2) ) = coordinates of the second point

Memorise this? ✅ YES – It’s the backbone of graph distance problems.

2. Horizontal/Vertical Distance (Shortcut)

Formula: - Horizontal distance = ( |x_2 - x_1| ) (absolute value) - Vertical distance = ( |y_2 - y_1| )

When to use: If the points share the same x-coordinate (vertical line) or y-coordinate (horizontal line), skip the distance formula—just subtract!

Example: - Points (5, 2) and (5, 7) → Vertical distance = ( |7 - 2| = 5 ).

3. Midpoint Formula (Bonus for Some Exams)

Formula: [ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) ]

Memorise this? ⚠️ Check your exam sheet – Some exams provide it, others expect you to know it.

Step-by-Step Method

Follow these steps exactly for every graph distance problem.

1. Label the points – Write down ( (x_1, y_1) ) and ( (x_2, y_2) ). Circle them to avoid mix-ups.
2. Check for shortcuts – Are the points aligned horizontally or vertically? If yes, subtract the coordinates.
3. Plug into the distance formula – Substitute the coordinates into ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ).
4. Simplify inside the square root – Calculate ( (x_2 - x_1)^2 ) and ( (y_2 - y_1)^2 ) first.
5. Add the squares – Sum the two results from Step 4.
6. Take the square root – Simplify ( \sqrt{\text{sum}} ) to get the final distance.
7. Check units – If the problem mentions units (e.g., "units" or "km"), include them in your answer.

Worked Examples

Example 1 – Basic

Problem: Find the distance between ( A(3, 4) ) and ( B(7, 1) ).

Step-by-Step Solution: 1. Label: ( A(3, 4) ) → ( (x_1, y_1) ), ( B(7, 1) ) → ( (x_2, y_2) ). 2. No shortcut (points aren’t aligned horizontally/vertically). 3. Plug into formula:
[ d = \sqrt{(7 - 3)^2 + (1 - 4)^2} ] 4. Simplify inside the square root:
[ d = \sqrt{(4)^2 + (-3)^2} ]
[ d = \sqrt{16 + 9} ] 5. Add the squares:
[ d = \sqrt{25} ] 6. Take the square root:
[ d = 5 ]

Answer: The distance is 5 units.

What we did and why: We used the distance formula because the points weren’t aligned. The key was squaring the differences first, then adding, then taking the square root.

Example 2 – Medium (Negative Coordinates)

Problem: Find the distance between ( C(-2, -5) ) and ( D(4, 3) ).

Step-by-Step Solution: 1. Label: ( C(-2, -5) ) → ( (x_1, y_1) ), ( D(4, 3) ) → ( (x_2, y_2) ). 2. No shortcut. 3. Plug into formula:
[ d = \sqrt{(4 - (-2))^2 + (3 - (-5))^2} ]
[ d = \sqrt{(4 + 2)^2 + (3 + 5)^2} ] 4. Simplify inside the square root:
[ d = \sqrt{(6)^2 + (8)^2} ]
[ d = \sqrt{36 + 64} ] 5. Add the squares:
[ d = \sqrt{100} ] 6. Take the square root:
[ d = 10 ]

Answer: The distance is 10 units.

What we did and why: Negative coordinates can be tricky—remember to subtract a negative (which becomes addition). The rest follows the same steps as Example 1.

Example 3 – Exam Style (Disguised Problem)

Problem: A treasure map shows a treasure at ( (5, -2) ) and a ship at ( (-3, 4) ). How far is the ship from the treasure? Give your answer to 1 decimal place.

Step-by-Step Solution: 1. Label: Treasure ( (5, -2) ) → ( (x_1, y_1) ), Ship ( (-3, 4) ) → ( (x_2, y_2) ). 2. No shortcut. 3. Plug into formula:
[ d = \sqrt{(-3 - 5)^2 + (4 - (-2))^2} ]
[ d = \sqrt{(-8)^2 + (6)^2} ] 4. Simplify inside the square root:
[ d = \sqrt{64 + 36} ] 5. Add the squares:
[ d = \sqrt{100} ] 6. Take the square root:
[ d = 10.0 ] (already exact, but problem asks for 1 decimal place)

Answer: The ship is 10.0 units from the treasure.

What we did and why: The problem was "disguised" with a story, but it’s the same math. Always label points first, then apply the formula. The decimal place requirement was a red herring here—10 is already exact.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in—tonight or right before your exam. Here’s the no-fail plan for graph distance problems:

1. Label your points – ( (x_1, y_1) ) and ( (x_2, y_2) ). Circle them so you don’t mix them up.
2. Check for shortcuts – Same x-coordinate? Subtract the y’s. Same y-coordinate? Subtract the x’s. Done.
3. No shortcut? Use the distance formula: ( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Square the differences, add them, then take the square root.
4. Watch for negatives – Parentheses are your best friend. ( (-3 - 5) ) is ( -8 ), not ( 2 ).
5. Simplify fully – If you get ( \sqrt{50} ), write ( 5\sqrt{2} ). If the problem asks for a decimal, round at the very end.

That’s it. No tricks, no guesswork. Label, check, plug in, simplify. You’ve got this—now go ace that exam!