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Distance Between Two Points

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Introduction

The Distance Between Two Points pattern teaches how to compute the straight-line distance (displacement) between two positions given their north-south and east-west movements or coordinates. This is fundamental in aptitude tests because many route and displacement problems reduce to finding the shortest (straight-line) distance.

Mastering this pattern helps solve map, navigation, and route-tracing questions quickly using coordinate thinking and the Pythagorean theorem.

Pattern: Distance Between Two Points

Pattern: Distance Between Two Points

Key concept: Treat movements as coordinate changes on perpendicular axes. The straight-line distance = √(Δx² + Δy²), where Δx and Δy are the net east-west and north-south displacements respectively.

Steps to use:

  1. Convert movements into signed coordinates (East/ North positive, West/ South negative - or choose any consistent sign convention).
  2. Sum components separately to get net Δx (east-west) and Δy (north-south).
  3. Apply Pythagoras: distance = √(Δx² + Δy²).
  4. Report both magnitude and direction (quadrant) using the signs of Δx and Δy.

Step-by-Step Example

Question

A person starts at point O, walks 6 km North, then 8 km East. What is his straight-line distance from O and in which direction?

Solution

  1. Step 1: Identify coordinate changes

    Taking O as (0,0): move 6 km North → Δy = +6; then 8 km East → Δx = +8.
  2. Step 2: Compute net displacements

    Net Δx = +8 (East), Net Δy = +6 (North).
  3. Step 3: Apply Pythagoras theorem

    Distance = √(Δx² + Δy²) = √(8² + 6²) = √(64 + 36) = √100 = 10 km.
  4. Step 4: Determine direction

    Δx > 0 and Δy > 0 → first quadrant → direction = North-East (specifically arctan(6/8) = 36.87° north of east).
  5. Final Answer:

    10 km North-East
  6. Quick Check:

    Recognize 6-8-10 as a Pythagorean triple (3-4-5 scaled by 2) → distance = 10 km ✅

Quick Variations

1. Movements include returns (e.g., North then South) - cancel opposite components first.

2. Use coordinates directly: given points (x₁,y₁) and (x₂,y₂), compute √((x₂-x₁)² + (y₂-y₁)²).

3. Mixed units or directions (km/m) - convert to same units before computing.

4. Angle asked: compute direction using tan⁻¹(|Δy/Δx|) and assign quadrant from signs.

Trick to Always Use

  • Step 1: Always write net Δx and Δy with signs (East = +, West = -).
  • Step 2: Cancel opposite movements before applying the formula.
  • Step 3: Look for known Pythagorean triples (3-4-5, 5-12-13, etc.) to speed computation.
  • Step 4: For direction, use θ = arctan(|Δy/Δx|) and adjust quadrant using sign of Δx, Δy.

Summary

  • Treat east-west and north-south movements as perpendicular components.
  • Always compute net displacement using √(Δx² + Δy²).
  • Cancel opposite directions before applying the distance formula.
  • Use direction ratios or tangent to specify angle when required.

Example to remember:
6 km North and 8 km East → Distance = 10 km North-East.

Practice

(1/5)
1. A person walks 3 km North and then 4 km East. What is his straight-line distance from the starting point and in which direction?
easy
A. 5 km North-East
B. 1 km East
C. 7 km North-East
D. √13 km North-East

Solution

  1. Step 1: Record component displacements

    North (Δy) = 3 km; East (Δx) = 4 km.
  2. Step 2: Apply distance formula

    Distance = √(Δx² + Δy²) = √(4² + 3²) = √(16 + 9) = √25 = 5 km.
  3. Step 3: Determine direction

    Δx > 0 and Δy > 0 → first quadrant → North-East.
  4. Final Answer:

    5 km North-East → Option A
  5. Quick Check:

    Recognise 3-4-5 Pythagorean triple → distance 5 km ✅
Hint: Look for 3-4-5 or other Pythagorean triples to get the distance quickly.
Common Mistakes: Adding component distances instead of using √(Δx²+Δy²).
2. From a point O a person walks 6 km North and then 8 km East. How far is he from O and what is the direction?
easy
A. 12 km North-East
B. 10 km North-East
C. 14 km East
D. √20 km North-East

Solution

  1. Step 1: Note component displacements

    Δy = 6 km (North), Δx = 8 km (East).
  2. Step 2: Compute squared components

    Δx² = 8² = 64; Δy² = 6² = 36.
  3. Step 3: Sum and root

    Distance = √(64 + 36) = √100 = 10 km.
  4. Step 4: Direction

    Both components positive → North-East.
  5. Final Answer:

    10 km North-East → Option B
  6. Quick Check:

    6-8-10 is a scaled 3-4-5 triple → 10 km ✅
Hint: Multiply or recognise common triples (3-4-5 scaled) to avoid heavy arithmetic.
Common Mistakes: Swapping Δx and Δy when stating direction (say East-North instead of North-East).
3. A person goes 8 m West, then 15 m South, then 8 m East. How far and in which direction is he from his starting point?
easy
A. 15 m South
B. 23 m South-West
C. √(8²+15²) m South-East
D. 7 m South

Solution

  1. Step 1: Compute net components

    Horizontal: West 8 + East 8 → they cancel → Δx = 0. Vertical: South 15 → Δy = -15 (South).
  2. Step 2: Distance

    Distance = √(Δx² + Δy²) = √(0² + (-15)²) = √225 = 15 m.
  3. Step 3: Direction

    Δx = 0, Δy < 0 → directly South.
  4. Final Answer:

    15 m South → Option A
  5. Quick Check:

    Horizontal cancel → only 15 m South remains ✅
Hint: Cancel equal opposite horizontal/vertical moves first to simplify result.
Common Mistakes: Using diagonal formula when one component is zero.
4. A runner goes 7 m South and then 24 m East. What is his straight-line distance from the start and which quadrant is he in?
medium
A. 25 m South-West
B. 31 m South-East
C. 25 m South-East
D. 17 m South-East

Solution

  1. Step 1: Components

    Δy = -7 m (South taken negative), Δx = +24 m (East positive).
  2. Step 2: Squares

    Δx² = 24² = 576. Δy² = 7² = 49.
  3. Step 3: Distance

    Distance = √(576 + 49) = √625 = 25 m.
  4. Step 4: Direction/quadrant

    Δx > 0 and Δy < 0 → East and South → South-East quadrant.
  5. Final Answer:

    25 m South-East → Option C
  6. Quick Check:

    7-24-25 is a Pythagorean triple → 25 m ✅
Hint: Watch signs: positive Δx, negative Δy → South-East quadrant.
Common Mistakes: Reporting South-West instead of South-East by ignoring sign of Δx.
5. Starting at P, a person walks 9 km East and then 40 km North. What is his straight-line distance from P?
medium
A. 41 km South-East
B. 31 km North-East
C. 41 km North-West
D. 41 km North-East

Solution

  1. Step 1: Record components

    Δx = +9 km (East), Δy = +40 km (North).
  2. Step 2: Square components

    Δx² = 9² = 81. Δy² = 40² = 1600.
  3. Step 3: Distance

    Distance = √(81 + 1600) = √1681 = 41 km.
  4. Step 4: Direction

    Both positive → North-East.
  5. Final Answer:

    41 km North-East → Option D
  6. Quick Check:

    9-40-41 is a Pythagorean triple → 41 km ✅
Hint: Memorise 9-40-41 and other larger triples to spot distances fast.
Common Mistakes: Forgetting to square before adding or mixing up signs for direction.