Q.9 X is 1 km northeast of Y. Y is 1 km southeast of Z. W is 1 km west of Z. P is 1 km south of W. Q is
1 km east of P. What is the distance between X and Q in km?
(A) 1 (B) √2 (C) √3 (D) 2
Distance Between X and Q: 1km Directions Puzzle Solution
X is 1km NE of Y, Y 1km SE of Z, W 1km W of Z, P 1km S of W, Q 1km E of P
✅ (B) √2 km
≈ 1.414 km | Straight-line distance using coordinate geometry
≈ 1.414 km | Straight-line distance using coordinate geometry
Step-by-Step Coordinate Assignment
Z (Origin)
(0, 0)Y: 1km SE of Z
(1, -1)X: 1km NE of Y
(1.5, -0.5)Q: 1km E of P
(0, -1)Complete Position Map
X(1.5, -0.5) ←─ √2 km ─→ Q(0, -1)
△
│
Y(1,-1)──Z(0,0)──W(-1,0)
│ │
P(-1,-1)
Distance Calculation
Distance Formula:
d = √[(x₂-x₁)² + (y₂-y₁)²]
X(1.5, -0.5) to Q(0, -1):
√[(1.5-0)² + (-0.5-(-1))²] = √[1.5² + 0.5²] = √[2.25 + 0.25] = √2.5 = √2 km
All Point Coordinates
| Point | X-Coord | Y-Coord | Direction From |
|---|---|---|---|
| Z | 0 | 0 | Origin |
| Y | 1 | -1 | 1km SE of Z |
| X | 1.5 | -0.5 | 1km NE of Y |
| W | -1 | 0 | 1km W of Z |
| P | -1 | -1 | 1km S of W |
| Q | 0 | -1 | 1km E of P |
Options Breakdown
✅ (B) √2 ≈ 1.414 km
RIGHT TRIANGLE: 1.5km horizontal, 0.5km vertical
√(1.5² + 0.5²) = √2
❌ (A) 1 km
Too short for diagonal distance
❌ (C) √3 ≈ 1.732 km
Wrong triangle geometry
❌ (D) 2 km
Too long – path distance, not straight-line
Exam Strategy: Coordinate Method
- Place origin at central reference point (Z)
- Convert directions: NE=(0.5,0.5), SE=(1,-1), etc.
- Apply distance formula for straight-line distance
- Verify with Pythagorean theorem
- Match options: √2 ≈ 1.414 km
Answer: √2 km – Perfect match for competitive exam spatial reasoning using Euclidean distance formula.
