Question

Let a, b, x, y be real number such that a² + b² = 25, x² + y² = 169, and ax + by = 65. If k = ay - bx, then 
1. k = 0
2. \(k = \frac{5}{13}\)
3. \(0< k \le \frac{5}{13}\)
4. \(k > \frac{5}{13}\)

Answer 1

Correct Answer - Option 1 : k = 0

Calculation:

a2 + b2 = 25      ----(1)

x2 + y2 = 169      ----(2)

ax + by = 65      ----(3)

k = ay - bx      ----(4)

Multiplying by equation (1) and equation (2):

(a2 + b2)(x2 + y2) = 169 × 25

⇒ (a2 + b2)(x2 + y2) = 6225

⇒ (a2x2 + a2y2 + b2x2 + b2y² )= 65²

By equation (3), we have ax + by = 65

⇒ (a2x2 + a2y2 + b2x2 + b2y2 )= (ax + by)2

⇒ (a2x2 + a2y2 + b2x2 + b2y2 )= a2x2 + b2y2 + 2axby

⇒ (a2y2 + b2x2 – 2axby) = 0

⇒ (ay - bx)² = 0

⇒ (ay - bx) = 0

By equation (4), we have (ay - bx) = k

⇒ k = 0

∴ The value of k is 0.