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Question 3 : Let a, b, x, y be real numbers 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. k = \\frac{5}{13})
  4. 0 < k ≤ \\frac{5}{13})

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Explanatory Answer

Given equations are a2 + b2 = 25 and x2 + y2 = 169
We know 52 = 25 and 132 = 169
Multiply both equations to get (a2 + b2) (x2 + y2) = 25 x 169
(a2 + b2) (x2 + y2) = 6225
We know, 6225 = 652
We also know that ax + by = 65
So, numerically (Not algebraically), (a2 + b2) (x2 + y2) = (ax + by)2
(a2 + b2) (x2 + y2) = (ax + by)2
Expanding the equation,
a2 x2 + a2 y2 + b2 x2 + b2 y2 = a2 x2 + b2 y2 + 2axby
a2 y2 + b2 x2 = 2axby
a2 y2 + b2 x2 - 2axby = 0
This is of the form, (a-b)2
(ay-bx)2 = 0
ay - bx = 0
K = 0


The question is "Let a, b, x, y be real numbers such that a2 + b2 = 25 , x2 + y2 = 169 and ax + by = 65. If k = ay - bx, then "

Hence, the answer is k = 0

Choice A is the correct answer.

 

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