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Question 3 : Let a, b, x, y be real numbers such that a^{2} + b^{2} = 25 , x^{2} + y^{2} = 169 and ax + by = 65. If k = ay - bx, then

- k = 0
- k > \\frac{5}{13})
- k = \\frac{5}{13})
- 0 < k ≤ \\frac{5}{13})

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Given equations are a^{2} + b^{2} = 25 and x^{2} + y^{2} = 169

We know 5^{2 }= 25 and 13^{2} = 169

Multiply both equations to get (a^{2} + b^{2}) (x^{2} + y^{2}) = 25 x 169

(a^{2} + b^{2}) (x^{2} + y^{2}) = 6225

We know, 6225 = 65^{2}

We also know that ax + by = 65

So, numerically (Not algebraically), (a^{2} + b^{2}) (x^{2} + y^{2}) = (ax + by)^{2}

(a^{2} + b^{2}) (x^{2} + y^{2}) = (ax + by)^{2}

Expanding the equation,

a^{2} x^{2 }+ a^{2} y^{2 }+ b^{2} x^{2 }+ b^{2} y^{2 }= a^{2} x^{2} + b^{2} y^{2 }+ 2axby

a^{2} y^{2 }+ b^{2} x^{2 }= 2axby

a^{2} y^{2 }+ b^{2} x^{2 }- 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 a ^{2} + b^{2} = 25 , x^{2} + y^{2} = 169 and ax + by = 65. If k = ay - bx, then " **

**Choice A** is the correct answer.

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