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03 Jan 2021, 00:30
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A
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Difficulty:
85%
(hard)
Question Stats:
56% (03:00) correct
44%
(03:04)
wrong
based on 91
sessions
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If a, b, x, and y are real numbers such that \(a^2\) + \(b^2\) = 25 , \(x^2\) + \(y^2\) = 169, and ax + by = 65. If k = ay - bx, then find the value of k.
A. k = 0
B. k = 1
C. k > 513
D. k = 513
E. 0 < k ≤ 513
A. k = 0
B. k = 1
C. k > 513
D. k = 513
E. 0 < k ≤ 513
03 Jan 2021, 00:55
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\(a^2\) + \(b^2\) = 25
\(x^2\) + \(y^2\) = 169
Multiply both equations, we get
\(a^2x^2+a^2y^2+b^2x^2+b^2y^2=5^2*13^2\)......(1)
ax + by = 65
squaring above equation, we get
\(a^2x^2+b^2y^2+2axby = 5^2*13^2\)........(2)
Subtract (1) and (2)
\(a^2y^2+b^2x^2- 2axby= 0\)
\((ay-bx)^2 = 0\)
\(ax-by=0\)
\(x^2\) + \(y^2\) = 169
Multiply both equations, we get
\(a^2x^2+a^2y^2+b^2x^2+b^2y^2=5^2*13^2\)......(1)
ax + by = 65
squaring above equation, we get
\(a^2x^2+b^2y^2+2axby = 5^2*13^2\)........(2)
Subtract (1) and (2)
\(a^2y^2+b^2x^2- 2axby= 0\)
\((ay-bx)^2 = 0\)
\(ax-by=0\)
DisciplinedPrep
General Discussion
08 Jun 2024, 07:34
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@DisciplinedPrepGiven: If a, b, x, and y are real numbers such that \(a^2\) + \(b^2\) = 25 , \(x^2\) + \(y^2\) = 169, and ax + by = 65.
Asked: If k = ay - bx, then find the value of k.
a^2 + b^2 = 25 = 5^2
x^2 + y^2 = 169 = 13^2
ax + by = 65:
k = ay - bx
(ax + by)^2 = a^2x^2 + b^2y^2 + 2abxy = 65^2 = 5^2*13^2
(a^2 + b^2)(x^2 + y^2) = a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2 = 5^2*13^2
b^2x^2 + a^2y^2 - 2abxy = 0
(ay - bx)^2 = 0
k = ay - bx = 0
IMO A
Asked: If k = ay - bx, then find the value of k.
a^2 + b^2 = 25 = 5^2
x^2 + y^2 = 169 = 13^2
ax + by = 65:
k = ay - bx
(ax + by)^2 = a^2x^2 + b^2y^2 + 2abxy = 65^2 = 5^2*13^2
(a^2 + b^2)(x^2 + y^2) = a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2 = 5^2*13^2
b^2x^2 + a^2y^2 - 2abxy = 0
(ay - bx)^2 = 0
k = ay - bx = 0
IMO A
