Let a, b, x, y be real number such a² + b² = 25, x² + y² = 169, and ax + by = 65. If k = ay - bx, then Option 1: k = 0 Option 2: k = 5/13 Option 3: 0 5/13

Answer: k = 0. Solution: We have three equations with four unknowns, but there's a beautiful pattern here that we can exploit. When you see equations like this, always check if Cauchy-Schwarz inequality applies! The Cauchy-Schwarz inequality tells us: (ax + by)^2 ≤ (a^2 + b^2)(x^2 + y^2) Let's substitute our values: - Left side: (ax + by)^2 = 65^2 = 4225 - Right side: (a^2 + b^2)(x^2 + y^2) = 25 x 169 = 4225 They're equal! When Cauchy-Schwarz has equality, it means the vectors (a,b) and (x,y) are proportional. If vectors are proportional: a/x = b/y Cross-multiplying: ay = bx Therefore: ay - bx = 0 So k = 0

Algebra

Inequalities

Hard

Let $a, b, x, y$ be real number such $a² + b² = 25, x² + y² = 169$ , and $ax + by = 65.$ If $k = ay - bx,$ then

$k = 0$

$k = \frac{5}{13}$

$0<k≤\frac{5}{13}$

$k>\frac{5}{13}$

✅ Correct Option: 1

We have three equations with four unknowns, but there's a beautiful pattern here that we can exploit.

When you see equations like this, always check if Cauchy-Schwarz inequality applies!

The Cauchy-Schwarz inequality tells us:

$(ax + by)^2 \leq (a^2 + b^2)(x^2 + y^2)$

Let's substitute our values:

Left side: $(ax + by)^2 = 65^2 = 4225$
Right side: $(a^2 + b^2)(x^2 + y^2) = 25 \times 169 = 4225$

They're equal! When Cauchy-Schwarz has equality, it means the vectors $(a,b)$ and $(x,y)$ are proportional.

If vectors are proportional:

$\dfrac{a}{x} = \dfrac{b}{y}$

Cross-multiplying: $ay = bx$

Therefore: $ay - bx = 0$

So $k = 0$

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Let a,b,x,ya, b, x, ya,b,x,y be real number such a2+b2=25,x2+y2=169a² + b² = 25, x² + y² = 169a2+b2=25,x2+y2=169, and ax+by=65.ax + by = 65.ax+by=65. If k=ay−bx,k = ay - bx,k=ay−bx, then

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Let $a, b, x, y$ be real number such $a² + b² = 25, x² + y² = 169$ , and $ax + by = 65.$ If $k = ay - bx,$ then