Question

Let $a,b,x,y$ be real numbers such that $a^2+b^2=81,x^2+y^2=121$ and $ax+by=99$. Then the set of all possible values of $ay-bx$ is:

• A. $(0,\frac{9}{11}]$
• B. $(0,\frac{9}{11})$
• C. $\left\{0\right\}$
• D. $[\frac{9}{11},\infty )$

Answer 1

The correct option is C $\left\{0\right\}$

$a^2+b^2=\mathrm{81....}\left(i\right)x^2+y^2=\mathrm{121...}\left(ii\right)ax+by=\mathrm{99....}\left(iii\right)$

Multiplying $\left(i\right)$ and $\left(ii\right)$

$(a^2+b^2)(x^2+y^2)=81\times 121a^2{x}^2+a^2{y}^2+b^2{x}^2+b^2{y}^2=81\times \mathrm{121....}\left(iv\right)$

Squaring $\left(iii\right)$

$a^2{x}^2+a^2{y}^2+2abxy=99\times \mathrm{99......}\left(v\right)$

Subtracting $\left(v\right)$ from $\left(iv\right)$

$a^2{y}^2+b^2{x}^2-2abxy=0{(ay-bx)}^2=0ay-bx=0$

So option $C$ is correct