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 $\left\{0\right\}$

Given $a^2+b^2=81\cdots \left(1\right)$ and $x^2+y^2=121\cdots \left(2\right)$

$\left(1\right)\times \left(2\right)⟹(a^2+b^2)(x^2+y^2)=81\times 121$

$⟹a^2{x}^2+a^2{y}^2+b^2{x}^2+b^2{y}^2=9\times 9\times 11\times 11$

$⟹a^2{x}^2+a^2{y}^2+b^2{x}^2+b^2{y}^2={99}^2\cdots \left(3\right)$

and also given $ax+by=99$

$⟹(ax+by{)}^2={99}^2$

$⟹a^2{x}^2+b^2{y}^2+2abxy={99}^2\cdots \left(4\right)$

$\left(3\right)-\left(4\right)⟹a^2{y}^2+b^2{x}^2-2abxy=0$

$⟹(ay{)}^2+(bx{)}^2-2\left(ay\right)\left(bx\right)=0$

$⟹(ay-bx{)}^2=0$

$⟹ay-bx=0$