Question

$$\begin{gathered} \frac{x}{a}=\frac{y}{b}, \\ ax+by=\left(a^2+b^2\right). \end{gathered}$$

Answer 1

The given equations may be written as:
$\frac{x}{a}-\frac{y}{b}=0$ ...(i)
ax + by − (a² + b²) = 0 ...(ii)
Here, a₁ = $\frac{1}{a}$, b₁ = $\frac{-1}{b}$, c₁ = 0, a₂ = a, b₂ = b and c₂ = −(a² + b²)
By cross multiplication, we have:

∴ $\frac{x}{\left[\left(-{\displaystyle \frac{1}{b}}\right)\times \left(-\left(a^2+b^2\right)\right)-0\right]}=\frac{y}{\left[0-{\displaystyle \frac{1}{a}}\times \left(-\left(a^2+b^2\right)\right)\right]}=\frac{1}{\left[{\displaystyle \frac{b}{a}}+{\displaystyle \frac{a}{b}}\right]}$
⇒ $\frac{x}{\left({\displaystyle \frac{a^2}{b}}+{\displaystyle \frac{b^2}{b}}\right)}=\frac{y}{\left({\displaystyle \frac{a^2}{a}}+{\displaystyle \frac{b^2}{a}}\right)}=\frac{1}{{\displaystyle \frac{b^2+a^2}{ab}}}$
⇒ $\frac{x}{\left[{\displaystyle \frac{\left(a^2+b^2\right)}{b}}\right]}=\frac{y}{\left[{\displaystyle \frac{\left(a^2+b^2\right)}{a}}\right]}=\frac{1}{\left[{\displaystyle \frac{\left(a^2+b^2\right)}{ab}}\right]}$
⇒ $x=\frac{\left[{\displaystyle \frac{\left(a^2+b^2\right)}{b}}\right]}{\left[{\displaystyle \frac{\left(a^2+b^2\right)}{ab}}\right]}=a,y=\frac{\left[{\displaystyle \frac{\left(a^2+b^2\right)}{a}}\right]}{\left[{\displaystyle \frac{\left(a^2+b^2\right)}{ab}}\right]}=b$
Hence, x = a and y = b is the required solution.