The solution of $\sqrt{5}x - \sqrt{7}y = 0$ and $\sqrt{3}y + \sqrt{13}x = 0$ is :

1. x = 0, y = 0

2. x = 0, y = 1

3. x = 1, y = 0

4. x = 1, y = -1

Given,

Equations: $\sqrt{5}x - \sqrt{7}y = 0, \sqrt{3}y + \sqrt{13}x = 0$

Solving first equation,

$\Rightarrow \sqrt{5}x - \sqrt{7}y = 0 \\[1em] \Rightarrow \sqrt{5}x = \sqrt{7}y \\[1em] \Rightarrow x = \dfrac{\sqrt{7}}{\sqrt{5}}y \text{ ….(1)}$

Substituting value of x from equation (1) in $\sqrt{3}y + \sqrt{13}x = 0$, we get :

$\Rightarrow \sqrt{3}y + \sqrt{13}x = 0 \\[1em] \Rightarrow \sqrt{3}y + \sqrt{13} \times \dfrac{\sqrt{7}}{\sqrt{5}}y = 0 \\[1em] \Rightarrow y \Big(\sqrt{3} + \dfrac{\sqrt{91}}{\sqrt{5}}\Big) = 0.$

Since, $\Big(\sqrt{3} + \dfrac{\sqrt{91}}{\sqrt{5}}\Big)$ is not equal to zero thus, y = 0.

Substituting value of y = 0 in equation (1), we get :

$\Rightarrow x = \dfrac{\sqrt{7}}{\sqrt{5}}y = \dfrac{\sqrt{7}}{\sqrt{5}} \times 0$ = 0.

Hence, option 1 is the correct option.