Charges $$Q_1$$ and $$Q_2$$ are at points A and B of a right-angled triangle OAB. The resultant electric field at point O is perpendicular to the hypotenuse, then $$Q_1/Q_2$$ is proportional to:

$$\text{Let } \angle OBA = \theta \implies \tan\theta = \frac{x_1}{x_2}$$

$$E_1 = \frac{Q_1}{4\pi\varepsilon_0 x_1^2} \quad (\text{along } -y)$$

$$E_2 = \frac{Q_2}{4\pi\varepsilon_0 x_2^2} \quad (\text{along } -x)$$

$$\text{Since } \vec{E} \perp AB\text{, the angle of } \vec{E} \text{ with the x-axis is } 90^\circ - \theta$$

$$\tan(90^\circ - \theta) = \frac{E_1}{E_2} \implies \cot\theta = \frac{E_1}{E_2}$$

$$\frac{x_2}{x_1} = \frac{\left(\frac{Q_1}{4\pi\varepsilon_0 x_1^2}\right)}{\left(\frac{Q_2}{4\pi\varepsilon_0 x_2^2}\right)} = \frac{Q_1}{Q_2} \times \frac{x_2^2}{x_1^2}$$

$$\frac{Q_1}{Q_2} = \frac{x_2}{x_1} \times \frac{x_1^2}{x_2^2} = \frac{x_1}{x_2}$$