1. Set up the geometry

Since $OAB$ is a right-angled triangle, let us take:

• $O$ at the right angle,
• $A$ on the $x$-axis with $OA=x_1$,
• $B$ on the $y$-axis with $OB=x_2$.

So the coordinates may be taken as: $O(0,0),\quad A(x_1,0),\quad B(0,x_2)$

The hypotenuse is $AB$.

2. Electric field at $O$ due to the charges

Let charge $Q_1$ be at $A$ and $Q_2$ be at $B$.

Magnitude of electric field at $O$ due to $Q_1$: $E_1 = \frac{k|Q_1|}{x_1^2}$ Direction is along $OA$.

Magnitude of electric field at $O$ due to $Q_2$: $E_2 = \frac{k|Q_2|}{x_2^2}$ Direction is along $OB$.

Since $OA \perp OB$, the resultant field has components along these two perpendicular directions.

Thus, taking components: $E_x \propto \frac{Q_1}{x_1^2}, \qquad E_y \propto \frac{Q_2}{x_2^2}$ (Signs depend on the nature of charges, but for proportionality we use the ratio condition.)

3. Condition for resultant to be perpendicular to hypotenuse

Slope of hypotenuse $AB$ is: $m_{AB} = \frac{x_2-0}{0-x_1} = -\frac{x_2}{x_1}$

A line perpendicular to $AB$ has slope: $m = \frac{x_1}{x_2}$

So the resultant electric field must satisfy: $\frac{E_y}{E_x} = \frac{x_1}{x_2}$

Now substitute the field components: $\frac{\dfrac{kQ_2}{x_2^2}}{\dfrac{kQ_1}{x_1^2}} = \frac{x_1}{x_2}$

Cancel $k$: $\frac{Q_2 x_1^2}{Q_1 x_2^2} = \frac{x_1}{x_2}$

Cross-multiplying: $Q_2 x_1^2 x_2 = Q_1 x_1 x_2^2$

Cancel $x_1x_2$: $Q_2 x_1 = Q_1 x_2$

Hence, $\frac{Q_1}{Q_2} = \frac{x_1}{x_2}$

4. Match with options

This corresponds to: $\boxed{\frac{x_1}{x_2}}$ So the correct option is C.

5. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.