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(D) Let the equation of the line $L_1$ passing through the origin be $y = mx$.Substituting $y = mx$ into the circle equation $x^2 + y^2 - x + 3y = 0$:

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$x - y = 0$ and $x + 7y = 0$

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(D) Let the equation of the line $L_1$ passing through the origin be $y = mx$.Substituting $y = mx$ into the circle equation $x^2 + y^2 - x + 3y = 0$:$x^2 + m^2x^2 - x + 3mx = 0$$x[x(1 + m^2) - (1 - 3m)] = 0$The points of intersection are $(0, 0)$ and $(\frac{1 - 3m}{1 + m^2}, \frac{m(1 - 3m)}{1 + m^2})$.The length of the intercept $I_1$ is $\sqrt{(\frac{1 - 3m}{1 + m^2})^2 + (\frac{m(1 - 3m)}{1 + m^2})^2} = |\frac{1 - 3m}{1 + m^2}| \sqrt{1 + m^2} = \frac{|1 - 3m|}{\sqrt{1 + m^2}}$.For the line $L_2: x + y = 1$,substitute $y = 1 - x$ into the circle equation:$x^2 + (1 - x)^2 - x + 3(1 - x) = 0$$x^2 + x^2 - 2x + 1 - x + 3 - 3x = 0$$2x^2 - 6x + 4 = 0 \Rightarrow x^2 - 3x + 2 = 0$.The roots are $x = 1, 2$. Corresponding $y$ values are $y = 0, -1$.The length of the intercept $I_2$ is $\sqrt{(1 - 2)^2 + (0 - (-1))^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.Equating $I_1^2 = I_2^2$: $\frac{(1 - 3m)^2}{1 + m^2} = 2$.$(1 - 3m)^2 = 2(1 + m^2) \Rightarrow 1 - 6m + 9m^2 = 2 + 2m^2$.$7m^2 - 6m - 1 = 0 \Rightarrow (7m + 1)(m - 1) = 0$.So,$m = 1$ or $m = -1/7$.The lines are $y = x$ $(x - y = 0)$ and $y = -\frac{1}{7}x$ $(x + 7y = 0)$.

Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight line $x + y = 1$. If the intercepts made by the circle $x^2 + y^2 - x + 3y = 0$ on $L_1$ and $L_2$ are equal,then which of the following equations can represent $L_1$?

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