Let L_1 be a line passing through the origin | vedclass

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

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$x - y = 0$

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(B) Let the equation of the line passing through the origin be $y = mx$.Substituting this into the circle equation $x^{2} + y^{2} - x + 3y = 0$:$x^{2} + (mx)^{2} - x + 3(mx) = 0$$x^{2}(1 + m^{2}) - x(1 - 3m) = 0$$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$$x^{2} + 1 - 2x + x^{2} - x + 3 - 3x = 0$$2x^{2} - 6x + 4 = 0 \implies x^{2} - 3x + 2 = 0$$(x - 1)(x - 2) = 0 \implies x = 1, 2$.The points are $(1, 0)$ and $(2, -1)$.The length of the intercept $I_2 = \sqrt{(2 - 1)^{2} + (-1 - 0)^{2}} = \sqrt{1^{2} + (-1)^{2}} = \sqrt{2}$.Given $I_1 = I_2$,so $\frac{(1 - 3m)^{2}}{1 + m^{2}} = 2$.$(1 - 3m)^{2} = 2(1 + m^{2}) \implies 1 - 6m + 9m^{2} = 2 + 2m^{2}$.$7m^{2} - 6m - 1 = 0 \implies (7m + 1)(m - 1) = 0$.$m = 1$ or $m = -\frac{1}{7}$.Thus,$y = x$ or $y = -\frac{1}{7}x$,which means $x - y = 0$ or $x + 7y = 0$.

Let $L_1$ be a line passing through the origin and $L_2$ be the 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 represents $L_1$?

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