Let the line L_1 : x + 3 = 0 intersect the li | vedclass

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(D) The line $L_1$ is $x = -3$.Intersection of $L_1$ and $L_2$ $(y = x)$ gives point $A(-3, -3)$.Intersection of $L_1$ and $L_3$ $(y = -3x)$ gives poi

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$3$ : $2$

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(D) The line $L_1$ is $x = -3$.Intersection of $L_1$ and $L_2$ $(y = x)$ gives point $A(-3, -3)$.Intersection of $L_1$ and $L_3$ $(y = -3x)$ gives point $B(-3, 9)$.The angle bisectors of $L_2$ $(x - y = 0)$ and $L_3$ $(3x + y = 0)$ are given by $\frac{x - y}{\sqrt{1^2 + (-1)^2}} = \pm \frac{3x + y}{\sqrt{3^2 + 1^2}}$, which simplifies to $\frac{x - y}{\sqrt{2}} = \pm \frac{3x + y}{\sqrt{10}}$, or $\sqrt{5}(x - y) = \pm (3x + y)$.Case $1$: $\sqrt{5}x - \sqrt{5}y = 3x + y \implies x(\sqrt{5} - 3) = y(1 + \sqrt{5}) \implies y = \frac{\sqrt{5} - 3}{\sqrt{5} + 1}x$.Case $2$: $\sqrt{5}x - \sqrt{5}y = -3x - y \implies x(\sqrt{5} + 3) = y(\sqrt{5} - 1) \implies y = \frac{\sqrt{5} + 3}{\sqrt{5} - 1}x$.Checking the obtuse angle bisector: The product of slopes $m_2 m_3 = (1)(-3) = -3$. Since $m_1 m_2 + 1 > 0$, the origin is in the acute angle. The obtuse bisector is the one that does not contain the origin. Substituting $x = -3$ into the bisector equations, we find point $C$. Calculating distances $AC$ and $BC$ and their ratio, we get $BC^2 : AC^2 = 3 : 2$.

Let the line $L_1 : x + 3 = 0$ intersect the lines $L_2 : x - y = 0$ and $L_3 : 3x + y = 0$ at the points $A$ and $B$, respectively. Let the bisector of the obtuse angle between the lines $L_2$ and $L_3$ intersect the line $L_1$ at the point $C$. Then $BC^2 : AC^2$ is equal to:

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