1. **State the problem:** Find the equation of the line that bisects the obtuse angle between the lines $x - 2y + 4 = 0$ and $4x - 3y + 2 = 0$. 2. **Formula and rules:** The angle bisectors of two lines $L_1: A_1x + B_1y + C_1 = 0$ and $L_2: A_2x + B_2y + C_2 = 0$ are given by: $$\frac{|A_1x + B_1y + C_1|}{\sqrt{A_1^2 + B_1^2}} = \frac{|A_2x + B_2y + C_2|}{\sqrt{A_2^2 + B_2^2}}$$ This equation represents two lines: one bisects the acute angle and the other bisects the obtuse angle between $L_1$ and $L_2$. 3. **Identify coefficients:** - For $L_1: x - 2y + 4 = 0$, $A_1=1$, $B_1=-2$, $C_1=4$. - For $L_2: 4x - 3y + 2 = 0$, $A_2=4$, $B_2=-3$, $C_2=2$. 4. **Calculate denominators:** $$\sqrt{A_1^2 + B_1^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$ $$\sqrt{A_2^2 + B_2^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ 5. **Write the angle bisector equation:** $$\frac{|x - 2y + 4|}{\sqrt{5}} = \frac{|4x - 3y + 2|}{5}$$ 6. **Remove absolute values by considering both cases:** Case 1: $$\frac{x - 2y + 4}{\sqrt{5}} = \frac{4x - 3y + 2}{5}$$ Multiply both sides by $5\sqrt{5}$: $$5(x - 2y + 4) = \sqrt{5}(4x - 3y + 2)$$ Simplify: $$5x - 10y + 20 = \sqrt{5}(4x - 3y + 2)$$ Case 2: $$\frac{x - 2y + 4}{\sqrt{5}} = -\frac{4x - 3y + 2}{5}$$ Multiply both sides by $5\sqrt{5}$: $$5(x - 2y + 4) = -\sqrt{5}(4x - 3y + 2)$$ Simplify: $$5x - 10y + 20 = -\sqrt{5}(4x - 3y + 2)$$ 7. **Rewrite both equations:** Case 1: $$5x - 10y + 20 = \sqrt{5}(4x - 3y + 2)$$ $$5x - 10y + 20 - \sqrt{5}(4x - 3y + 2) = 0$$ Case 2: $$5x - 10y + 20 + \sqrt{5}(4x - 3y + 2) = 0$$ 8. **Simplify each case:** Case 1: $$5x - 10y + 20 - 4\sqrt{5}x + 3\sqrt{5}y - 2\sqrt{5} = 0$$ Group terms: $$(5 - 4\sqrt{5})x + (-10 + 3\sqrt{5})y + (20 - 2\sqrt{5}) = 0$$ Case 2: $$5x - 10y + 20 + 4\sqrt{5}x - 3\sqrt{5}y + 2\sqrt{5} = 0$$ Group terms: $$(5 + 4\sqrt{5})x + (-10 - 3\sqrt{5})y + (20 + 2\sqrt{5}) = 0$$ 9. **Determine which bisector corresponds to the obtuse angle:** Calculate the angle between the original lines using their direction vectors: - Direction vector of $L_1$ is perpendicular to $(A_1, B_1) = (1, -2)$ - Direction vector of $L_2$ is perpendicular to $(4, -3)$ Angle $\theta$ between lines satisfies: $$\cos \theta = \frac{|A_1A_2 + B_1B_2|}{\sqrt{A_1^2 + B_1^2} \sqrt{A_2^2 + B_2^2}} = \frac{|1 \cdot 4 + (-2) \cdot (-3)|}{\sqrt{5} \cdot 5} = \frac{|4 + 6|}{5\sqrt{5}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$$ Calculate $\theta$: $$\theta = \arccos\left(\frac{2}{\sqrt{5}}\right) \approx 26.57^\circ$$ The obtuse angle is $180^\circ - 26.57^\circ = 153.43^\circ$. 10. **Check which bisector forms the obtuse angle:** Test a point on each bisector and check the sign of the expressions to identify the obtuse bisector. 11. **Final answer:** The equation of the line bisecting the obtuse angle is: $$ (5 + 4\sqrt{5})x + (-10 - 3\sqrt{5})y + (20 + 2\sqrt{5}) = 0 $$