Angle Bisectors of Two Lines

In an isosceles triangle, the angle bisector of the two equal sides is perpendicular to the third side (the base). Therefore, the slope of the third side ($$m$$) will be perpendicular to the slopes of the angle bisectors.
Let the given lines be $$L_1: -x + 2y - 4 = 0$$ and $$L_2: x + y - 4 = 0$$.
The equations of the angle bisectors are given by:
$$\frac{-x + 2y - 4}{\sqrt{(-1)^2 + 2^2}} = \pm \frac{x + y - 4}{\sqrt{1^2 + 1^2}}$$
$$\frac{-x + 2y - 4}{\sqrt{5}} = \pm \frac{x + y - 4}{\sqrt{2}}$$
Cross-multiplying and rearranging gives two bisector lines with slopes $$m_{b1}$$ and $$m_{b2}$$.
Since the third side is perpendicular to these bisectors, calculating its possible slopes yields:
$$m = 3 + \sqrt{10} \quad \text{and} \quad m = 3 - \sqrt{10}$$
Sum of possible values:
$$\text{Sum} = (3 + \sqrt{10}) + (3 - \sqrt{10}) = 6$$