Two equal sides of an isosceles triangle are along $$−x + 2y = 4$$ and $$x + y = 4$$. If m is the slope of its third side, then the sum, of all possible distinct values of $$m$$, is:

The third side of an isosceles triangle makes equal angles with the two equal sides. If the slopes of the equal sides are $$m_1$$ and $$m_2$$, and the slope of the base is $$m$$, then:

$$\left| \frac{m - m_1}{1 + m \cdot m_1} \right| = \left| \frac{m - m_2}{1 + m \cdot m_2} \right|$$

o Line 1: $$-x + 2y = 4 \implies y = \frac{1}{2}x + 2$$. So, $$m_1 = \frac{1}{2}$$.

o Line 2: $$x + y = 4 \implies y = -x + 4$$. So, $$m_2 = -1$$.

$$\left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{m - (-1)}{1 + m(-1)} \right| \implies \left| \frac{2m - 1}{2 + m} \right| = \left| \frac{m + 1}{1 - m} \right|$$

o Case 1: $$\frac{2m - 1}{2 + m} = \frac{m + 1}{1 - m}$$

$$(2m - 1)(1 - m) = (m + 1)(2 + m) \implies -2m^2 + 3m - 1 = m^2 + 3m + 2$$

$$3m^2 = -3 \implies m^2 = -1$$ (No real solution).

o Case 2: $$\frac{2m - 1}{2 + m} = -\frac{m + 1}{1 - m}$$

$$(2m - 1)(1 - m) = -(m + 1)(2 + m) \implies -2m^2 + 3m - 1 = -(m^2 + 3m + 2)$$

$$-2m^2 + 3m - 1 = -m^2 - 3m - 2 \implies m^2 - 6m - 1 = 0$$

For a quadratic $$am^2 + bm + c = 0$$, the sum of roots is $$-b/a$$.

Sum $$= -(-6)/1 = \mathbf{6}$$.