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

### Related Formula Angle theta$\theta$ between two lines with slopes m_1$m_1$ and m_2$m_2$: tantheta = left| fracm_1 - m_21 + m_1 m_2 right|$$\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ ### Core Logic Given lines for the equal sides: 1) -x + 2y = 4 implies y = frac12x + 2 implies m_1 = frac12$-x + 2y = 4 \implies y = \frac{1}{2}x + 2 \implies m_1 = \frac{1}{2}$ 2) x + y = 4 implies y = -x + 4 implies m_2 = -1$x + y = 4 \implies y = -x + 4 \implies m_2 = -1$ In an isosceles triangle, the third side makes equal angles theta$\theta$ with both equal sides. Let the slope of the third side be m$m$: left| fracm - 1/21 + m/2 right| = left| fracm - (-1)1 + m(-1) right|$$\left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{m - (-1)}{1 + m(-1)} \right|$$ left| frac2m - 12 + m right| = left| fracm + 11 - m right|$$\left| \frac{2m - 1}{2 + m} \right| = \left| \frac{m + 1}{1 - m} \right|$$ ### Step 1: Solve the Slope Equation Case 1 (Same sign): frac2m - 12 + m = fracm + 11 - m$$\frac{2m - 1}{2 + m} = \frac{m + 1}{1 - m}$$ (2m - 1)(1 - m) = (m + 1)(2 + m)$$(2m - 1)(1 - m) = (m + 1)(2 + m)$$ 2m - 2m^2 - 1 + m = m^2 + 3m + 2$$2m - 2m^2 - 1 + m = m^2 + 3m + 2$$ -2m^2 + 3m - 1 = m^2 + 3m + 2$$-2m^2 + 3m - 1 = m^2 + 3m + 2$$ 3m^2 + 3 = 0 implies m^2 = -1 quad (textNo real roots)$$3m^2 + 3 = 0 \implies m^2 = -1 \quad (\text{No real roots})$$ Case 2 (Opposite sign): frac2m - 12 + m = -fracm + 11 - m = fracm + 1m - 1$$\frac{2m - 1}{2 + m} = -\frac{m + 1}{1 - m} = \frac{m + 1}{m - 1}$$ (2m - 1)(m - 1) = (2 + m)(m + 1)$$(2m - 1)(m - 1) = (2 + m)(m + 1)$$ 2m^2 - 3m + 1 = m^2 + 3m + 2$$2m^2 - 3m + 1 = m^2 + 3m + 2$$ m^2 - 6m - 1 = 0$$m^2 - 6m - 1 = 0$$ ### Step 2: Sum of Roots The quadratic equation for m$m$ is m^2 - 6m - 1 = 0$m^2 - 6m - 1 = 0$. The sum of possible distinct values of m$m$ is given by the sum of roots of this quadratic: textSum of roots = -frac-61 = 6$$\text{Sum of roots} = -\frac{-6}{1} = 6$$ ### Pattern Recognition Instead of solving for the explicit values of the slopes (which involve radicals), using Vieta's relations directly on the quadratic equation m^2 - 6m - 1 = 0$m^2 - 6m - 1 = 0$ gives the final answer instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines