Two equal sides of an isosceles triangle are | vedclass

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(C) Let the slopes of the two equal sides be $m_1$ and $m_2$. From the equations $-x+2y=4$ and $x+y=4$,we get $m_1 = \frac{1}{2}$ and $m_2 = -1$. Let

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$6$

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(C) Let the slopes of the two equal sides be $m_1$ and $m_2$. From the equations $-x+2y=4$ and $x+y=4$,we get $m_1 = \frac{1}{2}$ and $m_2 = -1$. Let $m$ be the slope of the third side. Since the triangle is isosceles,the angle between the third side and the first side must be equal to the angle between the third side and the second side. Thus,$\left| \frac{m - m_1}{1 + m \cdot m_1} \right| = \left| \frac{m - m_2}{1 + m \cdot m_2} \right|$. Substituting the values,$\left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{m - (-1)}{1 + m(-1)} \right|$. $\left| \frac{2m - 1}{2 + m} \right| = \left| \frac{m + 1}{1 - m} \right|$. This gives two cases: Case $1$: $\frac{2m - 1}{2 + m} = \frac{m + 1}{1 - m} \implies (2m - 1)(1 - m) = (m + 1)(2 + m) \implies 2m - 2m^2 - 1 + m = 2m + m^2 + 2 + m \implies 3m^2 = -3 \implies m^2 = -1$ (No real solution). Case $2$: $\frac{2m - 1}{2 + m} = -\left( \frac{m + 1}{1 - m} \right) = \frac{m + 1}{m - 1} \implies (2m - 1)(m - 1) = (m + 1)(m + 2) \implies 2m^2 - 3m + 1 = m^2 + 3m + 2 \implies m^2 - 6m - 1 = 0$. The sum of the roots of this quadratic equation is given by $-\frac{b}{a} = -\frac{-6}{1} = 6$.

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:

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