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:
Step-by-Step Solution
Key Concept: The third side makes equal angles with the two equal sides. Use the condition that the angle of the third side's slope with $-x+2y=4$ equals the angle with $x+y=4$; this gives two equations for the slope $m$.
Slopes of equal sides: $m_1=1/2$ ($-x+2y=4$), $m_2=-1$ ($x+y=4$).
The angle condition gives two cases for slope $m$ of the third side:
$\tan\theta = \dfrac{m-m_1}{1+mm_1} = \pm\dfrac{m-m_2}{1+mm_2}$.
Case 1 (+): simplifying gives one value.
Case 2 (−): gives another value.
Sum of all distinct values $= M_{L_1}+M_{L_2} = \left[-\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-2\sqrt{2}}\right]+\left[-\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+2\sqrt{2}}\right] = 6$.
Correct Answer: 3