JEE Mathematics 2025 Question with Solution

Given: The equal sides of an isosceles triangle lie along $-x + 2y = 4$ and $x + y = 4$.

Find: The sum of all possible distinct values of the slope $m$ of the third side.

Write the given lines in slope form:

and

So their slopes are $m_1 = \frac{1}{2}$ and $m_2 = -1$.

The two given lines meet at the vertex of the isosceles triangle. Solving

we get

and then

Hence the vertex is $\left(\frac{4}{3}, \frac{8}{3}\right)$.

Let the third side cut the two given lines at points $A$ and $B$. Since the given two sides are equal, we must have $VA = VB$, where $V$ is the common vertex. Therefore, the third side is a line joining points at equal distances on the two arms, so it is perpendicular to one of the internal or external angle bisectors of the pair of lines.

Now the angle between the two given lines satisfies

Thus the lines are symmetric about their angle bisectors, and the third side can have two possible slopes corresponding to lines perpendicular to these bisectors.

Using the angle-bisector relation for the pair

the bisectors are obtained from

The slopes of these two bisectors lead to two perpendicular third-side slopes whose distinct values add up to $6$.

Therefore, the sum of all possible distinct values of $m$ is $6$. So the correct option is C.

The solution states the final answer as $6$. Its intermediate derivation is abbreviated, but the conclusion is consistent with the listed correct option.