Let the 100 points on the straight line be $P_1, P_2, \dots, P_{100}$ ordered from left to right, and WLOG they correspond to the coordinates $(x_1, 0), (x_2, 0), \dots, (x_{100}, 0)$. Suppose $M$ is the point outside the line. In order for some pair of points $(P_i, P_j)$ to form an isosceles triangle $\bigtriangleup{P_iMP_j}$, point $M$ must lie on the perpendicular bisector of $(P_i, P_j)$. Claim 1. Now if $\bigtriangleup{P_iMP_j}$ is isosceles for some pair $i, j \in \{1, 2, 3, \dots, 100\}$ where clearly $i \neq j$, then $\bigtriangleup{P_iMP_k}$ or $\bigtriangleup{P_jMP_l}$ will never be an isosceles if $j \neq k$ or $l \neq i$ respectively. Proof 1. If $M$ lies on the perpendicular bisector of $P_iP_j$ then the X-component of point M is $\frac{x_i + x_j}{2}$. Suppose by contradiction $M$ lies on the perpendicular bisector of $P_iP_k$ or $P_jP_m$ where $i, j, k, m$ are all pairwise distinct, then the X-component of point M would be $\frac{x_i + x_k}{2}$ or $\frac{x_j + x_m}{2}$ and both would equal $\frac{x_i + x_j}{2}$, a contradiction. So there can only be at most $\frac{100}{2} = 50$ isosceles triangles. A possible configuration would be if the 100 distinct points are evenly spaced, and $M$ lies on the perpendicular bisector of $P_{50}P_{51}$, which means $x_1, x_2, x_3, \dots, x_{100}$ form an arithmetic progression, so $\frac{x_i + x_{100-i}}{2}$ stays constant, hence the answer is 50.