$1 \cdot 66$ Given 5 points in a plane, no two lines connecting any two points coincide, are parallel, or are perpendicular. Through each point, draw perpendiculars to the lines connecting the other points. How many intersection points of these perpendiculars can there be, not counting the original 5 points, when the maximum number is considered?
[Solution] There are 10 lines connecting 5 given points pairwise. Every 3 points form a triangle, totaling $C_{5}^{3}=10$ triangles. Any 4 points can form 6 lines, and from the 5th point, 6 perpendiculars can be drawn to these 6 lines, thus there are a total of 30 such perpendiculars.

These 30 perpendiculars intersect pairwise, resulting in $C_{30}^{2}=435$ intersection points (including repeated and extra counts). For each of the 10 lines connecting the 5 points, the perpendiculars drawn from the other 3 points are parallel and do not intersect, so 30 non-existent intersection points must be subtracted from the total of 435. Additionally, the 3 altitudes of each of the 10 triangles are concurrent, so 20 extra intersection points from repeated counts must also be subtracted. Furthermore, since 6 perpendiculars are drawn from each of the 5 points, 75 extra intersection points from $5 C_{6}^{2}$ must also be subtracted. Thus, the remaining number of intersection points is
$$
435-30-20-75=310 \text {. }
$$

Clearly, the positions of the 5 points can be appropriately chosen so that the 310 intersection points are all distinct. Therefore, the maximum number of intersection points among these perpendiculars is 310.