We have 10 lines between these 5 points, and between any 4 points, there are 6 lines. Thus, from each point, we draw 6 perpendiculars, giving us 30 perpendiculars total. This gives us $\binom{30}{2}=435$ intersection points total. But clearly, we have overcounted. From every point, there are 6 lines emerging from it; we have counted $\binom{6}{2}=15$ intersections between these lines, when there is only 1. Thus we have an over count of 14 for each 5 points, giving us $435-70=365$ remaining intersections. We have also failed to account for parallel lines; each of the 10 lines has 3 perpendiculars which are parallel, giving an overcount of 30. We now have $365-30=335$ remaining points of intersection. Finally, we have failed to count when 3 perpendiculars (coming from distinct points) coincide. We know that this happens at each of the $\binom{5}{3}=10$ triangles formed from our original triangles, at their orthocenter. We count the intersection 3 times when there is only 1, giving an overcount of $20$, and reducing our maximum to $315$ points.