There are five points in a plane. From each point, perpendicular are drawn to the lines joining the other points. What is the maximum numbers of points of intersection of these perpendiculars.

Answer

To solve the problem of finding the maximum number of points of intersection of perpendiculars drawn from five points in a plane to the lines joining the other points, we can follow these steps:

### Step 1: Understanding the Setup
We have five points in a plane, which we can label as A, B, C, D, and E. From each point, we will draw perpendiculars to the lines formed by the other points.

### Step 2: Calculate the Number of Lines
From any three points, we can form a line. The number of ways to choose 3 points from 5 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of points and \( r \) is the number of points to choose.

\[ \text{Number of lines} = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 3: Drawing Perpendiculars
From each of the 5 points, we can draw perpendiculars to each of the lines formed by the other points. Each point can draw perpendiculars to the lines formed by the other 4 points taken 2 at a time.

The number of ways to choose 2 points from 4 (the remaining points after choosing one) is:

\[ \text{Number of lines from one point} = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

Since there are 5 points, the total number of perpendiculars drawn is:

\[ \text{Total perpendiculars} = 5 \times 6 = 30 \]

### Step 4: Points of Intersection
To find the maximum number of intersection points of these perpendiculars, we need to consider how many pairs of perpendiculars can intersect. Each pair of perpendiculars can intersect at most once.

The number of ways to choose 2 perpendiculars from 30 is given by:

\[ \text{Number of intersection points} = \binom{30}{2} = \frac{30 \times 29}{2} = 435 \]

### Step 5: Adjusting for Overlaps
However, we must consider that some of these intersection points may coincide, particularly if they are drawn from the same set of points. We need to account for the orthocenters of the triangles formed by the points.

Each set of 3 points can form an orthocenter, and there are \( \binom{5}{3} = 10 \) such triangles. Each triangle contributes one unique intersection point (the orthocenter).

### Final Calculation
Thus, the total number of unique intersection points is:

\[ \text{Total unique intersection points} = 435 - 10 = 425 \]

### Conclusion
Therefore, the maximum number of points of intersection of these perpendiculars is:

\[ \boxed{425} \]