Introduction:
We are given two points A(3,4) and B(7,13). We need to find a point P on the line y=x such that the product of the distances PA and PB is minimized. To solve this problem, we will use the concept of perpendicular distance.

Solution:

Step 1: Find the equation of the line y=x:
The equation of a line passing through the origin with a slope of 1 is y=x.

Step 2: Find the perpendicular distance between a point and a line:
The perpendicular distance between a point (x1, y1) and a line Ax + By + C = 0 is given by the formula:

d = |Ax1 + By1 + C| / sqrt(A^2 + B^2)

In our case, the equation of the line y=x can be rewritten as x - y = 0. So, A = 1, B = -1, and C = 0.

Step 3: Find the equation of the line passing through A and perpendicular to y=x:
The slope of the line y=x is 1. So, the slope of the line perpendicular to y=x is -1. We can use the point-slope form of a line to find the equation of the line passing through A:

y - 4 = -1(x - 3)
y - 4 = -x + 3
y = -x + 7

Step 4: Find the equation of the line passing through B and perpendicular to y=x:
Similarly, we can find the equation of the line passing through B:

y - 13 = -1(x - 7)
y - 13 = -x + 7
y = -x + 20

Step 5: Find the point of intersection of the two lines:
To find the point of intersection, we need to solve the two equations:

y = -x + 7
y = -x + 20

By solving these equations, we get x = 6 and y = 1.

Step 6: Conclusion:
The coordinates of the point P on the line y=x such that PA * PB is minimum are (6, 1).