Given:
- Point A has coordinates (3,4)
- Point B has coordinates (5,-2)
- Point P is such that PA = PB
- The area of triangle PAB is 10 square units

To find:
The coordinates of point P

Solution:
Step 1: Find the midpoint of line segment AB
- Midpoint formula: The coordinates of the midpoint M(x,y) between two points A(x1,y1) and B(x2,y2) is given by:
- x = (x1 + x2) / 2
- y = (y1 + y2) / 2

Using the given coordinates, the midpoint of AB can be calculated as follows:
- x = (3 + 5) / 2 = 8 / 2 = 4
- y = (4 + (-2)) / 2 = 2 / 2 = 1

So, the midpoint of AB is M(4,1).

Step 2: Find the equation of the perpendicular bisector of line segment AB passing through point M.
- The slope of line AB can be calculated using the formula:
- slope (m) = (y2 - y1) / (x2 - x1)

Using the given coordinates, the slope of AB is:
- m = (-2 - 4) / (5 - 3) = -6 / 2 = -3

- The slope of a line perpendicular to AB is the negative reciprocal of the slope of AB. So, the slope of the perpendicular bisector is:
- m_perpendicular = -1 / m = -1 / (-3) = 1/3

- The equation of a line with slope m and passing through point (x1,y1) is given by the point-slope form:
- y - y1 = m(x - x1)

Using the midpoint coordinates (4,1) and the slope of the perpendicular bisector (1/3), the equation of the line can be written as:
- y - 1 = (1/3)(x - 4)
- 3(y - 1) = x - 4
- 3y - 3 = x - 4
- x - 3y + 1 = 0

So, the equation of the perpendicular bisector is x - 3y + 1 = 0.

Step 3: Find the coordinates of point P
- The point P lies on the perpendicular bisector of AB and is equidistant from points A and B.
- The distance between a point (x1,y1) and a line Ax + By + C = 0 is given by the formula:
- distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2)

Using the equation of the perpendicular bisector (x - 3y + 1 = 0) and the coordinates of point A (3,4), the distance between P and A can be calculated as follows:
- distance_PA = |3(3) + (-3)(4) + 1| / sqrt(3^2 + (-3)^2)
- distance_PA = |-3 + (-12) + 1|