Gradient of AB is 4 - 0 / 8 - 2 = 4/6 = 2/3
Gradient of PM is -1 divide by 2/3 = -3/2
Solution for (1):
The midpoint of AB , M, is (2,0) + 1/2(8 - 2, 4 - 0)
(2,0) + (3,2) = 5,2
The gradient of PM then is (5 - 2) / (3 - 5) = -3/2
Therefore AB is perpendicular to PM since gradient of two perpendicular bisectors is -1
Solution for (2):
AM^2 + PM^2 = PA^2 , PM^2 + MB^2 = PB^2
AM^2 + PM^2 must be equal to PM^2 + MB^2
and since PM passes through the middle of AB then;
PA^2 = PB^2
(Finding the square root)
PA = PB