if A be(2,2), B be(5,-2) find all points on x-axis such that AP is perpendicular to PB

let P is (x,0) then

slope of AP = 2/x-2

slope of PB = 5/-2-x

if lines are perpendicular then product of slopes = -1

(2/x-2) (-5/x+2) = -1

x² - 4 = 10

x² = 14

x = +root14 & -root14

points are , (root14,0)  &  (-root14,0)

The points are A(2,2) and B(5,-2)

It is given that the point P is in the x-axis therefore lets assume it to be P(x,0)

it is also given that AP is perpendicular to BP

therefore if M1   is the slope of AP and M₂    is the slope of BP then we know that

               M_{1 x}  M_{2  = -1}

therefore (2-0)/2-x)*(-2-0)/5-x)=-1

you will get a quadratic equation x²-7x+6=0 

solving you will get x=1 or x=6

so the point required is P(1,0) or P(6,0)

HOPE THAT YOU GOT WHAT YOU NEEDED

BEST OF LUCK FOR YOUR EXAMINATIONS