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

```
9 years ago

```							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)
```
9 years ago
```							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
M1 x  M2  = -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

```
9 years ago
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