Question

TS is the perpendicular bisector of AB with coordinate A (0,4) and B (p,6) and the point S lies on the x axis. If x coordinate of s is an integer then the no. of integral value of p

Answer 1

There are two integral Values of p     if TS ⊥ AB  with A (0,4) and B (p,6)  and S lies on x axis  and x coordinate of s is an integer

Step-by-step explanation:

Slope of line AB  =   ( 6 - 4) /(p - 0)

= 2/p

TS ⊥ AB

=> slope of TS   =  - p/2

T =  ( 0 + p)/2 , ( 4 + 6)/2

=> T  = p/2  ,  5

S = ( x , 0)  lies on x axis

=> Slope of TS  =   (5-0)/ ( p/2 - x)  

=  10/ ( p - 2x)

- p/2  = 10/ ( p - 2x)

= p (2x - p)  = 20

10 =  1 * 20   ,   2 * 10    , 4 * 5  

p = 1  =>  2x - p = 20  =>  2x - 1 = 20  => x = 21/2 ( not integral)

p = 2 => 2x - 2 = 10  =>  2x = 12 => x  = 6

p = 4  =>  2x - 4 = 5  => x  = 9/2  ( not integral)

p = 5  =>  2x - 5 = 4  => x  = 9/2  ( not integral)

p = 10   =>  2x - 10 = 2  =>  x  = 6

p =20  =>  2x - p = 1  =>  2x - 20 = 1  => x = 21/2 ( not integral)

Integral value of p  = 2 &  6

There are two integral Values of p

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