Question

the distance between point (p,-5) and (2,7) is 13 unit then the value of is

Answer 1

AB=sq.root (x2-x2)^2 + (y2-y1)^2

5=sq.root (2-p)^2 + (7+5)^2

square both the side

5^2=(2-p)^2 + (7+5)^2

25=2^2-2p+p^2 + 12^2

25=4-2p+p^2 +24

25=26-2p+p^2

-1=-2p+p^2

0=p^2-2p+1

0=p^2-1p-1p+1

0=p (p-1) -1 (p-1)

0=(p-1)(p-1)

p-1=0

p=1

Answer 2

Answer:

The attachment as the steps.

The value of p is -3, 7