Respuesta :

TheHero
At first question unclear, that's what I understood is that you wanna  a  point that is far away from other by a distance from 50 to 60 

Let D be the distance between the two points , other Point (x,y)

D = sqrt[ (x-7)^2 + (y+2)^2)   => (1)

the distance is between 50 , 60:

50 < D < 60

From (1);

50 < sqrt((x-7)^2 + (y+2)^2) < 60   

Pick any x : let x = 7

50  < |y+2|< 60     (solve for y)

48 < y < 58

-48 > y > -58

Take any point in the range:

(7, 49) << This point will give you the distance between the two point is in range 50 to 60 

Let the required point be (a,b)

The distance of (a,b) from (7,-2) is

= [tex]\sqrt{(a-7)^2+(b+2)^2}[/tex]

But this distance needs to be betweem 50 & 60

So

[tex]50<\sqrt{(a-7)^2+(b+2)^2}<60[/tex]

Squaring all sides

2500 < (a-7)² + (b+2)² < 3600

Let a = 7

So we have

2500 < (b+2)² <3600

b+2 < 60    or  b+2 > -60   => b <58 or b > -62

Also

b+2 >50    or b + 2 < -50  => b >48 or B < -52

Let us take one value of b < 58 say b = 50

So now we have the point as (7, 50)

The other point is (7,-2)

Distance between them

= [tex]\sqrt{(7-7)^2+(50+2)^2}= \sqrt{(52)^2}=52[/tex]

This is between 50 & 60

Hence one point which has a distance between 50 & 60 from the point (7,-2) is (7, 50)


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