Question

Name a point that is between 50 and 60 units away from (7,-2) and state the distance between the two points.

Answer 1

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

Answer 2

Let the required point be (a,b)

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

=
`√((a-7)^2+(b+2)^2)`

But this distance needs to be betweem 50 & 60

So

`50<√((a-7)^2+(b+2)^2)<60`

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

=
`√((7-7)^2+(50+2)^2)= √((52)^2)=52`

This is between 50 & 60

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