Question

Point B has coordinates ​(3​,2​). The​ x-coordinate of point A is negative 9. The distance between point A and point B is 15 units. What are the possible coordinates of point​ A?

Answer 1

Answer:

Step-by-step explanation:The distance between points is given by the following equation:

 d = root ((x2-x1) ^ 2 + (y2-y1) ^ 2)

 Substituting we have:

 15 = root ((- 9 -3) ^ 2 + (a-2) ^ 2)

 We clear the value of a:

 (15) ^ 2 = (root ((- 12) ^ 2 + (a-2) ^ 2)) ^ 2

 225 = (- 12) ^ 2 + (a-2) ^ 2)

 Rewriting:

 225 = 144 + (a-2) ^ 2

 225-144 = (a-2) ^ 2

 81 = (a-2) ^ 2

 a-2 = +/- root (81)

 a = +/- 9 + 2

 The possible values are:

 a1 = 9 + 2 = 11

 a2 = -9 + 2 = -7

 Then, the possible coordinates of point a are:

 (-11, 11)

 (-11, -7)

 Answer:

 the possible coordinates of point a are:

 (-11, 11)

 (-11, -7)

Answer 2

Answer:

(-9,9) or (-9,-7) are the possible coordinates

Step-by-step explanation:

We can use the formula for the distance between two points to get this

Mathematically, this can be;

d = √(x2-x1)^2 + (y2-y1)^2

Now let our point A be (x1,y1) = (-9,n)

let’s say y1 is n for now

For point B, we have (x2,y2) = (3,2)

and our d is 15 units

Inputing the values, we have

15^2 = (3+9)^2 + (2-n)^2

225 = 144 + (2-n)^2

225-144 = (2-n)^2

(2-n)^2 = 81

(2-n) = √(81)

2-n = -9

or 2-n = 9

n = 11 or n = -7

Now the possible coordinate values of point A are;

(-9,9) or (-9,-7)