a. To find the possible coordinates of point A, we know that the x-coordinate of point A is 5 and the distance between point A and point B is 15 units. Since the x-coordinate of point B is -4, we can calculate the difference between the x-coordinates of point A and point B:
x-coordinate of point A - x-coordinate of point B = 5 - (-4) = 9
So, the possible x-coordinate of point A is 9.
Now, we can use the distance formula to find the possible y-coordinate of point A. The distance between point A (5, y) and point B (-4, -2) is given as 15 units:
√[(x2 - x1)^2 + (y2 - y1)^2] = 15
Plugging in the coordinates of the points:
√[(5 - (-4))^2 + (y - (-2))^2] = 15
Simplifying:
√[9^2 + (y + 2)^2] = 15
Squaring both sides:
81 + (y + 2)^2 = 225
Solving for (y + 2)^2:
(y + 2)^2 = 225 - 81
(y + 2)^2 = 144
Taking the square root:
y + 2 = ±12
Solving for y:
y = -2 ± 12
This gives us two possible y-coordinates:
y = -2 + 12 = 10
y = -2 - 12 = -14
Therefore, the possible coordinates of point A are (9, 10) and (9, -14).
b. If point B were moved to (-7, -2), we follow the same process as above.
The x-coordinate of point A is 5, and the x-coordinate of point B is -7. The difference in x-coordinates is:
x-coordinate of point A - x-coordinate of point B = 5 - (-7) = 12
So, the possible x-coordinate of point A is 12.
Using the distance formula, we have:
√[(12 - (-7))^2 + (y - (-2))^2] = 15
Simplifying:
√[19^2 + (y + 2)^2] = 15
Squaring both sides:
361 + (y + 2)^2 = 225
Solving for (y + 2)^2:
(y + 2)^2 = 225 - 361
(y + 2)^2 = -136
Since the square of a real number cannot be negative, there are no real solutions for (y + 2)^2. Therefore, there are no possible coordinates for point A if point B is moved to (-7, -2).
HOPE THIS HELP :)