If p is the point (a,6) and its distance to the point (-3,2) is 5, determine the value of A

1. Start with the distance formula:

\sqrt{(a + 3)^2 + (6 - 2)^2} = 5

2. Simplify inside the square root:

\sqrt{(a + 3)^2 + 4^2} = 5

3. Simplify:

\sqrt{(a + 3)^2 + 16} = 5

4. Square both sides to eliminate the square root:

(a + 3)^2 + 16 = 25

5. Subtract 16 from both sides to isolate the square term:

(a + 3)^2 = 9

6. Solve for ( a ) by taking the square root of both sides:

a + 3 = \pm 3

7. This gives two potential values for ( a ):

a + 3 = 3 \quad \Rightarrow \quad a = 0

a + 3 = -3 \quad \Rightarrow \quad a = -6

8. Therefore, the solutions are:

a = 0 \text{ or } a = -6