A point P (2, -1) is equidistant from the points (a,7) and (-3,a). Find a.

Answer

To find the value of \( a \) such that the point \( P(2, -1) \) is equidistant from the points \( A(a, 7) \) and \( B(-3, a) \), we will use the distance formula.

### Step-by-Step Solution:

1. **Set Up the Distance Equations**:
Since point \( P \) is equidistant from points \( A \) and \( B \), we can write:
\[ PA = PB \]
Using the distance formula, we have:
\[ PA = \sqrt{(a - 2)^2 + (7 - (-1))^2} \]
\[ PB = \sqrt{(-3 - 2)^2 + (a - (-1))^2} \]

2. **Simplify the Distances**:
First, simplify \( PA \):
\[ PA = \sqrt{(a - 2)^2 + (7 + 1)^2} = \sqrt{(a - 2)^2 + 8^2} = \sqrt{(a - 2)^2 + 64} \]
Now simplify \( PB \):
\[ PB = \sqrt{(-3 - 2)^2 + (a + 1)^2} = \sqrt{(-5)^2 + (a + 1)^2} = \sqrt{25 + (a + 1)^2} \]

3. **Set the Distances Equal**:
Now we set the two expressions equal to each other:
\[ \sqrt{(a - 2)^2 + 64} = \sqrt{25 + (a + 1)^2} \]

4. **Square Both Sides**:
To eliminate the square roots, square both sides:
\[ (a - 2)^2 + 64 = 25 + (a + 1)^2 \]

5. **Expand Both Sides**:
Expanding both sides gives:
\[ (a^2 - 4a + 4) + 64 = 25 + (a^2 + 2a + 1) \]
Simplifying further:
\[ a^2 - 4a + 68 = a^2 + 2a + 26 \]

6. **Combine Like Terms**:
Subtract \( a^2 \) from both sides:
\[ -4a + 68 = 2a + 26 \]
Rearranging gives:
\[ -4a - 2a = 26 - 68 \]
\[ -6a = -42 \]

7. **Solve for \( a \)**:
Divide by -6:
\[ a = \frac{42}{6} = 7 \]

### Final Answer:
The value of \( a \) is \( 7 \).

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