Find the point on the x-axis which is equidistant from `(2,\ -5)\ and\ (-2,\ 9)`

Options

A. `(-7,0)`
B. `(7,0)`
C. `(-2,0)`
D. `(-9,0)`

Explanation

To find the point on the x-axis that is equidistant from the points (2, -5) and (-2, 9), we can follow these steps:

### Step 1: Define the Point on the X-axis
Let the point on the x-axis be \( C(h, 0) \), where \( h \) is the x-coordinate we need to find.

### Step 2: Calculate the Distance from Point C to Point A (2, -5)
The distance \( AC \) from point \( C(h, 0) \) to point \( A(2, -5) \) can be calculated using the distance formula:
\[ AC = \sqrt{(h - 2)^2 + (0 + 5)^2} \]
This simplifies to:
\[ AC = \sqrt{(h - 2)^2 + 25} \]

### Step 3: Calculate the Distance from Point C to Point B (-2, 9)
Similarly, the distance \( BC \) from point \( C(h, 0) \) to point \( B(-2, 9) \) is:
\[ BC = \sqrt{(h + 2)^2 + (0 - 9)^2} \]
This simplifies to:
\[ BC = \sqrt{(h + 2)^2 + 81} \]

### Step 4: Set the Distances Equal
Since point \( C \) is equidistant from points \( A \) and \( B \), we can set the two distances equal:
\[ \sqrt{(h - 2)^2 + 25} = \sqrt{(h + 2)^2 + 81} \]

### Step 5: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (h - 2)^2 + 25 = (h + 2)^2 + 81 \]

### Step 6: Expand Both Sides
Expanding both sides gives:
\[ (h^2 - 4h + 4) + 25 = (h^2 + 4h + 4) + 81 \]
This simplifies to:
\[ h^2 - 4h + 29 = h^2 + 4h + 85 \]

### Step 7: Rearrange the Equation
Now, we can rearrange the equation to isolate \( h \):
\[ -4h - 4h + 29 - 85 = 0 \]
This simplifies to:
\[ -8h - 56 = 0 \]

### Step 8: Solve for h
Now, solving for \( h \):
\[ -8h = 56 \implies h = -7 \]

### Step 9: Write the Coordinates of Point C
Thus, the coordinates of point \( C \) are:
\[ C(-7, 0) \]

### Final Answer
The point on the x-axis which is equidistant from the points (2, -5) and (-2, 9) is \( (-7, 0) \).

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