Find the locus of a point which is equidistance from the points (1, 0) and (-1, 0).

Answer

To find the locus of a point that is equidistant from the points (1, 0) and (-1, 0), we can follow these steps:

### Step 1: Define the Points
Let the points be:
- A = (1, 0)
- B = (-1, 0)

Let the point P = (x, y) be the point whose locus we want to find.

### Step 2: Set Up the Distance Equations
Since point P is equidistant from points A and B, we can set up the equation based on the distance formula. The distance from P to A (denoted as PA) and the distance from P to B (denoted as PB) should be equal:

\[ PA = PB \]

Using the distance formula, we have:

\[ PA = \sqrt{(x - 1)^2 + (y - 0)^2} \]
\[ PB = \sqrt{(x + 1)^2 + (y - 0)^2} \]

### Step 3: Set the Distances Equal
Now, we can set these two distances equal to each other:

\[ \sqrt{(x - 1)^2 + y^2} = \sqrt{(x + 1)^2 + y^2} \]

### Step 4: Square Both Sides
To eliminate the square roots, we square both sides:

\[ (x - 1)^2 + y^2 = (x + 1)^2 + y^2 \]

### Step 5: Simplify the Equation
Now, we can simplify the equation. The \(y^2\) terms on both sides cancel out:

\[ (x - 1)^2 = (x + 1)^2 \]

Expanding both sides:

\[ x^2 - 2x + 1 = x^2 + 2x + 1 \]

### Step 6: Cancel Common Terms
Now, we can cancel \(x^2\) and \(1\) from both sides:

\[ -2x = 2x \]

### Step 7: Solve for x
Now, we can solve for \(x\):

\[ -2x - 2x = 0 \]
\[ -4x = 0 \]
\[ x = 0 \]

### Step 8: Conclusion
Since \(x = 0\), the locus of point P is a vertical line along the y-axis. Thus, the locus is:

\[ x = 0 \]

This means that the locus of the point which is equidistant from the points (1, 0) and (-1, 0) is the y-axis.

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