What is the point on y - axis which is equidistant from the points (2,3) and (-4,1)?

Options

A. (0,-1)
B. (0,1)
C. (0,2)
D. (0,-2)

Answer

To find the point on the y-axis that is equidistant from the points (2, 3) and (-4, 1), we can follow these steps:

### Step 1: Define the point on the y-axis
Let the point on the y-axis be \( P(0, y) \), where \( y \) is the y-coordinate we need to find.

**Hint:** Remember that any point on the y-axis has an x-coordinate of 0.

### Step 2: Use the distance formula
The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
We need to find the distances from point \( P(0, y) \) to the points \( (2, 3) \) and \( (-4, 1) \).

### Step 3: Calculate the distance from \( P(0, y) \) to \( (2, 3) \)
Using the distance formula:
\[ d_1 = \sqrt{(0 - 2)^2 + (y - 3)^2} = \sqrt{4 + (y - 3)^2} \]

### Step 4: Calculate the distance from \( P(0, y) \) to \( (-4, 1) \)
Using the distance formula:
\[ d_2 = \sqrt{(0 + 4)^2 + (y - 1)^2} = \sqrt{16 + (y - 1)^2} \]

### Step 5: Set the distances equal
Since the point \( P(0, y) \) is equidistant from both points, we set the distances equal:
\[ \sqrt{4 + (y - 3)^2} = \sqrt{16 + (y - 1)^2} \]

### Step 6: Square both sides to eliminate the square roots
\[ 4 + (y - 3)^2 = 16 + (y - 1)^2 \]

### Step 7: Expand both sides
Expanding the left side:
\[ 4 + (y^2 - 6y + 9) = y^2 - 6y + 13 \]
Expanding the right side:
\[ 16 + (y^2 - 2y + 1) = y^2 - 2y + 17 \]

### Step 8: Set the expanded equations equal
\[ y^2 - 6y + 13 = y^2 - 2y + 17 \]

### Step 9: Simplify the equation
Subtract \( y^2 \) from both sides:
\[ -6y + 13 = -2y + 17 \]
Rearranging gives:
\[ -6y + 2y = 17 - 13 \]
\[ -4y = 4 \]

### Step 10: Solve for \( y \)
Dividing both sides by -4:
\[ y = -1 \]

### Step 11: Write the coordinates of the point
Since \( x = 0 \) and \( y = -1 \), the coordinates of the point are:
\[ (0, -1) \]

### Final Answer
The point on the y-axis which is equidistant from the points (2, 3) and (-4, 1) is \( (0, -1) \).

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