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 represented as \( (0, y) \).

### 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 will set the distances from \( (0, y) \) to \( (2, 3) \) and \( (-4, 1) \) equal to each other.

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

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

### Step 5: Set the distances equal
Now we set \( d_1 \) equal to \( d_2 \):
\[ \sqrt{4 + (y - 3)^2} = \sqrt{16 + (y - 1)^2} \]

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

### Step 7: Expand both sides
Expanding both sides:
\[ 4 + (y^2 - 6y + 9) = 16 + (y^2 - 2y + 1) \]
This simplifies to:
\[ y^2 - 6y + 13 = y^2 - 2y + 17 \]

### Step 8: Cancel \( y^2 \) and simplify
Subtract \( y^2 \) from both sides:
\[ -6y + 13 = -2y + 17 \]
Rearranging gives:
\[ -6y + 2y = 17 - 13 \]
\[ -4y = 4 \]

### Step 9: Solve for \( y \)
Dividing both sides by -4:
\[ y = -1 \]

### Step 10: State the point
Thus, the point on the y-axis that is equidistant from (2, 3) and (-4, 1) is:
\[ (0, -1) \]