Find the point on y-axis which is equidistant from the points (5, - 2) and (- 3, 2).

Answer

To find the point on the y-axis that is equidistant from the points (5, -2) and (-3, 2), we can follow these steps:

### Step 1: Define the Point on the Y-axis
A point on the y-axis can be represented as (0, y), where y is the y-coordinate we need to find.

### Step 2: Use the Distance Formula
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step 3: Set Up the Equation for Distances
We need to find the distances from the point (0, y) to the points (5, -2) and (-3, 2) and set them equal to each other since the point is equidistant from both.

1. Distance from (0, y) to (5, -2):
\[ d_1 = \sqrt{(5 - 0)^2 + (-2 - y)^2} = \sqrt{25 + (y + 2)^2} \]

2. Distance from (0, y) to (-3, 2):
\[ d_2 = \sqrt{(-3 - 0)^2 + (2 - y)^2} = \sqrt{9 + (2 - y)^2} \]

### Step 4: Set the Distances Equal
Set \(d_1\) equal to \(d_2\):
\[ \sqrt{25 + (y + 2)^2} = \sqrt{9 + (2 - y)^2} \]

### Step 5: Square Both Sides
To eliminate the square roots, square both sides:
\[ 25 + (y + 2)^2 = 9 + (2 - y)^2 \]

### Step 6: Expand Both Sides
Expand the squares:
\[ 25 + (y^2 + 4y + 4) = 9 + (4 - 4y + y^2) \]
This simplifies to:
\[ 25 + y^2 + 4y + 4 = 9 + 4 - 4y + y^2 \]

### Step 7: Combine Like Terms
Combine like terms:
\[ 29 + 4y = 13 - 4y \]

### Step 8: Solve for y
Add \(4y\) to both sides:
\[ 29 + 8y = 13 \]
Subtract 29 from both sides:
\[ 8y = 13 - 29 \]
\[ 8y = -16 \]
Divide by 8:
\[ y = -2 \]

### Step 9: Write the Final Point
The point on the y-axis that is equidistant from (5, -2) and (-3, 2) is:
\[ (0, -2) \]