Find The Point On The Y-axis Which Is Equidistant From

Find the Point on the Y-Axis Equidistant from Two Given Points

Finding the point on the y-axis that's equidistant from two given points is a classic problem in coordinate geometry. It leverages the distance formula and the properties of the y-axis to arrive at a straightforward solution. This article will explore the problem in detail, providing a step-by-step solution, practical examples, and even consider extensions to the problem.

Understanding the Problem

Let's assume we have two points in a Cartesian coordinate system: Point A with coordinates (x₁, y₁) and Point B with coordinates (x₂, y₂). Our goal is to find the coordinates of a point, let's call it Point P, that lies on the y-axis and is equidistant from both Point A and Point B. Since Point P is on the y-axis, its x-coordinate will always be 0. Therefore, we only need to find its y-coordinate.

The Distance Formula: The Cornerstone of Our Solution

The distance between two points (x₁, y₁) and (x₂, y₂) in a Cartesian plane is given by the distance formula:

d = √((x₂ - x₁)² + (y₂ - y₁)²)

This formula is crucial for solving our problem. We'll use it to express the distances between Point P and both Point A and Point B. Since Point P is equidistant from A and B, these distances will be equal.

Step-by-Step Solution

1. Define Point P: Since Point P lies on the y-axis, its coordinates are (0, y). We need to find the value of 'y'.

2. Calculate the Distance PA: Using the distance formula, the distance between Point P(0, y) and Point A(x₁, y₁) is:

   PA = √((x₁ - 0)² + (y₁ - y)²) = √(x₁² + (y₁ - y)²)

3. Calculate the Distance PB: Similarly, the distance between Point P(0, y) and Point B(x₂, y₂) is:

   PB = √((x₂ - 0)² + (y₂ - y)²) = √(x₂² + (y₂ - y)²)

4. Equate the Distances: Since PA = PB, we can equate the two distance expressions:

   √(x₁² + (y₁ - y)²) = √(x₂² + (y₂ - y)²)

5. Solve for y: To solve for 'y', we can square both sides of the equation to eliminate the square roots:

   x₁² + (y₁ - y)² = x₂² + (y₂ - y)²

   Expanding the equation:

   x₁² + y₁² - 2y₁y + y² = x₂² + y₂² - 2y₂y + y²

   Notice that the y² terms cancel out:

   x₁² + y₁² - 2y₁y = x₂² + y₂² - 2y₂y

   Now, rearrange the equation to solve for 'y':

   2y₂y - 2y₁y = x₂² + y₂² - x₁² - y₁²

   2y(y₂ - y₁) = x₂² - x₁² + y₂² - y₁²

   y = (x₂² - x₁² + y₂² - y₁²) / (2(y₂ - y₁))

This is the general formula to find the y-coordinate of the point P on the y-axis that is equidistant from points A and B. Note that this formula assumes y₂ ≠ y₁. If y₂ = y₁, a different approach will be needed (discussed later).

Practical Examples

Let's work through a few examples to solidify our understanding.

Example 1:

Find the point on the y-axis equidistant from A(2, 3) and B(4, 1).

Using the formula:

y = (4² - 2² + 1² - 3²) / (2(1 - 3)) = (16 - 4 + 1 - 9) / (-4) = 4 / (-4) = -1

Therefore, the point on the y-axis is (0, -1).

Example 2:

Find the point on the y-axis equidistant from A(-1, 5) and B(3, -1).

Using the formula:

y = (3² - (-1)² + (-1)² - 5²) / (2(-1 - 5)) = (9 - 1 + 1 - 25) / (-12) = -16 / (-12) = 4/3

Therefore, the point on the y-axis is (0, 4/3).

Handling the Case Where y₁ = y₂

If y₁ = y₂, the denominator in our formula becomes zero, making it undefined. This situation implies that points A and B lie on a horizontal line parallel to the x-axis. In this case, the midpoint of the line segment AB will have the same y-coordinate as A and B. The point on the y-axis equidistant from A and B will lie on the perpendicular bisector of the line segment AB. The x-coordinate of this point is 0, and the y-coordinate is the average of y₁ and y₂ (which are equal).

Therefore, if y₁ = y₂, the point on the y-axis is (0, y₁) (or (0, y₂) since they are equal).

Geometric Interpretation

The solution represents the intersection of the y-axis with the perpendicular bisector of the line segment connecting A and B. The perpendicular bisector is the locus of all points equidistant from A and B.

Advanced Applications and Extensions

This fundamental concept finds applications in various fields, including:

• Robotics: Determining the position of a robot arm equidistant from two obstacles.
• Signal Processing: Locating the source of a signal based on its reception at two different points.
• Computer Graphics: Calculating the reflection of a point across a line.

Conclusion

Finding the point on the y-axis equidistant from two given points is a solvable problem using the distance formula and some algebraic manipulation. This article provides a comprehensive guide, including step-by-step instructions, examples, and a discussion of special cases. The understanding of this concept provides a strong foundation in coordinate geometry and opens doors to further explorations in related mathematical fields and practical applications. Remember to always consider the special case where the y-coordinates of the two given points are equal for a robust solution. The geometric interpretation of the solution enhances understanding and adds depth to the problem-solving process.