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 equidistant from two given points is a classic problem in coordinate geometry. This seemingly simple problem elegantly demonstrates the application of the distance formula and the power of algebraic manipulation. Understanding this problem builds a strong foundation for tackling more complex geometric challenges. This article will guide you through the solution, exploring different approaches and highlighting key concepts along the way.

Understanding the Problem

Let's start by defining the problem formally. We're given two points, A and B, with coordinates (x₁, y₁) and (x₂, y₂), respectively. Our goal is to determine the coordinates of a point, P, located on the y-axis such that the distance between P and A is equal to the distance between P and B. In other words, PA = PB. Since P lies on the y-axis, its x-coordinate is always 0. We need to find its y-coordinate.

Using the Distance Formula

The cornerstone of solving this problem lies in the distance formula. The distance between two points (x₁, y₁) and (x₂, y₂) in a Cartesian coordinate system is given by:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

We'll apply this formula twice: once to calculate the distance between P and A, and again to calculate the distance between P and B. Let's denote the coordinates of point P as (0, y).

• PA = √[(0 - x₁)² + (y - y₁)²] = √[x₁² + (y - y₁)²]
• PB = √[(0 - x₂)² + (y - y₂)²] = √[x₂² + (y - y₂)²]

Since PA = PB, we can set these two expressions equal to each other:

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

Solving for the Y-Coordinate

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

x₁² + (y - y₁)² = x₂² + (y - y₂)²

Expanding the squared terms, we get:

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

Notice that the 'y²' terms cancel out, simplifying the equation significantly:

x₁² - 2y₁y + y₁² = x₂² - 2y₂y + y₂²

Now, let's rearrange the equation to isolate 'y':

2y₂y - 2y₁y = x₂² - x₁² + y₂² - y₁²

2y(y₂ - y₁) = x₂² - x₁² + y₂² - y₁²

Finally, we can solve for 'y':

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

This formula provides the y-coordinate of the point P on the y-axis that is equidistant from points A and B. Note that this formula is only valid if y₂ ≠ y₁. If y₂ = y₁, the points A and B have the same y-coordinate, and the problem requires a different approach (discussed later).

Illustrative Example

Let's consider a concrete example to solidify our understanding. Suppose we have points A(2, 3) and B(6, 1). We can plug these coordinates into the formula we derived:

• x₁ = 2, y₁ = 3
• x₂ = 6, y₂ = 1

y = (6² - 2² + 1² - 3²) / (2(1 - 3))

y = (36 - 4 + 1 - 9) / (2(-2))

y = 24 / (-4)

y = -6

Therefore, the point on the y-axis equidistant from A(2, 3) and B(6, 1) is (0, -6).

Geometric Interpretation

The solution we obtained represents the perpendicular bisector of the line segment AB. The point P lies on this bisector. This geometric interpretation provides an alternative way to approach the problem, particularly helpful for visualizing the solution.

Handling the Case where y₁ = y₂

Our derived formula is undefined when y₁ = y₂ because it involves division by zero. This scenario arises when points A and B lie on the same horizontal line. In this case, the perpendicular bisector is a vertical line, and there is no point on the y-axis that is equidistant from A and B. The only exception is if A and B are symmetric around the y-axis, meaning x₁ = -x₂. In this specific case, any point on the y-axis would be equidistant from A and B.

Alternative Approaches

While the distance formula and algebraic manipulation provide a direct solution, other methods can be employed. These include:

• Geometric Construction: Using a compass and straightedge, you can construct the perpendicular bisector of the line segment AB and find its intersection with the y-axis.

• Vector Approach: This involves using vector operations to find the midpoint of AB and then projecting this midpoint onto the y-axis.

These methods provide alternative pathways to the same solution, offering different perspectives and enhancing understanding.

Advanced Applications and Extensions

The fundamental principles involved in finding the point on the y-axis equidistant from two given points extend to more complex scenarios. These include:

• Three-dimensional space: The problem can be extended to three dimensions, requiring the use of the three-dimensional distance formula.

• Finding equidistant points on other lines: Instead of the y-axis, we could seek the equidistant point on any other line. This might involve finding the equation of the line and using it to constrain the solution.

Conclusion

Finding the point on the y-axis equidistant from two given points is a problem that beautifully combines algebra and geometry. By using the distance formula and some algebraic manipulation, we can derive a concise formula to solve for the y-coordinate of this point. Understanding the geometric interpretation of the solution and considering special cases, such as when the given points have the same y-coordinate, completes our understanding of this fundamental problem in coordinate geometry. This problem, seemingly simple at first glance, serves as a stepping stone towards more advanced concepts in geometry and analytical mathematics. The techniques and principles discussed here are widely applicable in various fields, including computer graphics, physics, and engineering. Mastering these fundamental concepts paves the way for tackling complex problems involving distance, geometry, and coordinate systems with increased confidence and proficiency.