Deriving the Equation for a Point's Coordinates According to Given Distance Conditions

Introduction

In the field of geometric algebra and distance relations, finding the locus of points that satisfy certain conditions can be a challenging but rewarding task. This article explores a specific problem where the distance of a point $P(x, y)$ from point $A(3,2)$ is always twice its distance from point $B(-4,1)$. The article outlines the step-by-step process of deriving the equation that the coordinates of point $P$ must satisfy and demonstrates the resulting geometric locus.

Distance Relations

Given the distances from point $P(x, y)$ to point $A(3,2)$ and point $B(-4,1)$, we start with the equations: PA^2 (x - 3)^2 (y - 2)^2 PB^2 (x 4)^2 (y - 1)^2 Since $PA 2 cdot PB$, we have: PA^2 4 cdot PB^2 (x - 3)^2 (y - 2)^2 4 cdot [(x 4)^2 (y - 1)^2] Expanding and simplifying the equation, we get: (x - 3)^2 (y - 2)^2 4[(x 4)^2 (y - 1)^2] (x^2 - 6x 9) (y^2 - 4y 4) 4(x^2 8x 16 y^2 - 2y 1) 3x^2 38x 3y^2 - 4y -55 Further simplification yields: 3x^2 38x 3y^2 - 4y -55 This equation represents a circle in the coordinate plane. We can rewrite it in the standard form of a circle equation by completing the square:

Circle Equation

Rewriting the equation as a circle, we find the center and radius of the circle. The equation can be transformed to: 3(x^2 6x/3) 3(y^2 - 4y/3) -55 3[(x 3)^2 - 3^2/3] 3[(y - 2/3)^2 - 2^2/3] -55 3(x 3)^2 - 9 3(y - 2/3)^2 - 4/3 -55 3(x 3)^2 3(y - 2/3)^2 200/3 Dividing both sides by 3, we get: (x 3)^2 (y - 2/3)^2 200/9 This is the standard form of a circle equation with the center at $(-19/3, 2/3)$ and a radius of $10sqrt{2}/sqrt{3}$.

Verification

To verify the solution, we check the distance condition using a specific point. For instance, the point $P(-4, 4.763)$ satisfies the given condition:

PA^2 497.734 56.634 PB^2 3.763^2 14.16 PA/PB sqrt{56.634}/14.16 2 Thus, the derived equation correctly represents the locus of the point $P(x, y)$ that satisfies the given distance condition.

Conclusion

The equation of the locus for a point $P(x, y)$ in the plane, given that its distance from point $A(3,2)$ is always twice its distance from point $B(-4,1)$, is a circle with center at $(-19/3, 2/3)$ and a radius of $10sqrt{2}/sqrt{3}$. This problem provides a clear example of how algebraic geometry can be used to derive and verify the equation of a circle based on given distance conditions.

Key Terms

The key terms in this article are point coordinates (the coordinates of point $P(x, y)$), distance relations (the conditions involving distances between points), algebraic geometry (the use of algebra to analyze geometric shapes), circle equation (the equation representing the circle), and locus of points (the set of points satisfying a given set of conditions).