Locus of a Point: Find Equation for Equidistant Point P from (3,-1)

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SUMMARY

The discussion focuses on finding the equation of the locus of a point P that is equidistant from the y-axis and the point (3,-1). Participants utilize the distance formula, specifically the Euclidean distance, to set up the equation. The key equations derived are \sqrt{(x-3)^2 + (y+1)^2} = |x| and \sqrt{(x-0)^2 + (y-y)^2} = |x|. The final conclusion suggests that the equation of the locus is y = x^2 + 3.

PREREQUISITES
  • Understanding of Euclidean distance in a Cartesian plane
  • Familiarity with the concept of loci in geometry
  • Basic algebraic manipulation and expansion of equations
  • Knowledge of coordinate systems and points in 2D space
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  • Study the derivation of distance formulas in coordinate geometry
  • Learn about loci of points and their equations in geometry
  • Explore quadratic equations and their graphical representations
  • Investigate the properties of parabolas and their applications
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Students studying geometry, mathematics educators, and anyone interested in understanding the properties of loci and distance in coordinate systems.

zebra1707
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1. Find the equation of the locus of a point P which is equidistant from the y-axis and the point (3,-1)


2. I think that I need to use

SqRoot (x-x)sq + (y-y)sq = x
SqRoot (x-3)sq + (y+1)sq = x

Expanding is where I get stuck



Cheers
 
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If [tex](a,b)[/tex] is the point, you need work with

Distance from [tex](a,b)[/tex] to [tex](3,-1)[/tex] = distance from [tex](a,b)[/tex] to the [tex]y-[/tex] axis. Part of this equation is

[tex] \sqrt{(a-0)^2 + (b-b)^2} = \sqrt{a^2} = |a|[/tex]

What is the other part? remember your solution will be an equation, not a single number.
 
Hi there
There is no other part to this question. I think the equation should be y=x Sq + 3

Cheers (very confused).
 

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