How Do You Prove Point Equidistance and Find Loci in Geometry?

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SUMMARY

The discussion focuses on proving point equidistance in geometry, specifically demonstrating that if point P(x,y) is equidistant from points A(a+b,b-a) and B(a-b,a+b), then the relationship bx = ay holds true. Additionally, the locus of the variable point Z(a cos(θ), b sin(θ)) is derived, resulting in the equation x²/a² + y²/b² = 1, which represents an ellipse. Participants shared their approaches, emphasizing the use of the distance formula and the challenges of finding loci for variable points.

PREREQUISITES
  • Understanding of coordinate geometry
  • Familiarity with the distance formula
  • Knowledge of loci and their geometric representations
  • Basic trigonometric functions and their applications
NEXT STEPS
  • Study the derivation of loci in coordinate geometry
  • Explore the properties of ellipses and their equations
  • Learn advanced applications of the distance formula in geometry
  • Investigate the relationship between trigonometric functions and geometric shapes
USEFUL FOR

Students of geometry, mathematics educators, and anyone interested in the proofs and applications of point equidistance and loci in geometric contexts.

Kartik.
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1. A point P(x,y) is given equidistant from the points A(a+b,b-a) and B(a-b,a+b), then prove that bx = ay
also find the locus of the variable point Z(a cos (theta), b sin (theta)), where (theta) is a variable quantity.




2. T0 prove that ax = by



3. In an attempt towards implementing what's given, i applied the distance formula knowing the fact that it will be a lengthy solution but still i did not came up with an answer.
and the second part of the question, i am still doubtful about how to find a locus to a variable point. The value of theta is variable so, the x-coordinate and y-coordinate will oscillate from [-1,1].
 
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For the first one, could you show us what you've done? You can skip some trivial working out steps to save yourself time if you like, but the important parts would be what equality you set up by using the distance formula, and what answer you arrived at at the end.

For the second, it's quite similar to solving the locus of a circle (x,y)=(a\cos\theta,a\sin\theta) except now in this case we have a b constant as well, so simply using x^2+y^2=a^2 won't work.

But what about \frac{x^2}{a^2} can you see a way of using this and doing something similar for the y variable?
 
Solved 'em all. Thank you.
Not with the distance formula but deriving the locus of the point equidistant from A and B. for the second one its x^2/a^2+y^2/b^2 = 1
 

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