Pope
Lv 7
Pope asked in Science & MathematicsMathematics · 10 years ago

Locus of solutions to an equation?

Describe in detail the locus of points in the x-y plane satisfying this equation:

(x² + y² - 14x + 8y + 65)(7x² + 16xy - 15y² - 7x + 5y) = 0

3 Answers

Relevance
  • ?
    Lv 7
    10 years ago
    Favourite answer

    The first factor appears to be a circle, but completing the square yields:

    (x - 7)^2 + (y + 4)^2 = 0

    The radius is 0, so it is just a point, namely (7, -4).

    The second factor can be factored further to:

    (7x - 5y)(x + 3y - 1)

    Which describes two lines:

    7x - 5y = 0 and x + 3y - 1 = 0

    -5y = -7x and 3y = -x + 1

    y = 7x/5 and y = -x/3 + 1/3

    So the locus is the point (7, -4) and the lines y = 7x/5 and y = -x/3 + 1/3.

  • Ian H
    Lv 7
    10 years ago

    [(x -7)^2 + (y +4)^2]*[(x + 3y -1)(7x - 5y]= 0

    first expression is like circle with zero radius

    Is it two intersecting lines ?

    3y = 1 - x

    5y = 7x

    Not too sure

    Wolfram might clarify

    http://www.wolframalpha.com/examples/DifferentialE...

    Regards - Ian

  • 10 years ago

    I got two lines that cross near zero. One line goes thru zero with a positive slope and the other is a

    line with negative slope not thru zero but close.

    And then there's a spherical object at (7,-4)

    Source(s): Dreams and loss of sleep
Still have questions? Get answers by asking now.