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Is my locus eqn correct?

  1. Mar 23, 2005 #1
    Here is the question I think ive almost got the answer. Describe and create an equation for each locus.

    a.) THe points are equidistant from points A(1,5) and B(-2,6)

    I made another random point C(x,y)


    I wrote the distance formula for both AC and BC and set them equal to each other but im not sure how to simplify now

    square root[(x+2)^2 +(y-6)^2] = square root[(x-1)^2+(y-5)^2]

    I square both sides to get rid of the square root but after that i dont know what to do I got a final answer of 6x-2y+13=0 plz tell me is this correct?
  2. jcsd
  3. Mar 23, 2005 #2


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    Almost, just a small arithmetic error. It should be 6x-2y+14=0
    Geometrically you can see it should be a line. Moreover, it should pass through the midpoint of the line joining the points A and B and be perpendicular to it.
    Check that the line indeed satisfies these conditions.
  4. Mar 23, 2005 #3
    I got the equation but now I have to do the describing part how do I describe the locus? How will I check that the line is indeed perpendicular to points A and B?
  5. Mar 24, 2005 #4


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    It's not perpendicular to the "points" A and B. How can you be perpendicular to a point? :smile: The locus should be perpendicular to the line joining A and B, and to verify this, compare the slopes of the two lines. What can you say immediately about the slopes of two lines if they are perpendicular?
  6. Mar 24, 2005 #5
    oh the slope will be a negative reciprocal then ill know if it is perpendicular thanks for pointing that out lol ur right you cant be perpendicular to a point opps lol. :rofl:
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