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Math Help - equation of points

  1. #1
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    equation of points

    Find the equation of all points, P(x,y), which are aquidistant from the points A(-3,6) and B(2,1).

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    Quote Originally Posted by DINOCALC09 View Post
    Find the equation of all points, P(x,y), which are aquidistant from the points A(-3,6) and B(2,1).

    Thanks
    Call one of the points that you are seeking (x, y). Then we will derive a relationship between x and y such that you can find all such points.

    The distance between A and (x, y) is
    \sqrt{(x - -3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 6)^2}

    The distance between B and (x, y) is
    \sqrt{(x - 2)^2 + (y - 1)^2}

    We want (x, y) such that these are equal, so
    \sqrt{(x + 3)^2 + (y - 6)^2} = \sqrt{(x - 2)^2 + (y - 1)^2}

    I'll let you simplify it.

    -Dan
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    Hello, DINOCALC09!

    I bet you already know enough to solve this . . .


    Find the equation of all points, P(x,y), which are equidistant
    from the points A(-3,6) and B(2,1).

    You know the Distance Formula, right?

    Distance PA \:=\:\sqrt{(x+3)^2 + (y-6)^2}

    Distance PB \:=\:\sqrt{(x-2)^2 + (y-1)^2}

    Since these distances are equal: . \sqrt{(x+3)^2 + (y-6)^2} \;=\;\sqrt{(x-2)^2 + (y-1)^2}

    Square both sides: . (x+3)^2 + (y-6)^2\;=\;(x-2)^2 + (y-1)^2

    Expand: . x^2 + 6x + 9 + y^2 - 12y + 36 \;=\;x^2 - 4x + 4 + y^2 - 2y + 1

    And this simplifies to: . 10x - 10y \:=\:40\quad\Rightarrow\quad \boxed{y \:=\:x + 4}

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