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Math Help - Three Points Equidistant...

  1. #1
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    Three Points Equidistant...

    So I thought I knew how to solve this kind of question, but apparently not.

    Find a point equidistant from the three points, P(3, 2), Q(3, -3), R(-2, 2).

    My (wrong) answer: Mid-point of PQ is (3, -1/2), so the slope of the line from R to this point is (-5/2)/5 = -1/2. Going two to the right from R by this slope, we go down to 1, so the equation of the line is y = (-1/2)x + 1.

    Mid-point of PR is (1/2, 2), so the slop of the line from this point to Q is -5/(5/2) = -2. Going from this point, one-half unit to the left we go up one, so the equation of the line is y = -2x + 3.

    Now I want to find the point of intersection between these two lines, so I set them equal to each other and solve for x, giving me x = 4/3, which is wrong.
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  2. #2
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    Quote Originally Posted by ragnar View Post
    So I thought I knew how to solve this kind of question, but apparently not.

    Find a point equidistant from the three points, P(3, 2), Q(3, -3), R(-2, 2).

    My (wrong) answer: Mid-point of PQ is (3, -1/2), so the slope of the line from R to this point is (-5/2)/5 = -1/2. Going two to the right from R by this slope, we go down to 1, so the equation of the line is y = (-1/2)x + 1.

    Mid-point of PR is (1/2, 2), so the slop of the line from this point to Q is -5/(5/2) = -2. Going from this point, one-half unit to the left we go up one, so the equation of the line is y = -2x + 3.

    Now I want to find the point of intersection between these two lines, so I set them equal to each other and solve for x, giving me x = 4/3, which is wrong.
    the point equidistant from all three points P, Q, and R is the intersection of the perpendicular bisectors of PQ, QR and PR

    PQ is a vertical line ... the equation of the perpendicular bisector is y = -1/2

    PR is a horzontal line ... the equation of the perpendicular bisector is x = 1/2

    the point of intersection is (1/2, -1/2), which is equidistant from the three points.

    fyi ... the point equidistant from the three vertices of a right triangle is always the midpoint of the hypotenuse ... make a sketch on graph paper and check it out.
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  3. #3
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    Hello, ragnar!

    If you had made a sketch, you would have seen skeeter's FYI immediately.



    \text{Find a point equidistant from the points: }\;P(3, 2),\,Q(3, \text{-}3),\; R(\text{-}2, 2).
    Code:
                    |
         (-2,2)     |             (3,2)
            o - - - | - - - - - - - o P
               *    |               |
                  * |               |
      - - - - - - - |*- - - - - - - | - - -
                    |   ♥           |
                    |      *        |
                    |         *     |
                    |            *  |
                    |               * Q
                    |             (3,-3)


    \,PQR is a right triangle.

    Therefore, the circumcenter is the midpoint of the hypotenuse:

    . . . \left(\frac{1}{2},\:-\frac{1}{2}\right)
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