math

A(4,7) and B(-3,2) are points on a coordinate plane.Find the coordinates of a point C on the x-axis such that AC=BC .

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  1. If you make a sketch it should be easy to see that point C must be on the right-bisector of AB

    midpoint of AB = ( (4-3)/2 , (7+2)/2) = (1/2 , 9/2)
    slope o AB = (2-7)/(-3-4) = 5/7
    so the slope of the right-bisector is -7/5

    equation of right-bisector:
    y - 9/2 = (-7/5)(x - 1/2)
    for the x-intercept, let y = 0
    -9/2 = (-7/5)(x-1/2)
    -9/2 = -7x/5 + 7/10
    times 10
    -45 = -14x + 7
    14x = 52
    x = 52/14 = 26/7

    point C is (26/7, 0)

    or

    let the point be C(x,0)
    AC = BC
    √( (x-4)^2 + 7^2) = √( (x+3)^2 + 4^2)
    square both sides
    (x-4)^2 + 49 = (x+3)^2 + 4
    x^2 - 8x + 16 + 49 = x^2 + 6x + 9 + 4
    -14x = -52
    x = -52/-14 = 26/7 , just like before

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  2. THANKS A LOT !!:)

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