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Math Help - Coordinate Geometry

  1. #1
    Member maybeline9216's Avatar
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    Coordinate Geometry

    Please do part(c).

    I have found the coords. of point B which is (5/4,35/4).

    The answer to part (c) is (11,-2)
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  2. #2
    MHF Contributor
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    Quote Originally Posted by maybeline9216 View Post
    Please do part(c).

    I have found the coords. of point B which is (5/4,35/4).

    The answer to part (c) is (11,-2)
    I found the coordinates of B to be (1,8).

    The equation for AB is y = 3x +5

    As for the part (c), my way is way too long.
    The key is the given area of right triangle BAD = 40 sq.units
    That means 40 = (1/2)(AB)(AD)
    So find AB and AD.

    AB = sqrt[(1-(-1))^2 +(8-2)^2] = sqrt(40)

    AD = sqrt[(x -(-1))^2 +(y -2)^2] ......where (x,y) is corner D.

    So,
    40 = (1/2)sqrt(40)*sqrt[(x+1)^2 +(y-2)^2]
    40 / (1/2)sqrt(40) = sqrt[(x+1)^2 +(y-2)^2]
    2sqrt(40) = sqrt[(x+1)^2 +(y-2)^2]
    160 = (x+1)^2 +(y-2)^2 --------------(i)

    The equation of AD:
    slope = same slope as BC = -(1/3)
    using the (-1,2),
    (y -2) = (-1/3)(x -(-1))
    (y-2) = (-1/3)(x+1) -----------------(ii)
    Substitute that into (i),
    160 = (x+1)^2 +[(-1/3)(x+1)]^2
    160 = (x+1)^2 +(1/9)(x+1)^2
    160 = (x+1)^2 *[1 +1/9]
    160 = (x+1)^2 *(10/9)
    160 / (10/9) = (x+1)^2
    9*16 = (x+1)^2
    3*4 = x+1
    x = 12 -1 = 11 -----***

    And,
    (y-2) = (-1/3)(11 +1) = -4
    y = -4 +2 = -2

    Therefore, (x,y), the corner D, is (11,-2) ---------answer.
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  3. #3
    Member maybeline9216's Avatar
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    Ops sorry... i careless=(
    Last edited by maybeline9216; October 11th 2008 at 05:16 AM.
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