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Math Help - questions about binomial expansions, coordinate geometry

  1. #1
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    questions about binomial expansions, coordinate geometry

    1.
    Given that the expansion of (a+x)(1-2x)^n in ascending powers of x is 3-41x+bx^2+.....Find the values of the constants a, n and b.


    2.
    Given that the coefficient of x^2 in the expansion of (k+x)[2-(x/2)]^6 is 84, find the value of the constant k.


    3.
    Solutions to this question by accurate drawing will not be accepted. A parallelogram ABCD in which A(8,2) and B(2,6). The equation of BC is y=(1/2)x+5 and X is the point on BC such that AX is perpendicular to BC. Equation of AX is y=-2x+18. Coordinate of X is (5.2,7.6). Given also that BC = 5 BX, find the coordinates of C and D.
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  2. #2
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    Quote Originally Posted by wintersoltice View Post
    1.
    Given that the expansion of (a+x)(1-2x)^n in ascending powers of x is 3-41x+bx^2+.....Find the values of the constants a, n and b.
    (1 - 2x)^n \approx 1^n + n \cdot 1^{n - 1} \cdot (-2x)^1 + \frac{n(n - 1)}{2} \cdot 1^{n - 2} \cdot (-2x)^2

    So
    (a + x)(1 - 2x)^n \approx (a + x)(1 - 2nx + 2n(n - 1)x^2)

    = a + (1 - 2na)x + (2n(n - 1)a - 2n)x^2 (Keeping only terms to second order in x.)

    -Dan
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  3. #3
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    3.
    Solutions to this question by accurate drawing will not be accepted. A parallelogram ABCD in which A(8,2) and B(2,6). The equation of BC is y=(1/2)x+5 and X is the point on BC such that AX is perpendicular to BC. Equation of AX is y=-2x+18. Coordinate of X is (5.2,7.6). Given also that BC = 5 BX, find the coordinates of C and D.

    This is good, you have been given coordinates of X, and you know that the magnitude of vector BC is 5 times that of vector BX

    First find BX:

    BX= ((5.2-2),(7.6-6))=(3.2,1.6)

    now BC = 5BX= 5(3.2,1.6) = (16,8)

    So now you know that point C has position vector (16,8) relative to B,

    so that C=((16+2),(8+6)) = (18,14)

    Now to find D, note that vector CD is the same as vector BA,

    and BA= ((8-2),(2-6)) = (6,-4)

    so CD = (6,-4)

    and you find that D is at ((18+6),(14-4))= (24,10)

    Hope this helps?
    Last edited by Gaal Dornick; June 16th 2008 at 07:52 AM. Reason: numerical error
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  4. #4
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    Hello, wintersoltice!

    3. A parallelogram ABCD in which A(8,2) and B(2,6).
    The equation of BC is: .  y\:=\:\frac{1}{2}x+5
    X is the point on BC such that AX \perp BC
    Equation of AX is: . y\:=\:-2x+18 .
    . . . not needed
    Coordinates of X are (5.2,7.6)
    Given that BC = 5 BX, find the coordinates of C\text{ and }D.
    Code:
          |
          |                             o C
          |                        *
          |                   *
          |          X   *
          |          * (5.2,7.6)
          |   B o     *
          |   (2,6)    *
          |             *
          |              *
          |               o(8,2)
          |               A
        - + - - - - - - - - - - - - - - - - -
          |
    BC is five times BX.
    Moving from B to X, we move 3.2 right and 1.6 up.

    Moving from B to C, we will move 5 times as much.
    . . \begin{array}{ccccc}x &=& 2 + 5(3.2) &=& 18 \\ y &=& 6 + 5(1.6) &=& 14\end{array}\quad\Rightarrow\quad C(18,14)

    Moving from B to A, we move 6 right and 4 down.
    Moving from C to D, we will move the same.
    . . \begin{array}{ccccc}x &=& 18 + 6 &=& 24 \\y &=& 14 - 4 &=& 10\end{array} \quad\Rightarrow\quad D(24,10)

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