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    Can anyone help me?

    A curve has the equation y = x/2 + 3 - 1/x x isn't equal to 0

    The point A on the curve has x - coordinate 2.

    (i) Find the gradient of the curve at A.

    (ii) Show that the tangent to the curve at A has equation

    3x - 4y + 8 = 0

    The tangent to the curve at the point B is parallel to the tangent at A.

    (iii) Find the coordinates of B.


    The straight line l has gradient 3 and passes through point A (-6,4)

    (i) Find an equation for l in the form y = mx + c.

    The straight line m has the equation x - 7y + 14 = 0

    Given that m crosses the y axis at the point B and intersects l at the point C,

    (ii) Find the coordinates of B and C,

    (iii) Show that angle BAC = 90,

    (iv) Find the area of triangle ABC.

    Thanks =)
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    I'll do it one step at a time, as I know if do the whole thing someone will enter their answer just before mine
    i)
    To find the gradient, take dy/dx

    1/2 + 1/x^2

    We know that x = 2
    So 1/2 + 1/4
    =3/4

    Gradient = 3/4
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    ii) At A, x=2, so sub into the curve equuation

    y = 2/2 +3 -1/2
    y= 7/2

    The equation of a line is

    y = mx + c
    So
    7/2 = 3/4 * 2 + c
    14/4 - 6/4 = c
    c= 2

    So equation of the tangent is

    y = 3/4x + 2
    (*4) 4y = 3x + 8
    3x - 4y + 8 = 0
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    Sorry for the delay

    iii) Parallel, so dy/dx = 3/4

    Therefore 1/2 - 1/x^2 = 3/4
    1/x^2 = 1/4
    x = +/- 2

    As x = 2 for A, it must = -2 for B.

    Sub this into the orginal equation and you get
    y = -1 + 3 + 1/2
    = 5/2

    So B (-2, 5/2)
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    Question 2,

    i)
    Gradient 3, y = -6, x = 4, so

    -6 = 3*4 + c
    c = -6 - 12
    c = -18

    y= 3x -18
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    ii) At point B, x = 0 as the line crosses the y axis.
    As x = 0,
    7y = x + 14
    7y = 14
    y = 2


    As for C

    If it intersects l, you can combine the two equations, so

    y = 3x - 18 [1]
    7y = x + 14 [2]

    Divide [2] by 7

    y = x/7 + 2
    Now

    x/7 + 2 = 3x -18
    x + 14 = 21x - 126
    20x = 140
    x =7

    Therefore
    y = 21-18
    y = 3

    B (0, 2) C (7, 3)
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    Part iii) Is difficult to do here, but simply find AB, BC, CA, and then show that
    h^2 = a^2 + b^2 is true.
    The area can be found by the same principle using the appropriate formula
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    thank you thank you thank you thank you so much! =D
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    Np, just hope it all makes sense Feel free to PM me if you have any further problems
 
 
 
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Updated: February 1, 2006

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