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Math Help - Need Help with this quadratics stuff

  1. #1
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    Need Help with this quadratics stuff

    Diagram attatched
    Find:
    i) the equation of the line through points A and D

    ii) the coordinates of D

    iii) the equation of the perpendicular bisector of AB

    iv) the coordinates of C

    v) the area of triangle ADC

    vi) the area of the quadrilateral ABCD

    P.S Angle BAD is a right angle and C lies on the perpendicular bisector of AB. The equation of the line through points B and C is 3y=4x-14
    Please show working!!
    You dont have to answer all question any would be nice. Thanks
    Attached Thumbnails Attached Thumbnails Need Help with this quadratics stuff-quadrilateral.bmp  
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  2. #2
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    Coordinate Geometry

    Hello scubasteve94
    Quote Originally Posted by scubasteve94 View Post
    Diagram attatched
    Find:
    i) the equation of the line through points A and D

    ii) the coordinates of D

    iii) the equation of the perpendicular bisector of AB

    iv) the coordinates of C

    v) the area of triangle ADC

    vi) the area of the quadrilateral ABCD

    P.S Angle BAD is a right angle and C lies on the perpendicular bisector of AB. The equation of the line through points B and C is 3y=4x-14
    Please show working!!
    You dont have to answer all question any would be nice. Thanks
    Not quite sure why you've called it Quadratics, but here goes:

    (i) Gradient of AB = \frac{6-8}{8-2} = -\tfrac{1}{6}

    \Rightarrow gradient of AD = 6

    \Rightarrow equation of AD is y-8=6(x-2)

    i.e. y = 6x-4

    (ii) D is intercept of this line with y-axis; i.e. (0, -4)

    (iii) Perpendicular bisector of AB also has gradient 6, and passes through the mid-point of AB, i.e. (5, 7). So its equation is

    y -7=6(x-5)

    i.e. y = 6x -23

    (iv) The line in (iii) meets AC at C, where the equations are simultaneously true; i.e. where

    3(6x-23)=4x-14


    \Rightarrow 18x - 69 = 4x-14

    \Rightarrow 14x = 55

    \Rightarrow x=\frac{55}{14}


    I have to go now. Perhaps someone else can pick this up?

    Grandad
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  3. #3
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    Quote Originally Posted by scubasteve94 View Post
    i need to find the equation of the line through points A and D so..
    ...
    I need some more information, otherwise I can't help you:

    1. Which kind of quadrilateral do you have. (According to your sketch it could be a rectangle(?))

    2. Is the point D placed on the y-axis?
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  4. #4
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    It does not specify what kind of quadrilateral it is but i dont think it is a rectangle.

    Yes D is on the y axis
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  5. #5
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    Quote Originally Posted by scubasteve94 View Post
    It does not specify what kind of quadrilateral it is but i dont think it is a rectangle.

    Yes D is on the y axis
    The slope of the line passing through A and B is:

    m_{AB}=\dfrac{6-8}{8-2}=-\dfrac13

    Since AB \perp AD the slope of the line AD is:

    m_{AD}=-\dfrac1{m_{AB}}=3

    Now use the point-slope-formula of a straight line to determine the equation of the line AD:

    y-8=3(x-2)~\implies~\boxed{y=3x+2}

    And now you know that D(0, 2)
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  6. #6
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    I now have the equation of the perpendicular bisector of AB but i am unsure how to use this to find the coordinates of C. Grandad subbed it into the equation of BC whis is 3y=4x-14 but i am unsure why he did this?
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  7. #7
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    Quote Originally Posted by Grandad View Post
    [SIZE=3]Hello scubasteve94Not quite sure why you've called it Quadratics, but here goes:

    (i) Gradient of AB = \frac{6-8}{8-2} =\bold{\color{red} -\tfrac{1}{6}}

    ...
    I don't want to pick at you but ...
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  8. #8
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    Quote Originally Posted by scubasteve94 View Post
    It does not specify what kind of quadrilateral it is but i dont think it is a rectangle.

    Yes D is on the y axis
    If the quadrilateral is actually a rectangle then the midpoint of BD must be the midpoint of AC too:

    M_{BD}\left(\frac{6+0}2\ ,\ \frac{8+2}2\right) that means:  M_{BD}\left(3, 5\right)

    M_{BD} = M_{AC}\left(\frac{x_C+2}2\ ,\ \frac{y_C+8}2\right)

    Therefore x_C=4, y_C=2 that means the point C is C(4, 2)
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  9. #9
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    i no i also picked up on that and have made the correction to my notes!!
    but im still unsure how to find the coordinates of C
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  10. #10
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    Quote Originally Posted by earboth View Post
    If the quadrilateral is actually a rectangle then the midpoint of BD must be the midpoint of AC too:
    yes that is true but it does not specify that it is a rectangle so..
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  11. #11
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    Quote Originally Posted by scubasteve94 View Post
    i no i also picked up on that and have made the correction to my notes!!
    but im still unsure how to find the coordinates of C
    1. Calculate the equation of the perpendicular bisector of AB:
    <br />
M_{AB}(5, 7) Slog of the perpendicular bisector is m = 3 (compare my previous post)

    Then b: y - 7 = 3(x-5)~\implies~\boxed{y=3x-8}

    C is situated on the bisector and the line BC: 3y=4x-14

    Calculate the intersection point:

    3(3x-8)=4x-14~\implies~5x=10~\implies~x=2 . Plug in this value into the first equation: y = 3\cdot 2 - 8 = -2

    Thus C(2, -2)
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  12. #12
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    Coordinate Geometry

    Hello again -

    Sorry for my silly error earlier. Of course 6 - 8 = -2, not -1.

    (While I was teaching, I always used to tell my students that 99% of computational errors were either due to a minus sign or a factor of 2. So it's a factor of 2, then, this time!)

    Anyway, let's complete this answer now.

    Of course, as earboth has shown, D is (0, 2) and C is (2, -2). So the diagram is like the one I have attached. So:

    (v) The area of triangle ADC = \tfrac{1}{2}AC\cdot DE = \tfrac{1}{2}10\cdot 2 = 10 sq units.

    (vi) The area of triangle ABC = \tfrac{1}{2}AC\cdot BF = \tfrac{1}{2}10\cdot 6 = 30 sq units. So area of quadrilateral ABCD = 40 sq units.

    Grandad
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