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Math Help - the problems about equation of a straight line!!!

  1. #1
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    Exclamation the problems about equation of a straight line!!!

    ok.. as you have suggested, here are the problems:
    1.The equations of the sides of a quadrilateral are x+4y-6=0 , 4x-y+4=0,
    x+4y+2=0, 8x-2y+13=0. Show that it is a rectangle and find its area..

    2.Show that the lines x-y-8=0, 3x+2y+1=0, x-y-3=0 and 2x+3y+24=0 are the sides of an isosceles trapezoid and find its area.

    weee... i do hope you will help me pls... thanks!!!
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  2. #2
    Lord of certain Rings
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    Tell me, the properties of a quadrilateral that uniquely determine a rectangle and an isosceles trapezoid.
    WIKI says: A trapezoid (in North America) or trapezium (in Britain and elsewhere) is a quadrilateral, which is defined as a shape with four sides, that has one set of parallel sides. In an isosceles trapezoid, the base angles are equal, and so are the pair of non-parallel opposite sides.

    WIKI says: In geometry, a rectangle is defined as a quadrilateral where all four of its angles are right angles.


    For area, once you do show the above, you can calculate required distances. For example length and breadth are enough for a rectangle.
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  3. #3
    Newbie JoFaSs's Avatar
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    for the rectangle you should note that the product of the lines(that intersect) of the 2 gradients is equal to -1, that is m_{1}\times m_{2}=-1
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by april29 View Post
    sorry i didnt know much about that formula M1x M2=-1...can you tell me all about it..my problem is that i really dont have an idea how to get its area...
    We do not really have an idea, about what you know and don't. So it will be a good idea to post your background.
    So, in your notes, do you have a formula for calculating area? Perhaps you know distance formula OR you know area of triangle given vertices OR quadrilateral area given vertices....

    So which one do you know? And tell us how much have you tried?
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  5. #5
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    Unhappy ???

    sorry i didnt know much about that formula M1x M2=-1...can you tell me all about it..my problem is that i really dont have an idea how to get its area and also, how can i get its distances since the given is only the 4 equations. the finding of the corners(points) itself is hard for me..i already tried solving for the intersection points of those equations..but then i got mixed up when i plotted those intersections,it didnt seem to be a rectangle.. well thanks a lot for your replies...
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  6. #6
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    Arrow hi!

    i did know about the distance formula,straight line formulas...i had a little background in analytic geometry..
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  7. #7
    Lord of certain Rings
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    Quote Originally Posted by april29 View Post
    i did know about the distance formula,straight line formulas...i had a little background in analytic geometry..
    There are a variety of ways to do it.
    First you should know that iff lines have this form
    L_1:AX+BY+C_1=0  \\\ L_2:AX+BY+C_2=0
    then L1 and L2 are parallel.

    Using this prove your lines form a parallelogram. What additional property(compared to a parallelogram) does a rectangle have??

    Now, say you proved it is a rectangle, then calculate Area = length x breadth
    To calculate length and breadth, use the following fact:
    Say you know this
    L_1:AX+BY+C_1=0  \\\ L_2:AX+BY+C_2=0 then d =\frac{|C_1 - C_2|} { \sqrt{A^2+B^2}}
    where 'd' is the distance between parallel lines
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by april29 View Post
    sorry i didnt know much about that formula M1x M2=-1...can you tell me all about it..
    Given two lines with slopes M1 and M2, we know that they are perpendicular if M1 times M2 is -1.

    -Dan
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