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Math Help - Algebra 2 help please

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    Question Algebra 2 help please

    hello,

    i have to show that if the midpoints of consecutive sides of any quadrilateral are connected the result is a parallelogram.

    i know the answer but do not know how to get it, help with explaining the process would be appreciated.

    First point: (0,0)
    Second Point: (a,0)
    Third Point: (b,c)
    Forth Point: (d,e)

    I think the midpoints are right but i am not sure, i got them myself.

    Midpoint of line A: (a/2),0)
    Midpoint of line B: ((a+b)/2),(c/2)
    Midpoint of line C: ((b+d)/2),((c+e)/2))
    Midpoint of line D: (d/2),(e/2)

    Here is where i get stuck, i do not know how to prove that the slopes of opposite sides are equal.

    thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    hello,

    i have to show that if the midpoints of consecutive sides of any quadrilateral are connected the result is a parallelogram.

    i know the answer but do not know how to get it, help with explaining the process would be appreciated.

    First point: (0,0)
    Second Point: (a,0)
    Third Point: (b,c)
    Forth Point: (d,e)

    I think the midpoints are right but i am not sure, i got them myself.

    Midpoint of line A: (a/2),0)
    Midpoint of line B: ((a+b)/2),(c/2)
    Midpoint of line C: ((b+d)/2),((c+e)/2))
    Midpoint of line D: (d/2),(e/2)

    Here is where i get stuck, i do not know how to prove that the slopes of opposite sides are equal.

    thank you.
    well, first you need to construct lines going from point to point, and show that the slopes of the lines that are opposite sides are the same. and the notation you use for the coordinates is totally off
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    reply

    i do not know what you by notation. Here is there pictureAlgebra 2 help please-image-1.jpg. i hope this helps.

    Thanks.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    i do not know what you by notation. Here is there pictureClick image for larger version. 

Name:	image-1.jpg 
Views:	22 
Size:	12.3 KB 
ID:	4036. i hope this helps.

    Thanks.
    coordinates are of the form (x,y), that is not what the midpoints of you're answers look like.


    recall that the equation of a line is of the form: y = mx + b

    where m is the slope and b is the y-intercept.

    forget making lines though, just find the slopes of lines connecting the midpoints.

    Remember, if two points on a line are (x_1,y_1) and (x_2,y_2), then:

    m = \frac {y_2 - y_1}{x_2 - x_1}


    show that the m's for opposite sides are equal
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  5. #5
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    Hello, OnMyWayToBeAMathProffesor!

    You should be familiar with the Slope Formula by now.
    . . But you can also use the Distance Formula . . .


    Show that if the midpoints of consecutive sides of any quadrilateral
    are connected, the result is a parallelogram.

    \begin{array}{cc}\text{First point:} & (0,\,0) \\<br />
\text{Second point:} & (a,\,0) \\<br />
\text{Third point:} & (b,\,c) \\<br />
\text{Fourth point:} & (d,\,e) \end{array}


    I think the midpoints are right. . . . . they are!

    \begin{array}{cc}\text{Midpoint of line A:} & P\left(\frac{a}{2},\,0\right) \\<br />
\text{Midpoint of line B:}& Q\left(\frac{a+b}{2},\,\frac{c}{2}\right) \\<br />
\text{Midpoint of line C:} & R\left(\frac{b+d}{2},\,\frac{c+e}{2}\right) \\<br />
\text{Midpoint of line D:} & S\left(\frac{d}{2},\,\frac{e}{2}\right)\end{array}

    Theorem .If opposite sides of a quadrilateral are equal,
    . . . . . . . . the quadrilateral is a parallelogram.

    You can use the Distance Formula to show that: . PQ = RS .and . QR = PS.

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  6. #6
    Member OnMyWayToBeAMathProffesor's Avatar
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    thanxs

    so i just use

    d= sqrt((x_1-x_2)^2+(y_1-y_2)^2)

    thank you very much, i got it now.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    so i just use

    d= sqrt((x_1-x_2)^2+(y_1-y_2)^2)

    thank you very much, i got it now.
    yes. and if you are going to use LaTex, you might as well use it properly, type \sqrt {(x_1 - x_2)^2 + (y_1 - y_2)^2} between the math tags to get \sqrt {(x_1 - x_2)^2 + (y_1 - y_2)^2}
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  8. #8
    Math Engineering Student
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    Quote Originally Posted by Jhevon View Post
    and if you are going to use LaTex, you might as well use it properly, type \sqrt {(x_1 - x_2)^2 + (y_1 - y_2)^2}
    Since you use MathType, this generates a lot of space when you're typin', so to type this formula it suffices to write

    \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Since you use MathType, this generates a lot of space when you're typin', so to type this formula it suffices to write

    \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
    i don't use MathType anymore. i haven't used it ever since LaTex came back, typing code here is a lot faster than using MathType templates. I type with spaces here because I like to see it that way. it helps me to better visualize what the code will look like when i'm done. i feel weird typing a+b i must type a + b to distinguish between entities
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  10. #10
    Math Engineering Student
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    Is it that true?

    Wow... I didn't think that writin' all closer was botherin' you

    --

    I'm not agree with you anyway, 'cause MathType generates the codes quickly (the handicap it's that you can't align of give more presentation in your post), as this one makes lots of spaces, then reduces the capacity (which supports 400 characters), therefore, it is useless on this forum.

    --

    Nevertheless, there're people which writes all closer, so, it depends of them.
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