• anonymous
Find the value of x for which ABCD must be a parallelogram.
  • Stacey Warren - Expert
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  • katieb
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  • anonymous
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  • Owlcoffee
A parallelogram has a very interesting property which is a little theorem (if you want the proof for you it, I can link you to my tutotial on verctorial spaces where I tackle this). But something important to note is that a parallelogram has two diagonals and these two diagonals actually intersect eachother on their mid-points. What does that imply? Let's call the point of intersection of your parallelogram as "M" so this means that: \[dist(A,M)=dist(M,C)\] Ain't that a charm? and we have these distances, they are \(2x+8\) and \(6x\) so it's just a matter of plugging them in: \[dist(A,M)=dist(M,C) \iff 2x+8=6x\] And now, you have a first degree equality, which I have to suppose you can solve and thus obtaining the desired value of "x".

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