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@iambatman have a look here please.... help him...
If a quadrilateral is a parallelogram, then the diagonals bisect each other. To force the given quadrilateral to be a parallelogram, you must find values of x and y that will do that.
So, solve this system for x and y: @Night-Watcher 3x - 6 = x + 12 y + 12 = 6y - 8 ===========
I see that there is one variable here: y + 12 = 6y - 8 So, solve that for y first.
Does y= 4?
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And does x=9?
I'm feelin really confident about these...
I agree with (9,4). @Night-Watcher
Okay, Thanks!
You are welcome.
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