Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (anonymous):

Given the points on the graph below, what are the coordinates of another point that would create a parallelogram? (6, 3) (6, 0) (6, 1) (6, 2)

OpenStudy (anonymous):

OpenStudy (accessdenied):

A parallelogram has opposite congruent sides and also opposite parallel sides. So, (2,4) - (2,0) is parallel to (6,6) - (x,y) (we don't know it yet). (6,6)-(x,y) is a vertical line segment, and so their x-values will be the same. (6,6)-(6,y) The length of (2,0)-(2,4) is also equal to the length of (6,6)-(6,y). The distance between (2,0) and (2,4) is 4 units. So, we could move either up 4 units or down 4 units from (6,6) for our (6,y). So, two points we could use for the last vertex would be (6,6+4) and (6,6-4). (6,10) or (6,2).

OpenStudy (anonymous):

i still dont get it , ughhh...

OpenStudy (accessdenied):

hmm... something you don't understand, exactly?

OpenStudy (anonymous):

yea, i guess all this i try but its not coming to me at all

OpenStudy (accessdenied):

Basically, we're just using a few properties of parallelograms. - Opposite sides are congruent/ equal - Opposite sides are parallel / have same slope We're looking for a point that will make both of these properties true for our figure.

OpenStudy (anonymous):

well i kinda get that part, guess i'll be studying more on this lesson then

hero (hero):

You probably just need to grab some graph paper to visually observe where the points are and where the missing point should go.

OpenStudy (anonymous):

well idk the picture is there and everything

hero (hero):

I see, well, let me help you out..

OpenStudy (anonymous):

ok

hero (hero):

Directrix (directrix):

@Hero --> Would you post the graph for the other point derived by AccessDenied? Thanks. (6,10) or (6,2)

OpenStudy (accessdenied):

there's actually another point that also makes a parallelogram, except that one and (6,10) aren't on the graph provided. We could find it in a similar fashion to finding the others, only we'll be using the slope of a nonvertical/nonhorizontal line I graphed all three parallelograms that are made on one graph here, hopefully its not too messy.

Directrix (directrix):

Thanks for graphing these. I have spatial viewing difficulties. I can see two of the parallelograms but not the third because I was following the colors of the sides of the parallelograms. When some of the segments coincided, the colors changed, and for me, the entire set of graphs flipped out into a 3-D optical illusion. I'll just need to graph them separately. If you don't mind saying, where are you doing these graphs?

OpenStudy (accessdenied):

i just used paint + some graph paper from online.

OpenStudy (accessdenied):

hopefully this turns out well, im uploading the same file twice but saved different times

OpenStudy (accessdenied):

aww, it turned out the same... well, there's the second one two times

OpenStudy (accessdenied):

Directrix (directrix):

@AccessDenied --> Thanks for your time and effort. After viewing the graphs individually, I am able to see them collectively.

hero (hero):

@AccessDenied , your efforts to help @JahEmpress in this case may only confuse her further.

OpenStudy (anonymous):

yep your right

OpenStudy (accessdenied):

well, i'll attempt to write up a better explanation later if you're still uncertain.

OpenStudy (falco276):

The answer actually is (6,3). Why? Because you can clearly tell that the shape forms into a parallelogram. There. :D

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!