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Mathematics 18 Online
OpenStudy (anonymous):

Help remembering how to solve a problem like this? (posting question below)

OpenStudy (anonymous):

OpenStudy (dtan5457):

any coordinates?

OpenStudy (dtan5457):

@Catseyeglint911

OpenStudy (anonymous):

That's all that I was given.

OpenStudy (dtan5457):

ok give me a second.

OpenStudy (anonymous):

alright

OpenStudy (dtan5457):

well the only way i see a numerical answer is if you make up the coordinate points but make them true throughout the rest of the points. so for example you can put |dw:1419033468796:dw| so based on the coordinate grid i can say c=1,1 and b=7,3

OpenStudy (dtan5457):

then i would go ahead and use \[distance=\sqrt{(x2-x1)^2+(y2-y1)^2}\]

OpenStudy (dtan5457):

plug it all in and you get the square root of 40. which rounds to 6.3 to the nearest tenth.

OpenStudy (dtan5457):

make sense?

OpenStudy (anonymous):

That does make sense when you add the coordinates, but is there any way to know that those are the exact coordinates?

OpenStudy (dtan5457):

Well distance is distance. It wouldn't matter what the exact coordinates are, the distance will always be the same.

OpenStudy (dtan5457):

I mean if point c was 60,61. as long as you move one point per line point b will end up as 66,63. the distance is the same still if you use the distance formula

OpenStudy (dtan5457):

\[\sqrt{(66-60)^2+(63-61)^2} = \sqrt{(6^2)+(2)^2}\]

OpenStudy (dtan5457):

still the square root of 40

OpenStudy (anonymous):

Ahh I understand now. Thank you so much for your help!

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