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Mathematics 7 Online
OpenStudy (anonymous):

what is the distance formula?

OpenStudy (anonymous):

suppose you want to find the distance between two points (x1,y1) and (x2,y2) the distance formula is =sqrt((x1-x2)^2+(y1-y2)^2)

OpenStudy (anonymous):

\[d = \sqrt{(x2-x1)^{2} +(y2-y1)^{2} }\]

OpenStudy (chaise):

Which in essence is simply Pythagoras theorem, if you form an 'imaginary' triangle between the two points.

OpenStudy (anonymous):

okay i have to find the distance between (5,5) and (10,2)

OpenStudy (anonymous):

then take (5,5) as (x1,y1) and (10,2) as (x2,y2) then put these values in require formula

OpenStudy (anonymous):

is it 14?

OpenStudy (anonymous):

how did you get that?

OpenStudy (anonymous):

idkk.im not sure what the answer is

OpenStudy (anonymous):

did you just guess at 14? the instructions provided by the two posters are pretty clear...

OpenStudy (anonymous):

no i just didnt plug it in right.whats the answer then?

OpenStudy (anonymous):

i can't directly give you the answer... but i can make you give me the answer....

OpenStudy (anonymous):

k?

OpenStudy (anonymous):

what's \(\large x_2-x_1= \) and \(\large y_2-y_1= \)

OpenStudy (anonymous):

2 and -3?

OpenStudy (anonymous):

no... \(\large x_2-x_1= \) refers to subtracting the x coordinates... what are the x coordinates?

OpenStudy (anonymous):

arent they 5 and 10?

OpenStudy (anonymous):

yes... now subtract those two x values...

OpenStudy (anonymous):

-5

OpenStudy (anonymous):

good. now \(\large (x_2-x_1)^2=(-5)^2=25 \), agreed?

OpenStudy (anonymous):

yess

OpenStudy (anonymous):

good... i want you to do the exact same thing with the y-coordinates.... find me \(\large (y_2-y_1)^2= \)

OpenStudy (anonymous):

its 9 right?

OpenStudy (anonymous):

correct... so in the distance formula: \(\large d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{25+9} \) can you simplify it from here?

OpenStudy (anonymous):

is it 5.83?

OpenStudy (anonymous):

if that's what your calculator gave you for \(\large \sqrt{34} \), then yes...

OpenStudy (anonymous):

yupp it did

OpenStudy (anonymous):

the exact way to write your answer is \(\large \sqrt{34} \)

OpenStudy (anonymous):

wasn't that better than guessing? :)

OpenStudy (anonymous):

it tells me to round my answer to the nearest tenth though

OpenStudy (anonymous):

oh.. ok....:)

OpenStudy (anonymous):

thank you so much:)

OpenStudy (anonymous):

yw...:)

OpenStudy (anonymous):

can you help me find the midpoint of (7,9) and (12,11)

OpenStudy (anonymous):

it is equal to( x1+x2)/2,(y1+y2)/2. here (x1,y1) is (7,9) and (x2,y2) is (12,11)

OpenStudy (anonymous):

midpoint =(7+12)/2,(9+11)/2

OpenStudy (anonymous):

9.5,10?

OpenStudy (anonymous):

yes

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