Ask your own question, for FREE!
Geometry 15 Online
OpenStudy (anonymous):

Find the distance between (-1,-3);(11,7)

OpenStudy (anonymous):

I am not sure how to do this :(

OpenStudy (espex):

Use the distance formula \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

OpenStudy (anonymous):

ok i have that formula but i don't understand the little 1 and 2 under the x and y i am used to exponents...

OpenStudy (whpalmer4):

note that it doesn't matter which point you choose as (x1,y1) — you just have to be consistent and use the values from the same point as the "1s"

OpenStudy (whpalmer4):

x1 = -1, y1 = -3 x2 =11, y2 = 7

OpenStudy (espex):

Think of your first pair as _1 and your second pair as _2

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

hang on let me work it out and you can check me

OpenStudy (anonymous):

I believe it is 4?

OpenStudy (espex):

How did you get that?

OpenStudy (whpalmer4):

no. x2 = 11, x1 = -1, x2-x1=11-(-1) = 12. 12^2 = 144. do the same with y. retry your work :-)

OpenStudy (anonymous):

oh i didnt noticed each parentheses were squared

OpenStudy (espex):

\[d = \sqrt{(11-(-1))^2+(7-(-3))^2}\]

OpenStudy (anonymous):

hmm i am confused when i do it all through i get like 15 and some change lol

OpenStudy (espex):

Good, nice work.

OpenStudy (anonymous):

um but how do i work out that remainder since it is not a perfect square?

OpenStudy (espex):

What makes you think it is a perfect square?

OpenStudy (anonymous):

i said it's not -_-

OpenStudy (espex):

Ah, my bad. You already have the decimal form of your root, do you not?

OpenStudy (anonymous):

no i think its supposed to be like x times the square root of x

OpenStudy (espex):

Well, if you work it out, \(d = \sqrt{(11-(-1))^2+(7-(-3))^2}\) into \(d = \sqrt{(11+1))^2+(7+3))^2}\), you get \(d = \sqrt{(12)^2+(10))^2}\rightarrow d = \sqrt{144+100} \rightarrow \sqrt{244}\)

OpenStudy (anonymous):

yes got all that on paper

OpenStudy (espex):

Okay, so what does 244 break into, I like to start with 2s if it's not obvious. \(\sqrt{2*122}\) => \(\sqrt{2*2*61}\) so I already have a couple of terms I can remove.

OpenStudy (espex):

\(2\sqrt{61}\) and I know of no other reduction for this short of a calculator to simplify into a decimal.

OpenStudy (whpalmer4):

yeah, that's as far as it goes (and where I'd leave it, unless a decimal was requested)

OpenStudy (anonymous):

ah wait i see they will take an answer form of square root of 244

OpenStudy (espex):

@Jhannybean that's not correct.

OpenStudy (jhannybean):

No?..

OpenStudy (espex):

11+1 is not 18. :)

OpenStudy (jhannybean):

ahaha My bad.. yes i see it.

OpenStudy (anonymous):

lol i got it thanks guys

OpenStudy (jhannybean):

Oh my goodness.

OpenStudy (whpalmer4):

@Jhannybean probably your 12 looked like an 18...

OpenStudy (espex):

You're welcome @countonme123

OpenStudy (jhannybean):

It probably did....

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!