Find the distance between (-1,-3);(11,7)
I am not sure how to do this :(
Use the distance formula \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
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...
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"
x1 = -1, y1 = -3 x2 =11, y2 = 7
Think of your first pair as _1 and your second pair as _2
oh
hang on let me work it out and you can check me
I believe it is 4?
How did you get that?
no. x2 = 11, x1 = -1, x2-x1=11-(-1) = 12. 12^2 = 144. do the same with y. retry your work :-)
oh i didnt noticed each parentheses were squared
\[d = \sqrt{(11-(-1))^2+(7-(-3))^2}\]
hmm i am confused when i do it all through i get like 15 and some change lol
Good, nice work.
um but how do i work out that remainder since it is not a perfect square?
What makes you think it is a perfect square?
i said it's not -_-
Ah, my bad. You already have the decimal form of your root, do you not?
no i think its supposed to be like x times the square root of x
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 \(d = \sqrt{(12)^2+(10))^2}\rightarrow d = \sqrt{144+100} \rightarrow \sqrt{244}\)
yes got all that on paper
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.
\(2\sqrt{61}\) and I know of no other reduction for this short of a calculator to simplify into a decimal.
yeah, that's as far as it goes (and where I'd leave it, unless a decimal was requested)
ah wait i see they will take an answer form of square root of 244
@Jhannybean that's not correct.
No?..
11+1 is not 18. :)
ahaha My bad.. yes i see it.
lol i got it thanks guys
Oh my goodness.
@Jhannybean probably your 12 looked like an 18...
You're welcome @countonme123
It probably did....
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