Question

Help & show work: Find the length of the following segment created by these pairs of endpoints: (0, 1), (6, 6)

Answer 1

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

Answer 2

over 6 (from zero to one) up 5
use pythagoras
\[c^2=6^2+5^2=25+36=61\] so distance is
\[\sqrt{61}\]

Answer 3

just replace x1;y1 coordiaantes of the 1st point and x2;y2 coordinates of the 2nd point

Answer 4

A: 7.8
B: 9.1
C: 2.5
D: 6

Answer 5

7.8

Answer 6

Well how do i solve this?

Answer 7

i gave u the formula

Answer 8

can you brake it down? a bit more plz

Answer 9

the steps are:

subtract one point from the other:

(6, 6)
-(0, 1)
------
6 ,5 ; use these as the lengths of the legs of a right triangle, with the hypotenuse being the distance between the points

Answer 10

d^2 = 6^2 + 5^2

d^2 = 36 + 25

d^2 = 61 ; to find d sqrt it all

d = sqrt(61) , and simplifiy

Answer 11

Umm you lost me at 6,5 :S

Answer 12

maybe i need more sleep :(

Answer 13

ok , what does 6 and 5 represent as far as distances between th epoints?

Answer 14

do you see how I got 6 and 5?

Answer 15

yes but once i get that what do i do simple math plz

Answer 16

the simplest math possible is the pythagorean thrm for this ..

Answer 17

o jod

Answer 18

once you know the distance across the base (the difference in xs) and the distance up the side (the difference in ys) you have all the parts needed to use the pythag thrm

Answer 19

so XS is 6 or the 5 in 6, 5

Answer 20

and ys is the 6 or the 5 in 6,5

Answer 21

at this point it doesnt matter does it?

does: 6^2 + 5^2 = 5^2 + 6^2 ??

Answer 22

in other words, does 36+25 = 25+36 ?? if so, then its mathematically moot

Answer 23

:'( i'm lost sorry i think i need more sleep

Answer 24

get some sleep and look at it again with fresh buggery eyes :)

Answer 25

if you draw a picture it might help to place things