Question

Need some geometry help please. Could someone help me with this problem

Use the Distance Formula and the x-axis of the coordinate plane. Show why the distance between two points on a number line (the x-axis) is | a – b |, where a and b are the x-coordinates of the points.


Thanks

Answer 1

\(\Large\color{black}{ \displaystyle {\rm D} =\sqrt{ (\color{blue}{{\rm y}_1}-\color{red}{{\rm y}_2})^2+(\color{green}{{\rm x}_1}-\color{darkgoldenrod}{{\rm x}_2})^2 }}\)


where
\(\Large\color{black}{ \displaystyle {\rm D} }\) is the distance.

\(\Large\color{black}{ \displaystyle (\color{green}{{\rm x}_1}~,~~\color{blue}{{\rm y}_1}) }\) and \(\Large\color{black}{ \displaystyle (\color{darkgoldenrod}{{\rm x}_2}~,~~\color{red}{{\rm y}_2}) }\) are your two points.

Answer 2

your number line is the x, axis so your points are (a,0) and (b,0)

Answer 3

And the definition of an absolute value for any function f(x) is

\(\large\color{black}{ \displaystyle \sqrt{\left( f(x) \right)^2} }\)

Answer 4

you are 15 and i am 22
the difference in our ages (the distance between 15 and 22) is \(|22-15|\) or \(|15-22|\)

Answer 5

well we have to prove using distance formula.

Answer 6

which I just (hopefully) showed how to do

Answer 7

If there are questions, please ask. Don't leave or wait for snow/

Answer 8

Sorry having site connection issues wont let me stay connected for more then a few seconds for some reason

Answer 9

take your time:)

Answer 10

So it would be starting like
\[D = \sqrt{(0_{1}-0_{1})^{2}+ (a_{2}-b _{2})^{2} }\]

Answer 11

well, not exactly, you don't need the sub symbols.
it would just be \(\large\color{blue}{ \displaystyle {\rm D}=\sqrt{(0-0)^2+(a-b)^2} }\)

where your points are \(\large\color{black}{ \displaystyle (a,~0) }\) and \(\large\color{black}{ \displaystyle (b,~0) }\).

Answer 12

what can you simplify the expression inside the distance formula to?

Answer 13

what is \(\large\color{blue}{ \displaystyle (0-0)^2 }\) equal to?

Answer 14

It's 0

Answer 15

yes

Answer 16

So, \(\large\color{blue}{ \displaystyle \sqrt{(0-0)^2+(a-b)^2} }\) is same as ?

Answer 17

(same as square root of what ? )

Answer 18

\[\sqrt{0 +ab } ?\]

Sorry, really have not good with letters

Answer 19

if you have struggle putting in the code in the equation editor you can always use the drawing tool:)

Answer 20

No that's not it. Just not good with letters in math problems

Answer 21

\({ \large\color{blue}{ \displaystyle {\rm D}=\sqrt{\color{red}{(0-0)^2}+(a-b)^2} } }\)

\({ \large\color{blue}{ \displaystyle {\rm D}=\sqrt{\color{red}{0}+(a-b)^2} } }\)

and you just remain with

\({ \large\color{blue}{ \displaystyle {\rm D}=\sqrt{(a-b)^2} } }\)

Answer 22

are we clear on this part (i posted just now) ?

Answer 23

Yes

Answer 24

\(\large\color{blue}{{ \displaystyle \left|~\color{red}{f(x)}~\right|=\sqrt{(~\color{red}{f(x)}~)^2} } }\)

Answer 25

this is the definition of the absolute value of any function
(in fact this is how we differentiate it... later on in math)

now, you have instead of \(\large\color{blue}{{ \displaystyle \color{red}{f(x)}} }\), the \(\large\color{blue}{{ \displaystyle \color{red}{a-b} } }\).

Answer 26

so what is the expression in the root equivalent to?

Answer 27

\[\left| a-b \right| = \sqrt{(a-b)^{2}}\]

Answer 28

yes

Answer 29

and, wouldn't it be logical to say that \(\large\color{blue}{ \left| a-b\right|= \left| b-a\right| }\) ?

Answer 30

oh, nvm you don't need that

Answer 31

we got \(\large\color{blue}{ \left| a-b\right| }\) already

Answer 32

so is everything clear so far, want a recap/review, have questions?

Answer 33

I think I got it. So the answer would be
\[D = \sqrt{(0-0)^{2} + (a - b)^{2}}\]
\[D=\sqrt{0+(a-b)^{2}}\]
\[D=\sqrt{(a-b)^{2}}\]
\[\left| a-b \right|\]

Answer 34

yes, those are the proper steps !!

Answer 35

Thank you so much for your help

Answer 36

yw