anonymous
  • anonymous
i need help with a couple questions is anyone willing to help?
Geometry
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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calculusxy
  • calculusxy
Sorry, but can you provide me the type of question that you are willing to ask? It can enhance your help for me to help you.
anonymous
  • anonymous
okay find the length of the segment.then find the coordinate of the midpoint of the segment.
calculusxy
  • calculusxy
Lets check it out, shall we?

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anonymous
  • anonymous
okay
anonymous
  • anonymous
|dw:1378771557737:dw| srry its kinda weird
anonymous
  • anonymous
cause the points are suposed to be on the negative four and then on the line between the positive 2 and 4 is that clear enough for u i can redo it if its not
calculusxy
  • calculusxy
Don't worry I can perfectly see it. So what is your question?
anonymous
  • anonymous
basicaly how do i find the lenghth of the segment and the coordinate of the midpoint of the segment?
calculusxy
  • calculusxy
Please wait. I am trying to solve your question. Your paitence will be appreciated.
anonymous
  • anonymous
kk :)
calculusxy
  • calculusxy
There are several different types of method. Would you like for me to explain the methods?
anonymous
  • anonymous
shure
calculusxy
  • calculusxy
Method 1: This method is consisted of counting from one point to the other. Lets say we wish to know how long is (-4, 2). So what we would do is we would just simply count and see how many points far is from -4 to 2 (counting by 1s) . So it would be (-4, 2)=6.
calculusxy
  • calculusxy
However, it doesn't work with diagonal segments.
anonymous
  • anonymous
..... what do u mean by it doesnt work with diagonal segments is the thing i drew a diagonal segment? and what is method #2
calculusxy
  • calculusxy
|dw:1378772500826:dw|No your segment isn't diagonal, your one is horizontal.
calculusxy
  • calculusxy
Method 2: You have to use the Pythgorean Theorem. This theorem consists of \[a^{2}+b^{2}=c^{2}\] Use this example for your understandings.
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calculusxy
  • calculusxy
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  • calculusxy
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anonymous
  • anonymous
thank u these help!!!!!

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