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surryyy

  • one year ago

Select the equation of the line graphed below: y + 7 = 1(x + 6) y – 7 = 1(x – 6) y + 6 = 1(x + 7) y – 6 = 1(x – 7) https://static.k12.com/eli/bb/1211/4_68960/2_33580_12_68972/e1120da006e352e437dc8fe29146b21aef604445/media/79f8ace6ffc61f82a31f6c5ce5ecbce34a8a0321/mediaasset_958465_1.jpg

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  1. imqwerty
    • one year ago
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    select any 2 points that lie on the line for example - http://prntscr.com/8bp02w and then apply 2 point form which is - \[\frac{ y-y _{1} }{ y_{2}-y_{1} }=\frac{ x-x_{1} }{ x_{2}-x_{1} }\] where (x1,y2) and (x2,y2) are the cordinates of the 2 points :)

  2. Michele_Laino
    • one year ago
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    your line passes at point (7,6). For example if I substitute x=7 and y=6, into the secondequation, I get: \[\Large \begin{gathered} 6 - 7 = 1 \cdot \left( {7 - 6} \right) \hfill \\ \hfill \\ - 1 = 1 \hfill \\ \end{gathered} \] which is a false statement, so the second option can not be the right option. Please do the same with the other options

  3. imqwerty
    • one year ago
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    or u can do as @Michele_Laino said :) select any point the lies on the line an put its x and y cordinates in the options nd the one which satisfies is the answer :)

  4. imqwerty
    • one year ago
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    but sometimes it may happen that more than 1 equation satisfy the reason is that, that point may be the solution to those lines :) |dw:1441188655630:dw|

  5. surryyy
    • one year ago
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    would it be c?

  6. Michele_Laino
    • one year ago
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    I think no, since if I substitute x=7 and y=6 into the third equation, I get: \[\Large \begin{gathered} 6 + 6 = 1 \cdot \left( {7 + 7} \right) \hfill \\ \hfill \\ 12 = 14 \hfill \\ \end{gathered} \] which, again, is a flase statement

  7. Michele_Laino
    • one year ago
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    false*

  8. surryyy
    • one year ago
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    so a

  9. Michele_Laino
    • one year ago
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    that's right!

  10. surryyy
    • one year ago
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    thnx

  11. Michele_Laino
    • one year ago
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    :)

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