## 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

1. imqwerty

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

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

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

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

would it be c?

6. Michele_Laino

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

false*

8. surryyy

so a

9. Michele_Laino

that's right!

10. surryyy

thnx

11. Michele_Laino

:)