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anonymous

  • one year ago

passes through A(-3, 0) and B(-6, 5). What is the equation of the line that passes through the origin and is parallel to ?

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  1. Michele_Laino
    • one year ago
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    hint: equation for the line which passes at point A and at point B, is: \[\Large \frac{{x - x1}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}}\] where A=(x1,y1) and B=(x2,y2) Now you have to substitute the coordinates of both points A and B, in order to to write the corresponding equation

  2. anonymous
    • one year ago
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    i have no idea im clueless

  3. Michele_Laino
    • one year ago
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    your exercise asks for the equation of the line parallel to the line which passes at points A and B, right?

  4. anonymous
    • one year ago
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    yes

  5. Michele_Laino
    • one year ago
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    so, we have to write the equation which passes at point A and B, first

  6. anonymous
    • one year ago
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    ok

  7. Michele_Laino
    • one year ago
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    in order to do that, you have to substitute the coordinates of your points A and B into the equation above

  8. anonymous
    • one year ago
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    ok

  9. Michele_Laino
    • one year ago
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    please try!

  10. anonymous
    • one year ago
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    haha im trying but i dont understand the equation

  11. anonymous
    • one year ago
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    ive never seen it like that

  12. anonymous
    • one year ago
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    the question asks line ab passes through A(-3, 0) and B(-6, 5). What is the equation of the line that passes through the origin and is parallel to line ab ?

  13. Michele_Laino
    • one year ago
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    ok! Here is the procedure: we have this: \[\Large \begin{gathered} {x_1} = - 3,\quad {x_2} = - 6 \hfill \\ {y_1} = 0,\quad {y_2} = 5 \hfill \\ \end{gathered} \]

  14. anonymous
    • one year ago
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    ok

  15. anonymous
    • one year ago
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    i need like help fast class ends in 5 mins and i have 1 more question after so i wrote the equation from your thing now what?

  16. Michele_Laino
    • one year ago
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    now, I substitute those value into my equation above, and I get: \[\Large \frac{{x - \left( { - 3} \right)}}{{ - 6 - \left( { - 3} \right)}} = \frac{{y - 0}}{{5 - 0}}\]

  17. anonymous
    • one year ago
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    yes i got that

  18. Michele_Laino
    • one year ago
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    now I can simplify as below: \[\Large \begin{gathered} \frac{y}{5} = \frac{{x + 3}}{{ - 6 + 3}} \hfill \\ \hfill \\ \frac{y}{5} = \frac{{x + 3}}{{ - 3}} \hfill \\ \hfill \\ y = - \frac{5}{3}\left( {x + 3} \right) \hfill \\ \end{gathered} \]

  19. anonymous
    • one year ago
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    parallel to that is?

  20. Michele_Laino
    • one year ago
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    no, the requested line, being parallel to that line has to have the same slope, in other words the slope of the requested parallel line is: \[\Large m = - \frac{5}{3}\]

  21. Michele_Laino
    • one year ago
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    since parallel lines have the same slope

  22. Michele_Laino
    • one year ago
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    |dw:1440003243267:dw|

  23. Michele_Laino
    • one year ago
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    both lines have the same slope m= -5/3

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