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anonymous

  • one year ago

Write an equation of the line perpendicular to the given line that contains P. y space equals space 1 fifth x space plus space 1; P(2, -7)

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  1. anonymous
    • one year ago
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    help me pleaase!!!

  2. IrishBoy123
    • one year ago
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    lmLtfy! "y space equals space 1 fifth x space plus space 1" is: \(y = \frac{1}{5}x+ 1\) and hint: if the lines \(y = mx + c\) and \(y = m' x + c'\) are perp, then \(m' = -\frac{1}{m}\) and \(m = -\frac{1}{m'}\)

  3. anonymous
    • one year ago
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    im osrry i still dont understand

  4. IrishBoy123
    • one year ago
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    what is the gradient of your line?

  5. anonymous
    • one year ago
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    there is no line

  6. anonymous
    • one year ago
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    is the answer y=-1/5x-33/5??

  7. IrishBoy123
    • one year ago
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    go back to your question you have a line

  8. IrishBoy123
    • one year ago
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    i did the Latex for you

  9. IrishBoy123
    • one year ago
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    no m = -1 / m' m is slope of original line m' is slope of line that is perp to original line

  10. anonymous
    • one year ago
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    oh right so then would it be y=1/5x+3

  11. IrishBoy123
    • one year ago
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    i will keep asking this :p what is the slope of your original line, the one that you posted??

  12. IrishBoy123
    • one year ago
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    do you know what a slope is? \(\frac{rise}{run}\) ??

  13. anonymous
    • one year ago
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    1/5?

  14. IrishBoy123
    • one year ago
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    great so m' = -1 / m m is slope of original line m' is slope of line that is perp to original line

  15. anonymous
    • one year ago
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    so then it would be -1/5?

  16. IrishBoy123
    • one year ago
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    |dw:1435948825065:dw|

  17. IrishBoy123
    • one year ago
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    you're just "minusing" it. you need to invert and minus it

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spraguer (Moderator)
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