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SephI

  • 3 years ago

Help with graphing, please :P

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  1. SephI
    • 3 years ago
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    Choose the slope-intercept equation of the line that passes through the point (-5, -1) and is perpendicular to y = 5/2x + 2

  2. SephI
    • 3 years ago
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    They have the same slope since they are perpendicular. so I know (-5, -1) with a slope of 5/2 has something to do with the equation, but what happens next?

  3. akaisha
    • 3 years ago
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    they dont have the same slope, that is when they are parallel. For perpendicular the slope is opposite

  4. akaisha
    • 3 years ago
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    its a negative reciprocal, so can you figure it out what the slope is from there?

  5. hba
    • 3 years ago
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    @SephI User Please Respond.

  6. SephI
    • 3 years ago
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    Oh, ok. So I would have a slope of 2/5, and then what?

  7. akaisha
    • 3 years ago
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    close.. m=-2/5 then you use your point-interecpt form: y-y=m(x-x)

  8. SephI
    • 3 years ago
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    y - (-1) = -2/5[(x - (-5)] y + 1 = -2/5x + 5 y = -2/5x - 4? ^ That's not one of my options. So I did it wrong...

  9. akaisha
    • 3 years ago
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    thats close but you need to distribute the (-2/5) to the 5 as well

  10. SephI
    • 3 years ago
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    Why is that?

  11. akaisha
    • 3 years ago
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    that is because the x+5 is in parentheses. so the -2/5 applies to both pieces

  12. akaisha
    • 3 years ago
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    or (x-(-5)) however you want to think about it

  13. SephI
    • 3 years ago
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    y - (-1) = -2/5[(x - (-5)] y - (-1) = -2/5x + ??? How would I distribute the 5? Wouldn't it be -2/5*5 Which is -10/5 = -2?

  14. akaisha
    • 3 years ago
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    yes! perfect. so you would have (y-(-1))=-2/5x-2 then you just need to y by itself

  15. SephI
    • 3 years ago
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    y + 1 = -2/5x - 2 y = -2/5x - 3 - 1 y = -2/5x - 3?

  16. akaisha
    • 3 years ago
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    yep! That should be your answer.

  17. SephI
    • 3 years ago
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    It is. thanks

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