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anonymous

  • 4 years ago

parallel and perpendicular equations for the point (-5,-10) 2x + 5y -12 = 0

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  1. amistre64
    • 4 years ago
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    parallel have same equation with a difference constat; perps switch up the xy coeffs and negate one; and change the constant too

  2. amistre64
    • 4 years ago
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    // 2x + 5y + c = 0 ; calibrate with the given point L 5x - 2y + c = 0 ; calibrate with given point

  3. anonymous
    • 4 years ago
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    so if you calculate it out as in slope intercept form, and got y=-2/5 = 12/5 perpendicular would just be 5/2+ whatever else?

  4. anonymous
    • 4 years ago
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    that was supposed to be + 12/5

  5. amistre64
    • 4 years ago
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    looks to be a good set up yes

  6. anonymous
    • 4 years ago
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    thank you

  7. amistre64
    • 4 years ago
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    slope is -2/5 for parallel \\ slope is 5/2 for perp L

  8. amistre64
    • 4 years ago
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    then ignore the 12 and use the point fo find a new constant

  9. amistre64
    • 4 years ago
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    y = -2/5 x + c -10 = -5(-2/5) + c -10 = 2 + c -12 = c y = -2/5 - 12 ..... that worked out kinda cool lol

  10. amistre64
    • 4 years ago
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    forgot the "x" in there tho ... bad amistre, bad!! :)

  11. anonymous
    • 4 years ago
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    lol thank you:)

  12. amistre64
    • 4 years ago
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    yw :) the perp is the same concept; ad it looks like your competent enough for that one

  13. anonymous
    • 4 years ago
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    I know math, I just doubt myself. Whenever I get my work back it's always STUPID little mistakes

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