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anonymous

  • one year ago

For what value of k are the two lines 2x +ky =3 and x+y=1 (a)parallel? (b)perpendicular?

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  1. jdoe0001
    • one year ago
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    if two lines are parallel, that means their "slope" is?

  2. anonymous
    • one year ago
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    the same

  3. jdoe0001
    • one year ago
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    right so.... let us put those two fellows in slope-intercept form so \(\bf \begin{cases} 2x+ky=3\to ky=-2x+3\to y=\cfrac{-2x+3}{k}\to &y=-\cfrac{2}{k}+\cfrac{3}{k} \\ \quad \\ x+y=1\to y=-x+1\to &y=-1x+1 \end{cases}\) notice the slopes of each of them

  4. anonymous
    • one year ago
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    yeah the slop of the first one is -2/k and the second one is -1

  5. jdoe0001
    • one year ago
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    so the 1st one has a slope of -2/k the 2nd one, has a slope of -1 what should "k" need to be, for them to be equal? well \(\bf -\cfrac{2}{k}=-1\implies -\cfrac{2}{-1}=k\implies 2=k\) meaning, that, if "k" is 2, then, the EQUATion is true, thus the slopes are equal and thus both lines are parallel, when "k" is 2

  6. jdoe0001
    • one year ago
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    now... for two lines to be perpendicular their slopes need to be the NEGATIVE RECIPROCAL, that is \(slope=\cfrac{a}{{\color{blue}{ b}}}\qquad negative\implies -\cfrac{a}{{\color{blue}{ b}}}\qquad reciprocal\implies - \cfrac{{\color{blue}{ b}}}{a}\)

  7. jdoe0001
    • one year ago
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    and we can negativize and reciprocalize either for that and then find "k"

  8. anonymous
    • one year ago
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    so i do the same thing but just have the slopes as a reciprocal

  9. jdoe0001
    • one year ago
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    right.. .one sec

  10. jdoe0001
    • one year ago
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    \(\bf -\cfrac{1}{{\color{blue}{ 1}}}\qquad negative\implies \cfrac{1}{{\color{blue}{ 1}}}\qquad reciprocal\implies \cfrac{{\color{blue}{ 1}}}{1}\implies 1 \\ \quad \\ \textit{thus if we equate the 1st one, with THAT negative reciprocal} \\ \quad \\ -\cfrac{2}{k}=1\implies -\cfrac{2}{1}=k\implies 2=k\)

  11. anonymous
    • one year ago
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    oh ok thnx Cx i get it now

  12. jdoe0001
    • one year ago
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    hmmm actually, should be -2, shoot, missing the negative there \(\bf -\cfrac{1}{{\color{blue}{ 1}}}\qquad negative\implies \cfrac{1}{{\color{blue}{ 1}}}\qquad reciprocal\implies \cfrac{{\color{blue}{ 1}}}{1}\implies 1 \\ \quad \\ \textit{thus if we equate the 1st one, with THAT negative reciprocal} \\ \quad \\ -\cfrac{2}{k}=1\implies -\cfrac{2}{1}=k\implies -2=k\)

  13. anonymous
    • one year ago
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    so i just had to do the reciprocal? i thought you had to change the signs too

  14. anonymous
    • one year ago
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    because the -2/k was already negative so wouldnt it be k/2 =1 @jdoe0001

  15. anonymous
    • one year ago
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    nevermind

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