anonymous
  • anonymous
For what value of k are the two lines 2x +ky =3 and x+y=1 (a)parallel? (b)perpendicular?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jdoe0001
  • jdoe0001
if two lines are parallel, that means their "slope" is?
anonymous
  • anonymous
the same
jdoe0001
  • jdoe0001
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

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anonymous
  • anonymous
yeah the slop of the first one is -2/k and the second one is -1
jdoe0001
  • jdoe0001
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
jdoe0001
  • jdoe0001
now... for two lines to be perpendicular their slopes need to be the NEGATIVE RECIPROCAL, that is \(slope=\cfrac{a}{{\color{blue}{ b}}}\qquad negative\implies -\cfrac{a}{{\color{blue}{ b}}}\qquad reciprocal\implies - \cfrac{{\color{blue}{ b}}}{a}\)
jdoe0001
  • jdoe0001
and we can negativize and reciprocalize either for that and then find "k"
anonymous
  • anonymous
so i do the same thing but just have the slopes as a reciprocal
jdoe0001
  • jdoe0001
right.. .one sec
jdoe0001
  • jdoe0001
\(\bf -\cfrac{1}{{\color{blue}{ 1}}}\qquad negative\implies \cfrac{1}{{\color{blue}{ 1}}}\qquad reciprocal\implies \cfrac{{\color{blue}{ 1}}}{1}\implies 1 \\ \quad \\ \textit{thus if we equate the 1st one, with THAT negative reciprocal} \\ \quad \\ -\cfrac{2}{k}=1\implies -\cfrac{2}{1}=k\implies 2=k\)
anonymous
  • anonymous
oh ok thnx Cx i get it now
jdoe0001
  • jdoe0001
hmmm actually, should be -2, shoot, missing the negative there \(\bf -\cfrac{1}{{\color{blue}{ 1}}}\qquad negative\implies \cfrac{1}{{\color{blue}{ 1}}}\qquad reciprocal\implies \cfrac{{\color{blue}{ 1}}}{1}\implies 1 \\ \quad \\ \textit{thus if we equate the 1st one, with THAT negative reciprocal} \\ \quad \\ -\cfrac{2}{k}=1\implies -\cfrac{2}{1}=k\implies -2=k\)
anonymous
  • anonymous
so i just had to do the reciprocal? i thought you had to change the signs too
anonymous
  • anonymous
because the -2/k was already negative so wouldnt it be k/2 =1 @jdoe0001
anonymous
  • anonymous
nevermind

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