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kenneyfamily Group Title

Find the distance between P(-2, 1) and the line with equation x - 2y + 4 = 0. a. (-4√5)/5 b. 0 c. 5/4 d. (4√5)/5

  • one year ago
  • one year ago

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  1. wio Group Title
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    I can't believe they're making this a multiple choice question. What a wingspan move.

    • one year ago
  2. kenneyfamily Group Title
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    can you help me?

    • one year ago
  3. kenneyfamily Group Title
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    @wio

    • one year ago
  4. wio Group Title
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    Okay, well the distance from it is given my distance formula:\[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]

    • one year ago
  5. wio Group Title
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    In this case those... \(x_1=-2\), \(x_2=x\), \(y_1=1\), \(y_2=y\).

    • one year ago
  6. wio Group Title
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    I would do \(x_2=x=2y-4\) so that you have it all in one variable.

    • one year ago
  7. wio Group Title
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    So we get \[ d(y) = \sqrt{((2y-4)-(-2))^2+((y)-(1))^2}=\sqrt{(2y-2)^2+(y-1)^2} \]

    • one year ago
  8. kenneyfamily Group Title
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    ok

    • one year ago
  9. wio Group Title
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    That needs to be simplified a bit.

    • one year ago
  10. wio Group Title
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    \[ \sqrt{4y^2-8y+4+y^2-2y+1}=\sqrt{5y^2-10y+5} \]

    • one year ago
  11. wio Group Title
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    \[ 5y^2-5y-5y+5 = 5y(y-1)-5(y-1)=(5y-5)(y-1)=5(y-1)(y-1) \]Nice how these problems seem to work out for us...

    • one year ago
  12. wio Group Title
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    \[ d(y)=\sqrt{5(y-1)^2}=\sqrt{5}(y-1) \]Now we want to minimize this distance function....

    • one year ago
  13. kenneyfamily Group Title
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    ok how do we do that?

    • one year ago
  14. wio Group Title
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    We take the derivative and find some critical numbers!!

    • one year ago
  15. wio Group Title
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    \[d'(y)=\sqrt{5} \]Wait what...

    • one year ago
  16. wio Group Title
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    So there are no critical points... which means that there is no minimum.... except for the fact that the distance can't be negative.... So the minimum distance must be 0.

    • one year ago
  17. kenneyfamily Group Title
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    so the answer is 0?

    • one year ago
  18. kenneyfamily Group Title
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    @wio

    • one year ago
  19. wio Group Title
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    We can check by putting the point into the line.\[ (-2)+-2(1)+4=0\implies -2-2+4=0\implies -4+4=0\implies 0=0 \]So the point is actually on the line! So the distance is 0!

    • one year ago
  20. kenneyfamily Group Title
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    awesome thanks!

    • one year ago
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