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kenneyfamily

  • 2 years ago

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

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  1. wio
    • 2 years ago
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    I can't believe they're making this a multiple choice question. What a wingspan move.

  2. kenneyfamily
    • 2 years ago
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    can you help me?

  3. kenneyfamily
    • 2 years ago
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    @wio

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

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

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

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

  8. kenneyfamily
    • 2 years ago
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    ok

  9. wio
    • 2 years ago
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    That needs to be simplified a bit.

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

  11. wio
    • 2 years ago
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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...

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

  13. kenneyfamily
    • 2 years ago
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    ok how do we do that?

  14. wio
    • 2 years ago
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    We take the derivative and find some critical numbers!!

  15. wio
    • 2 years ago
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    \[d'(y)=\sqrt{5} \]Wait what...

  16. wio
    • 2 years ago
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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.

  17. kenneyfamily
    • 2 years ago
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    so the answer is 0?

  18. kenneyfamily
    • 2 years ago
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    @wio

  19. wio
    • 2 years ago
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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!

  20. kenneyfamily
    • 2 years ago
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    awesome thanks!

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