anonymous
  • anonymous
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
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
I can't believe they're making this a multiple choice question. What a wingspan move.
anonymous
  • anonymous
can you help me?
anonymous
  • anonymous

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anonymous
  • anonymous
Okay, well the distance from it is given my distance formula:\[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]
anonymous
  • anonymous
In this case those... \(x_1=-2\), \(x_2=x\), \(y_1=1\), \(y_2=y\).
anonymous
  • anonymous
I would do \(x_2=x=2y-4\) so that you have it all in one variable.
anonymous
  • anonymous
So we get \[ d(y) = \sqrt{((2y-4)-(-2))^2+((y)-(1))^2}=\sqrt{(2y-2)^2+(y-1)^2} \]
anonymous
  • anonymous
ok
anonymous
  • anonymous
That needs to be simplified a bit.
anonymous
  • anonymous
\[ \sqrt{4y^2-8y+4+y^2-2y+1}=\sqrt{5y^2-10y+5} \]
anonymous
  • anonymous
\[ 5y^2-5y-5y+5 = 5y(y-1)-5(y-1)=(5y-5)(y-1)=5(y-1)(y-1) \]Nice how these problems seem to work out for us...
anonymous
  • anonymous
\[ d(y)=\sqrt{5(y-1)^2}=\sqrt{5}(y-1) \]Now we want to minimize this distance function....
anonymous
  • anonymous
ok how do we do that?
anonymous
  • anonymous
We take the derivative and find some critical numbers!!
anonymous
  • anonymous
\[d'(y)=\sqrt{5} \]Wait what...
anonymous
  • anonymous
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.
anonymous
  • anonymous
so the answer is 0?
anonymous
  • anonymous
anonymous
  • anonymous
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!
anonymous
  • anonymous
awesome thanks!

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