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anonymous
 4 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
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can't believe they're making this a multiple choice question. What a wingspan move.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, well the distance from it is given my distance formula:\[ d = \sqrt{(x_2x_1)^2+(y_2y_1)^2} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In this case those... \(x_1=2\), \(x_2=x\), \(y_1=1\), \(y_2=y\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would do \(x_2=x=2y4\) so that you have it all in one variable.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So we get \[ d(y) = \sqrt{((2y4)(2))^2+((y)(1))^2}=\sqrt{(2y2)^2+(y1)^2} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That needs to be simplified a bit.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \sqrt{4y^28y+4+y^22y+1}=\sqrt{5y^210y+5} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ 5y^25y5y+5 = 5y(y1)5(y1)=(5y5)(y1)=5(y1)(y1) \]Nice how these problems seem to work out for us...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ d(y)=\sqrt{5(y1)^2}=\sqrt{5}(y1) \]Now we want to minimize this distance function....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok how do we do that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We take the derivative and find some critical numbers!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[d'(y)=\sqrt{5} \]Wait what...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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
 4 years ago
Best ResponseYou've already chosen the best response.0We can check by putting the point into the line.\[ (2)+2(1)+4=0\implies 22+4=0\implies 4+4=0\implies 0=0 \]So the point is actually on the line! So the distance is 0!
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