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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
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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wioBest ResponseYou've already chosen the best response.2
I can't believe they're making this a multiple choice question. What a wingspan move.
 one year ago

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

wioBest ResponseYou've already chosen the best response.2
In this case those... \(x_1=2\), \(x_2=x\), \(y_1=1\), \(y_2=y\).
 one year ago

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

wioBest ResponseYou've already chosen the best response.2
So we get \[ d(y) = \sqrt{((2y4)(2))^2+((y)(1))^2}=\sqrt{(2y2)^2+(y1)^2} \]
 one year ago

wioBest ResponseYou've already chosen the best response.2
That needs to be simplified a bit.
 one year ago

wioBest ResponseYou've already chosen the best response.2
\[ \sqrt{4y^28y+4+y^22y+1}=\sqrt{5y^210y+5} \]
 one year ago

wioBest ResponseYou've already chosen the best response.2
\[ 5y^25y5y+5 = 5y(y1)5(y1)=(5y5)(y1)=5(y1)(y1) \]Nice how these problems seem to work out for us...
 one year ago

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

kenneyfamilyBest ResponseYou've already chosen the best response.0
ok how do we do that?
 one year ago

wioBest ResponseYou've already chosen the best response.2
We take the derivative and find some critical numbers!!
 one year ago

wioBest ResponseYou've already chosen the best response.2
\[d'(y)=\sqrt{5} \]Wait what...
 one year ago

wioBest ResponseYou've already chosen the best response.2
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

kenneyfamilyBest ResponseYou've already chosen the best response.0
so the answer is 0?
 one year ago

wioBest ResponseYou've already chosen the best response.2
We 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!
 one year ago
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