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kenneyfamily
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
I can't believe they're making this a multiple choice question. What a wingspan move.
Okay, well the distance from it is given my distance formula:\[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]
In this case those... \(x_1=-2\), \(x_2=x\), \(y_1=1\), \(y_2=y\).
I would do \(x_2=x=2y-4\) so that you have it all in one variable.
So we get \[ d(y) = \sqrt{((2y-4)-(-2))^2+((y)-(1))^2}=\sqrt{(2y-2)^2+(y-1)^2} \]
That needs to be simplified a bit.
\[ \sqrt{4y^2-8y+4+y^2-2y+1}=\sqrt{5y^2-10y+5} \]
\[ 5y^2-5y-5y+5 = 5y(y-1)-5(y-1)=(5y-5)(y-1)=5(y-1)(y-1) \]Nice how these problems seem to work out for us...
\[ d(y)=\sqrt{5(y-1)^2}=\sqrt{5}(y-1) \]Now we want to minimize this distance function....
ok how do we do that?
We take the derivative and find some critical numbers!!
\[d'(y)=\sqrt{5} \]Wait what...
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.
so the answer is 0?
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!