Question

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

Answer 1

I can't believe they're making this a multiple choice question. What a wingspan move.

Answer 2

can you help me?

Answer 3

@wio

Answer 4

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

Answer 5

In this case those... \(x_1=-2\), \(x_2=x\), \(y_1=1\), \(y_2=y\).

Answer 6

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

Answer 7

So we get \[
d(y) = \sqrt{((2y-4)-(-2))^2+((y)-(1))^2}=\sqrt{(2y-2)^2+(y-1)^2}
\]

Answer 8

ok

Answer 9

That needs to be simplified a bit.

Answer 10

\[
\sqrt{4y^2-8y+4+y^2-2y+1}=\sqrt{5y^2-10y+5}
\]

Answer 11

\[
5y^2-5y-5y+5 = 5y(y-1)-5(y-1)=(5y-5)(y-1)=5(y-1)(y-1)
\]Nice how these problems seem to work out for us...

Answer 12

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

Answer 13

ok how do we do that?

Answer 14

We take the derivative and find some critical numbers!!

Answer 15

\[d'(y)=\sqrt{5} \]Wait what...

Answer 16

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.

Answer 17

so the answer is 0?

Answer 18

@wio

Answer 19

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!

Answer 20

awesome thanks!