## kenneyfamily Group Title 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

1. wio Group Title

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

2. kenneyfamily Group Title

can you help me?

3. kenneyfamily Group Title

@wio

4. wio Group Title

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

5. wio Group Title

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

6. wio Group Title

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

7. wio Group Title

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

8. kenneyfamily Group Title

ok

9. wio Group Title

That needs to be simplified a bit.

10. wio Group Title

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

11. wio Group Title

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

12. wio Group Title

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

13. kenneyfamily Group Title

ok how do we do that?

14. wio Group Title

We take the derivative and find some critical numbers!!

15. wio Group Title

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

16. wio Group Title

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.

17. kenneyfamily Group Title

so the answer is 0?

18. kenneyfamily Group Title

@wio

19. wio Group Title

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!

20. kenneyfamily Group Title

awesome thanks!