## kenneyfamily 2 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

1. wio

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

2. kenneyfamily

can you help me?

3. kenneyfamily

@wio

4. wio

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

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

6. wio

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

7. wio

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

8. kenneyfamily

ok

9. wio

That needs to be simplified a bit.

10. wio

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

11. wio

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

12. wio

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

13. kenneyfamily

ok how do we do that?

14. wio

We take the derivative and find some critical numbers!!

15. wio

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

16. wio

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

so the answer is 0?

18. kenneyfamily

@wio

19. wio

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

awesome thanks!