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## anonymous 4 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

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1. anonymous

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

2. anonymous

can you help me?

3. anonymous

@wio

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

ok

9. anonymous

That needs to be simplified a bit.

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

ok how do we do that?

14. anonymous

We take the derivative and find some critical numbers!!

15. anonymous

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

16. anonymous

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. anonymous

so the answer is 0?

18. anonymous

@wio

19. anonymous

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. anonymous

awesome thanks!

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