Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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 got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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!
awesome thanks!

Not the answer you are looking for?

Search for more explanations.

Ask your own question