Find the distance from (2,-1) to the line y=2x+3. The answer is 8/5*root5 by the way, but I want to know how to solve. I used distance formula. sqrt ( (x-2)^2 + (y+1)^2) sqrt ( (x-2)^2 + (2x+3+1)^2) sqrt ( (x-2)^2 + (2x+4)^2) sqrt (5x^2 + 12x + 20) When I enter this on my calculator to solve for x, it cannot do it. What should I do next?

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Find the distance from (2,-1) to the line y=2x+3. The answer is 8/5*root5 by the way, but I want to know how to solve. I used distance formula. sqrt ( (x-2)^2 + (y+1)^2) sqrt ( (x-2)^2 + (2x+3+1)^2) sqrt ( (x-2)^2 + (2x+4)^2) sqrt (5x^2 + 12x + 20) When I enter this on my calculator to solve for x, it cannot do it. What should I do next?

Mathematics
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you don't use X
What kind of calculator are you using? Are you using the solve feature?
Yes I am usin the solve feature.

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Other answers:

Why don't you use graph paper instead?
I have to use algebraic methods and show my work.
graphing on graph paper is a way to show your work
Is there a way to answer the problem algebraically?
Well, the line to the other linen willl be perpendicular to it
Assuming you want the shortest distance
using graph paper is a valid method, then you can count the number of units
Find the equation of a line perpendicular to y=2x+3 and use the distance formula on the intersection
You could use a point from the equation. I would graph it to decide what point to use. You need two ponts for the distance formula
Have you studied the formula for finding the perpendicular to a line through a point?
Well, I'm in Calculus right now, but I don't know how to continue this problem.
Yep calculus would study that Do the normal line
Wouldn't the normal slope be -1/2 ?
-1/2*
\[y+1=\frac{-1}{2}(x-2)\]
\[y=-\frac{1}{2}x\]
You could put the equation in the form Ax+By=C=0 and use this formula: \[d=\frac{|Am+Bn+C|}{\sqrt{A ^{2}+B ^{2}}}\] The point is (m,n)
Now set 2x+3=-(1/2)x to find the intersection, then use the distance formula
x = -6/5
Get y now, and use this point in your distance formula
Mertsj what formula did you post? Do you know the name of it? Just curious
Perpendicular distance from a Point to a Line
ok I have not seen that one before. Good to know
Okay I now have two methods of approaching this problem. Thank you all for helping me. :)

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