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(I just want to note that I made up this problem. I'm not trying to cheat on a test or anything.)

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- Oct 14th 2008, 01:34 PMBenBartFinding a point a certain distance from a line.
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(I just want to note that I made up this problem. I'm not trying to cheat on a test or anything.) - Oct 14th 2008, 01:44 PMPeritus
$\displaystyle \sqrt {(x - 1)^2 + (y - 1)^2 } = 2$

- Oct 14th 2008, 01:44 PMicemanfan
The point satisfies $\displaystyle \sqrt{(x-2)^2 + (y-1)^2} = 2$ and $\displaystyle y = 2x - 3$.

Substitution yields:

$\displaystyle \sqrt{(x-2)^2 + (2x - 4)^2} = 2$

$\displaystyle \sqrt{(x-2)^2 + 4(x-2)^2} = 2$

$\displaystyle 5(x-2)^2 = 4$

$\displaystyle (x-2)^2 = \frac{4}{5}$

$\displaystyle x-2 = \sqrt{\frac{4}{5}}$

$\displaystyle x = \sqrt{\frac{4}{5}} + 2$

Now you plug x back into the line equation to find y. - Oct 14th 2008, 03:58 PMBenBartClarity
I just want to clarify that what I'm looking for is the point where that circle is intersecting the line. I'm not looking to solve for the X coordinate.

So I'm not looking for that answer that's around (4,5) I'm looking for the one that's around (2.9, 2.8)ish – I thought I would need pi.

If this is in fact the correct solution then I guess I need a little help figuring out what the 1's and 2's represent. - Oct 14th 2008, 06:49 PMBenBartHow to solve.
Problems get bumped fairly quickly in this section. Is there a more appropriate place to post this? Geometry, pre-calc?